Answer:
f = 7.57 Hz
Explanation:
To find the frequency of the damping oscillator, you first use the following formula for the angular frequency:
[tex]\omega=\sqrt{\omega_o-(\frac{b}{2m})^2}=\sqrt{\frac{k}{m}-(\frac{b}{2m})^2}\\\\[/tex] (1)
k: spring constant = 2.65*10^4 N/m
m: mass = 11.7 kg
b: damping coefficient = 4.50 Ns/m
You replace the values of k, m and b in the equation (1):
[tex]\omega=\sqrt{\frac{2.65*10^4N/m}{11.7kg}-(\frac{4.50Ns/m}{2(11.7kg)})^2}\\\\\omega=47.59\frac{rad}{s}[/tex]
Finally, you calculate the frequency:
[tex]f=\frac{\omega}{2\pi}=\frac{47.59}{2\pi}Hz=7.57\ Hz[/tex]
hence, the frequency of the oscillator is 7.57 Hz
A. Suppose the laser light has wavelength 400 nm, and that the two small slits, which act as point sources of light, are separated by 0.1 mm. (Recall that 1 nm = 10-9 meters.) Approximately how many nodal lines would be present in the pattern of overlapping light waves when the laser light emerges from the two slits in the mask?
Answer:
200 nodal lines
Explanation:
To find the number of lines you first use the following formula for the condition of constructive interference:
[tex]dsin\theta=m\lambda[/tex] (1)
d: distance between slits = 0.1mm = 0.1*10^-3 m
θ: angle between the axis of the slits and the m-th fringe of interference
λ: wavelength of light = 400 nm = 400*10^-9 m
You obtain the max number of lines when he angle is 90°. Then, you replace the angle by 90° and solve the equation (1) for m:
[tex]dsin90\°=m\lambda\\\\d=m\lambda\\\\m=\frac{d}{\lambda}=\frac{0.1*10^{-3}m}{500*10^{-9}m}=200[/tex]
hence, the number of lines in the interference pattern are 200
Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the velocityof a proton is straight up(thumb pointing up) then RHR2 shows that the force points to the left. What would the direction of the force be if the velocitywere a) down b) to the rightc) to the leftd) into the pagee) out of the page
Answer:
Explanation:
We shall show all given data in vector form and calculate the direction of force with the help of following formula
force F = q ( v x B )
q is charge , v is velocity and B is magnetic field.
Given B = - Bk ( i is right , j is upwards and k is straight up the page )
v = v j
F = q ( vj x - Bk )
= -Bqvi
The direction is towards left .
a ) If velocity is down
v = - v j
F = q ( - vj x - bk )
= qvB i
Direction is right .
b ) v = v i
F = q ( vi x - Bk )
= qvB j
force is upwards
c ) v = - vi
F = q ( -vi x - Bk )
= -qvBj
force is downwards
d ) v = - v k
F = q( - vk x -Bk )
= 0
No force will be created
e ) v = v k
F = q( vk x -Bk )
= 0
No force will be created
Which angle is the angle of reflection? 1, 2, 0R 3?
Answer:
The answer is angle 2.
Explanation:
Angle 2 is the Reflected Ray of angle 1.
Angle 3 is the Refracted angle.
Angle 2 is the angle of reflection. the light incident will bounce to the opposite end after striking is called the reflection.
What is the law of reflection?The law of reflection specifies that upon reflection from a downy surface, the slope of the reflected ray is similar to the slope of the incident ray.
The reflected ray is consistently in the plane determined by the incident ray and perpendicular to the surface at the point of reference of the incident ray.
When the light rays descend on the smooth surface, the angle of reflection is similar to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in a similar plane.
In the reflection, the light incident will bounce to the opposite end after striking. Angle 2 satisfies the properties of the reflection.
Hence angle 2 is the angle of reflection.
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You are to drive to an interview in another town, at a distance of 310 km on an expressway. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 42.0 km/h for 42.0 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview
Answer:
Explanation:
Time to cover first 100 km = 1 hour.
time remaining = 3.15 - 1 = 2.15 hour .
Time to cover next 42 km = 1 hour .
Time remaining = 2.15-1 = 1.15 hour.
Distance to be covered = 310 - 142
= 168 km
least speed needed = distance remaining / time remaining
= 168 / 1.15
= 146.08 km / h .
Consider two sinusoidal waves traveling along a string, modeled as y1(x, t) = (0.4 m)sin[(3 m−1)x + (2 s−1)t] and y2(x, t) = (0.8 m)sin[(6 m−1)x − (5 s−1)t]. What is the vertical displacement (in m) of the resultant wave formed by the interference of the two waves at the position x = 0.9 m at time t = 0.4 s? (Indicate the direction with the sign of your answer.)
Answer:
y(x,t)=-0.395m
Explanation:
The wave function modeling waves are:
[tex]y_1(x,t)[/tex]= (0.4 m)sin[(3 [tex]m^-^1[/tex])x + (2 [tex]s^-^1[/tex])t]
[tex]y_2(x,t)[/tex]=(0.8 m)sin[(6 [tex]m^-^1[/tex])x − (5 [tex]s^-^1[/tex])t]
The principle of superposition can be defined as the resultant of [tex]y_1[/tex] and [tex]y_2[/tex] is equal to their algebraic sum
[tex]y(x,t)=y_1(x,t)+y_2(x,t)[/tex]
y(x,t)= (0.4 m)sin[(3 [tex]m^-^1[/tex])x + (2 [tex]s^-^1[/tex])t] + (0.8 m)sin[(6 [tex]m^-^1[/tex])x − (5 [tex]s^-^1[/tex])t]
Substitute 0.9m for x and 0.4s for t as required
y(x,t)= (0.4 )sin[(3 [tex]m^-^1[/tex])(0.9) + (2 [tex]s^-^1[/tex])(0.4)] + (0.8 m)sin[(6 [tex]m^-^1[/tex])(0.9)− (5 [tex]s^-^1[/tex])(0.4)]
y(x,t)= -0.14 - 0.255
y(x,t)=-0.395m
The vertical displacement of the resultant wave formed by the interference of the two waves is -0.395m
Interference of waves:Two sinusoidal waves are given such that:
y₁(x, t) = (0.4 m)sin[(3 m⁻¹)x + (2s⁻¹)t] and
y₂(x, t) = (0.8 m)sin[(6 m⁻¹)x − (5s⁻¹)t]
The superposition of the two waves gives the outcome of the interference at x = 0.9 m and t = 0.4 s.
y(x,t) = y₁(x, t) + y₂(x, t)
y(0.9,0.4) = y₁(0.9, 0.4) + y₂(0.9, 0.4)
y(0.9,0.4) = (0.4 m)sin[(3 m⁻¹)0.9 + (2s⁻¹)0.4] + (0.8 m)sin[(6 m⁻¹)0.9 − (5s⁻¹)0.4]
y(0.9,0.4) = -0.395m
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A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before coming to rest. Assume that the acceleration of the bullet is constant. What force is exerted by the wood on the bullet
Answer:
The force exerted by the wood on the bullet is 399.01 N
Explanation:
Given;
mass of bullet, m = 0.0021 kg
initial velocity of the bullet, u = 497 m/s
final velocity of the bullet, v = 0
distance traveled by the bullet, S = 0.65 m
Determine the acceleration of the bullet which is the deceleration.
Apply kinematic equation;
V² = U² + 2aS
0 = 497² - (2 x 0.65)a
0 = 247009 - 1.3a
1.3a = 247009
a = 247009 / 1.3
a = 190006.92 m/s²
Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;
F = ma
F = 0.0021 x 190006.92
F = 399.01 N
Therefore, the force exerted by the wood on the bullet is 399.01 N
When time is measured in days, the decay constant for a particular radioactive isotope is 0.16. Determine the time required for a confined sample of the isotope to fall to 80% of its original value.
Answer:
The time take is [tex]t = 1.3964 \ days[/tex]
Explanation:
From the question we are told that
The decay constant is [tex]\lambda = 0.16[/tex]
The percentage fall is [tex]c = 0.80[/tex]
The equation for radioactive decay is mathematically represented as
[tex]N(t) = N_o * e^{- \lambda t }[/tex]
Where is [tex]N(t)[/tex] is the new amount of the new the isotope while [tex]N_o[/tex] is the original
At initial [tex]N_o = 100[/tex]% = 1
At [tex]N(t ) = 80[/tex]% = 0.80
[tex]0.80 = 1 * e^{- 0.16 t }[/tex]
=> [tex]-0.223 = -0.16 t[/tex]
=> [tex]t = 1.3964 \ days[/tex]
Answer:
t = 1.4 days
Explanation:
The law of radioactive decay gives the amount of radioactive substance, left after a certain amount of time has passed. The formula of law of radioactive decay is given as follows:
N = N₀ (e)^-λt
where,
λ = decay constant = 0.16
N₀ = Initial Amount of the Substance
N = The Amount of Substance Left after Decay = 80% of N₀ = 0.8 N₀
t = Time Required by the Substance to decay to final value = ?
Substituting these values in the law of radioactive decay formula, we get:
0.8 N₀ = N₀ (e)^-0.16 t
0.8 = (e)^-0.16 t
ln (0.8) = -0.16 t
t = - 0.2231/-0.16
t = 1.4 days
A transmission line consisting of two concentric circular cylinders of metal, with radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0. A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0. The cylinders are symmetric about the z-axis. Calculate the time-averaged power flow, P, along the line.
Answer:
P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.
Explanation:
So, we will be making use of the data or parameters given in the question above;
=>" radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0."
=> " A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0."
So, we will be making use of the two equations below;
Ë = ( λ/ 2πEP) × P'. --------------------------(1)..
B' = √ μE × ( λ/ 2πEP) × P' --------------(2).
Where equation (1) and (2) represent Gauss' law and magnetic field equation respectively.
Jo = 1/√ μE × λ/2 × π × a.
When we solve for charge per unit length, we have;
λ = 2 × π × Jo × a × √ μE.
The energy flux,s = E' × J'= √μE× |Jo(Z) |^2 × a^2/b^2 { cos^2 kz - wt + Avg Jo} Z'.
Hence, the time. Average power flux = 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z'.
Therefore, P = ∫z' . <s> da
P = ∫ ∫ 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z' pd pd p Θ.
(Take limit on the first at second integration as : 2π,0 and b,a).
P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.
The kinetic energy of a ball with a mass of 0.5 kg in a velocity of 10 meterss
Answer:
The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is J.
Explanation:
Thus the kinetic energy of the ball is 25 J. The unit Joule (J) is the same as kgm^2/s^2 and is the SI unit for kinetic energy.
Hope this helps.
A ball is dropped from the top of a 91-m-high building. What speed does the ball have in falling 3.2 s?
Answer:
The speed of the ball is 28.4m/s.
Explanation:
Given that the formula of speed is Speed = Distance/Time. So you have to substitute the values into the formula :
[tex]speed = distance \div time[/tex]
Let distance = 91m,
Let time = 3.2s,
[tex]speed = 91 \div 3.2[/tex]
[tex]speed = 28.4 \: metrepersecond[/tex]
A 58.0 kg snow skier is on the top of 351 m high hill. After she has gone down a vertical distance of 142 m, what is her mechanical energy? Explain your answer (CER)!
Answer:
Explanation:
Initially skier is at a height of 351 m . Her kinetic enery will be nil because she is at rest . Her potential energy will be calculated as follows
potential energy = mgh where m is mass , h is height and g is acceleration due to gravity
potential energy = 58 x 9.8 x 351
= 199508 .4 J
Total mehanical energy = potential energy + kinetic energy
= 199508.4 J
According to conservation of mechanical energy , at the height of 142 m also total mechanical energy will be same . At this height some potential energy will be converted into kinetic energy but total of potential and kinetic energy will be same.
At the same temperature, two wires made of pure copper have different resistances. The same voltage is applied at the ends of each wire. The wires may differ in:________
a. length.b. cross-sectional area.c. resistivity.d. amount of electric current passing through them
Answer:
A and B are true
Explanation:
C and D are false
Answer:
The answer is the length, cross-sectional area, and the amount of electric current passing through them.
So,
a
b
d
are the correct answers.
Explanation:
I hope this helps future Physics students who are also struggling...
If you have any questions let me know!!
This answer is 100% correct
When solving problems involving forces and Newton's laws, the following summary of things to do will start your mind thinking about getting involved in the problem at hand.
When solving problems involving forces and Newton'
Apply these steps
Use the steps outlined above to find the magnitude of the acceleration a of a chair and the magnitude of the normal force FN acting on the chair: Yusef pushes a chair of mass m = 50.0 kg across a carpeted floor with a force Fp (the subscript 'p' here is lowercase and throughout the question) of magnitude Fp = 168 N directed at ? = 35.0 degrees below the horizontal. The magnitude of the kinetic frictional force between the carpet and the chair is Fk= 99.7 N .
Part A
Identify and sketch all the external forces acting on the chair. Because the chair can be represented as a point particle of mass m, draw the forces with their tails centered on the black dot in the middle of the chair. Be certain to draw your forces so that they have the correct orientation.
Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded.
Answer:
Explanation:
There is no acceleration in vertical direction
Fn = mg + Fp sin 35
Frictional force = 99.7 N
Net force in forward direction = Fp cos 35 - 99.7
= 168 cos 35 - 99.7
= 137.61-99.7
= 37.91 N
net acceleration of chair = net force / mass
= 37.91 / 50
= .7582 m / s² .
As you are waiting at the Willard airport for a friend coming from Chicago, suddenly something goes wrong with the loudspeakers, and they start making a low pitch hum. You realized that you are standing exactly in the middle between the two loudspeakers on the opposite walls. Since the sound from both loudspeakers has the same intensity, you are standing in the interference maximum As you move toward one of the loudspeakers, the intensity decreases. When you have moved by 0.5 m , you hit the first point of zero intensity. Assuming that the speed of sound is v = 330 m/s , what is the frequency f of the sound in Hz ?
Answer:
The frequency is [tex]f = 165 Hz[/tex]
Explanation:
From the question we are told that
The position of zero intensity is [tex]L = 0.5 \ m[/tex] from the center
Now the wavelength of the sound is mathematical represented as
[tex]\lambda = 4 L[/tex]
[tex]\lambda = 4 * 0.5[/tex]
[tex]\lambda = 2 \ m[/tex]
Now the frequency of the sound is mathematically represented as
[tex]f = \frac{v}{\lambda}[/tex]
substituting values
[tex]f = \frac{330}{ 2}[/tex]
[tex]f = 165 Hz[/tex]
Bats are capable of navigating using the earth's field-a plus for an animal that may fly great distances from its roost at night. If, while sleeping during the day, bats are exposed to a field of a similar magnitude but different direction than the earth's field, they are more likely to lose their way during their next lengthy night flight. Suppose you are a researcher doing such an experiment in a location where the earth's field is 50muT at a 60 degree angle below horizontal. You make a 70 cm diameter, 100-turn coil around a roosting box; the sleeping bats are at the center of the coil. You wish to pass a current through the coil to produce a field that, when combined with the earth's field, creates a net field with the same strength and dip angle (60 degree below horizontal) as the earth's field but with a horizontal component that points south rather than north.What is the proper orientation of the coil? What is the necessary current? Express your answer to two significant figures and include the appropriate units.
Answer:
The correct magnitude of the coil's magnetic field= =50μT
Explanation :
The magnetic field takes place as a result of movement of charge I e current, can also occurs from magnetised material, magnetic field as a result of charge movement can be deducted using right hand grip rule.
At the equator magnetic field lines are parallel towards the earth's surface and the angle of inclination of the magnetic lines of force at the horizontal position is referred to as the angle of dip at the point.
As the current is produced then the varying magnetic field is opposed ,then there is induced current when the coil is positioned at varying magnetic field.
Given from the question,
angle below horizontal θ=60-degree
The Earth's magnetic field B=50μT
The horizontal magnetic field can be expressed in terms of the formula below;
BH=Bcosθ
B(H) = earth's horizontal component of magnetic field
Ø is the angle between coil's field
B=the magnetic field in Tesla
Then,
BH=50μT×cos60∘
=50μT× 0.5
As the current is passed through the coil to produce a field , when combined with the earth's field, which creates a net field with the same strength and dip angle (60 degrees below horizontal) as the earth's field.
It can be deducted that B has the same magnitude and angle which makes the those vertical component to cancel each other since they are the same.
For the magnetic field to be pointed out at North direction, we can calculate the corrected magnetic field using the formula below
Bc=2BH
Bc=2×50μT× 0.5=50μT
The correct magnitude of the coil's magnetic field= =50μT
If an object is in equilibrium, which of the following statements is not true? A) The velocity is constant. B) The object must be at rest. C) The net force acting on the object is zero. D) The acceleration of the object is zero. E) The speed of the object remains constant.
When the object is in equilibrium so the statement i.e. not true should be that object should be at rest.
Newton second law of motion:
In the case when the net force of an object should be zero at the time when it is in equilibrium. In the case when the net force should be zero the acceleration should also be zero. And, if the acceleration if zero so the speed and velocity should remain the same. So for maintaining the equilibrium, the object should not have to be only at rest.
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A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the base of the cylinder. The gas is now compressed and 0.1 kJ of boundary work is added to the gas. The temperature of the gas remains constant during this process.
a. How much heat was transferred to/from the gas?
b. What is the final volume and pressure in the cylinder?
c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?
Answer:
ΔQ = 0.1 kJ
[tex]\mathbf{v_f = 1.445*10^{-3} m^3}[/tex]
[tex]\mathbf{P_f = 156.5 \ kPa}[/tex]
ΔS = -0.337 J/K
The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.
Explanation:
GIven that:
Diameter of the piston-cylinder = 12 cm
Pressure of the piston-cylinder = 100 kPa
Temperature =24 °C
Length of the piston = 20 cm
Boundary work ΔW = 0.1 kJ
The gas is compressed and The temperature of the gas remains constant during this process.
We are to find ;
a. How much heat was transferred to/from the gas?
According to the first law of thermodynamics ;
ΔQ = ΔU + ΔW
Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.
Now
ΔQ = ΔU + ΔW
ΔQ = 0 + 0.1 kJ
ΔQ = 0.1 kJ
Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ
b. What is the final volume and pressure in the cylinder?
In an isothermal process;
Workdone W = [tex]\int dW[/tex]
[tex]W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}[/tex]
Since the gas is compressed ; then [tex]v_f< v_i[/tex]
However;
[tex]W =- nRT \ In \dfrac{V_f}{V_i}[/tex]
[tex]W =- P_1V_1 \ In \dfrac{V_f}{V_i}[/tex]
The initial volume for the cylinder is calculated as ;
[tex]v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3[/tex]
Replacing over values into the above equation; we have :
[tex]100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}[/tex]
The final pressure can be calculated by using :
[tex]P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f[/tex]
[tex]P_f =\dfrac{P_iV_i}{V_f}[/tex]
[tex]P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}[/tex]
[tex]P_f = 1.565*10^2 \ kPa[/tex]
[tex]\mathbf{P_f = 156.5 \ kPa}[/tex]
c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?
The change in entropy of the gas is given by the formula:
[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]
where
T = 24 °C = (24+273)K
T = 297 K
[tex]\Delta S=\dfrac{-100 \ J}{297 \ K}[/tex]
ΔS = -0.337 J/K
The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.
Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of −7.50 nC? electrons (b) How many electrons must be removed from a neutral object to leave a net charge of 0.580 µC? electrons
Answer:
a) 4.681*10^10 electrons
b) 3.67*10^12 electrons
Explanation:
The amount of electrons in a charge of 1C is:
[tex]1C=6.2415*10^{18}\ electrons[/tex]
You use the previous equality as a conversion factor.
a) The sing of the charge is not important in the calculation of the number electrons, so, you use the absolute value of the charge
[tex]7.50nC=7.50*10^{-9}C*\frac{6.2415*10^{18}}{1C}=4.681*10^{10}\ electrons[/tex]
In 7.50nC there are 4.61*10^18 electrons
b)
[tex]0.580\mu C=0.580*10^{-6}C*\frac{6.2415*10^{18}}{1C}=3.67*10^{12}\ electrons[/tex]
To obtain a charge of 0.580 µC in a neutral object you need to take out 3.67*10^12 electrons
Why intermediate elements has negative packing fraction?
Explanation:
The negative value of the packing fraction indicates that the actual isotopic mass is less the mass number.
g 16.0 kg sled is pulled on a horizontal frictionless ice rink with a light spring of force constant 220 N/m. If the sled has an acceleration of 2.0 when pulled horizontally, how much does the spring stretch (in meters)
Answer:
Spring stretches by 0.145m
Explanation:
We are given;
Mass of sled;m = 16 kg
Force constant;k = 220 N/m
Acceleration;a = 2 m/s²
Now, we know that from Newton's second law of motion, Force is given by the expression ; F = ma
Thus,
F = 16 x 2
F = 32 N
Now from Hooke's law, we know that this force is expressed as;
F = -kx
Where x is the distance that the spring stretches.
The minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. But we'll take the magnitude and therefore ignore the negative sign to get F = Kx
Thus,
32 = 220x
x = 32/220
x = 0.145 m
Answer:
[tex]x=-0.145m[/tex]
Explanation:
Information we have:
Mass of the sled: [tex]m=16kg[/tex]
Spring constant: [tex]k=220N/m[/tex]
acceleration of the sled: [tex]a=2m/s^2[/tex]
We use the formula for the force on a spring (Hooke's Law)
[tex]F=-kx[/tex]
where k is the spring constant and x the distance the mass moved from the equilibrium
and we also use Newton's second Law of motion:
[tex]F=ma[/tex]
we combine these two equations:
[tex]-kx=ma[/tex]
we solve the last equation for the distance x:
[tex]x=-\frac{ma}{k}[/tex]
and substitute the values:
[tex]x=-\frac{(16kg)(2m/s^2)}{220N/m}\\ x=-0.145m[/tex]
the negative number means that the mass was moved in the opposite direction that the force, this because the force in a spring is restorative and points towards the equilibrium point
One of the advantages of skates is that they allow you to push against ice, which typically has low friction. Let’s consider an hockey player on frictionless ice who is going to shoot a slap shot. The player weighs Mp = 100kg. He is going to fire the puck, which weighs Mpuck = 0.5kg with a velocity of v = 50(m/s). How fast does the hockey player recoil in the direction opposite to the puck’s trajectory. If the force he exerts on the puck occurs over a time, ∆t = 0.1s what is the average force needed to create the impulse that accelerates the puck to that velocity?
Answer:
a) v1f = 0.25 m/s
b) F = 50000N
Explanation:
a) In order to calculate the speed of the player after he fires the puck, you use the conservation of momentum law. Before the puck is in motion and after the total momentum of both player and puck must conserve:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex] (1)
m1: mass of the player = 100kg
v1i: initial velocity of the player = 0m/s
v2f: final velocity of the puck = ?
m2: mass of the puck = 0.5kg
v2i: initial velocity of the puck = 0 m/s
v2f: final velocity of the puck = 50 m/s
You replace these values into equation (1) and you solve for v1f (final velocity of player):
[tex]0kgm/s+0kgm/s=(100kg)v_{1f}+(0.5kg)(50m/s)\\\\v_{1f}=-\frac{(0.5kg)(50m/s)}{100kg}\\\\v_{1f}=-0.25\frac{m}{s}[/tex]
The minus sign means that player moves in a opposite direction to the motion of the puck
The velocity of the player after he fires the puck is 0.25 m/s
b) The force needed is given by the change in time , of the momentum of the player, which is given by:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex]
The change on the velocity of the puck is 50m/s, and the time interval is 0.1s.
[tex]F=(100kg)\frac{50m/s}{0.1s}=50000N[/tex]
The force needed to create the needed impulse, to accelerate the puck is 50000N
The sound of a tuba is very low. Why?
0
O When you blow into a tuba the air vibrates very strongly.
) O When you blow into a tuba the air vibrates very quickly.
O When you blow into a tuba the air vibrates very softly.
O When you blow into a tuba the air vibrates very slowly.
Answer:
When you blow into a tuba the air vibrates very slowly.
Explanation:
Tuba is a buzz instrument ie sound is produced in it with the help of lip vibration . It is the lowest pitched musical instrument in the brass family .
Due to absence of resonance in it , it produces music of lowest pitch , So when one blows into it the air column of the instrument vibrates very slowly producing low pitched sound.
Sometimes earthquakes occur at a boundary between the Indian and Eurasian Plates. The collisions of these two continental plates formed the Himalayan mountain range.
Based on this information, which type of boundary exists between the Indian and Eurasian Plates?
continental-continental convergent
continental-continental divergent
oceanic-continental divergent
oceanic-continental convergent
Answer:
Continental Continental convergent.
Explanation:
Continental Continental convergent is a type of earthquakes that occur between boundaries of two continents in which the tectonic plates move closer to each other or converge.
Eurasian plates refers to tectonic plates that is found in the Continent of Eurasia which include Asian and Europe excluding India subcontinent.
From the question, the earthquake occur between boundaries of two continents India and Eurasia and this converging and collision of the two continental plates formed the Himalayan mountain range.
Therefore, it is continental continental convergent.
Answer:
A
Explanation:
Four cubes of the same volume are made of different materials: lead (density 11,300 kg/m3), aluminum (density 2700 kg/m3), wood (density 800 kg/m3), and Styrofoam (density 50 kg/m3). You place the cubes in a large container filled with water.Rank the buoyant forces that the water exerts on the cubes from largest to smallest.Rank from largest to smallest. To rank items as equivalent, overlap them.
Answer:
Lead > Aluminium > Wood > Styrofoam
Explanation:
Buoyant force is described by the Archimedes's Principle, which states that buoyant force is equal to the weight of the fluid displaced by the submerged object. By Newton's Laws, the buoyant force is represented by the following equation of equilibrium:
[tex]\Sigma F = F_{D} - W_{cube} = 0[/tex]
[tex]F_{D} = W_{cube}[/tex]
[tex]F_{D} = \rho_{cube} \cdot g \cdot V_{cube}[/tex]
Where:
[tex]\rho_{cube}[/tex] - Density of the cube, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]V_{cube}[/tex] - Volume of the cube, measured in cubic meters.
Let suppose that volume of the cube is known. Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the buoyant force is computed for each material:
Lead ([tex]\rho_{cube} = 11,300\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(11,300\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 110,819.1V_{cube}[/tex]
Aluminium ([tex]\rho_{cube} = 2,700\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(2,700\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 26478.9V_{cube}[/tex]
Wood ([tex]\rho_{cube} = 800\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(800\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 7845.6V_{cube}[/tex]
Styrofoam ([tex]\rho_{cube} = 50\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(50\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 490.35V_{cube}[/tex]
Therefore, the buoyant forces that the water exerts on the cubes from largest to smallest corresponds to: Lead > Aluminium > Wood > Styrofoam.
The inhabitants of a planet in another galaxy have their eyes at the exact center of their 4.0-m-long bodies. How long must a plane mirror be for such a creature to be able to see all of its body in the mirror
Answer:
The correct answer will be "2 m".
Explanation:
As we know,
⇒ [tex]\frac{Mirror's \ size}{Body's \ height} =\frac{1}{2}[/tex]
Now,
⇒ [tex]Size \ of \ a \ mirror = \frac{n}{2}[/tex]
Then,
⇒ [tex]Size \ of \ mirror = \frac{4}{2}[/tex]
⇒ [tex]=2 \ meter[/tex]
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a specified tile surface. A typical athletic shoe has a coefficient of 0.870. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.20 m on a tile surface if she is wearing the following footwear?
(a) footwear meeting the Postal Service minimum
(b) a typical athletic shoe
Answer:
(a) 1.14 s
(b) 0.87 s
Explanation:
A person moves by the help of frictional force, as a result of gtound reaction. So, the formula for frictional force is:
F = μR
where,
F = frictional force
μ = coefficient of friction
R = Normal Reaction = Weight of Body = W = mg
Therefore,
F = μmg
but, from Newton's 2nd Law of Motion:
F = ma
Comparing both equations, we get:
μmg = ma
a = μg ---------- equation (1)
Now, to calculate the distance moved by a body, we use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
using equation (1), we get:
s = (Vi)(t) + (0.5)μgt²
where,
s = distance moved by body
Vi = initial velocity of body
t = time taken to cover the distance
g = acceleration due to gravity
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.5
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)
t = √1.30612 s²
t = 1.14 s
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.5
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)
t = √1.30612 s²
t = 1.14 s
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.87
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.87)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.87)(9.8 m/s²)
t = √0.75 s²
t = 0.87 s
The speed of sound
Medium Air (0C) Air (20°C) Helium (0°C) Ethyl alcohol Water Human tissue (ultrasound Lead Aluminum Granite Diamond
Speed (m/s) 331 343 970 1170 1480 1540 1200 5100 6000 12.000
Oil explorers set off explosives to make loud sounds, then listen for the echoes from underground oil deposits. Geologists suspect that there is oil under 480-m-deep Lake Physics. It's known that Lake Physics is carved out of a granite basin. Explorers detect a weak echo 0.930 ss after exploding dynamite at the lake surface. Part A If it's really oil, how deep will they have to drill into the granite to reach it
Answer:
Explanation:
Let the required depth be d .
Sound will first travel trough 480 m deep lake . Then it will enter granite layer . Sound travelling through granite will reach oil level , get reflected and come back to the surface of lake as echo . Total time taken by sound to travel total distance is .93 s
Total distance = 2d + 2 x 480 m
= 2d + 960 m
speed of sound in granite is given as 6000 m / s and speed through water is 1480 m /s
total time taken
= 2d / 6000 + 960 / 1480 = .93
2d / 6000 + .6486 = .93
2d / 6000 = .2814
d= 844.2 m
You are throwing a stone straight-up in the absence of air friction. If the stone is caught at the same height it was thrown from, which of the following is true of the stones motion?
1. the acceleration is zero at the top point of the motion.
2. the acceleration is minimum at the top point of the motion.
3. acceleration and velocity always point in opposite directions.
4. the time going up is equal to the time coming down the acceleration is directly proportional to the velocity.
Answer:
4. the time going up is equal to the time coming down the acceleration is directly proportional to the velocity.
Explanation:
1. FALSE
The acceleration at the top point is equal to the acceleration due to gravity (9.8 m/s²). It is the velocity of the stone that becomes zero for a moment, at the top point.
2. FALSE
The acceleration of the body in a free fall motion always remains constant, and its value is equal to the acceleration due to gravity (9.8 m/s²)
3. FALSE
The velocity and acceleration point in opposite direction, during upward motion only (Velocity points up and acceleration points down). But they point in same direction during the downward motion (Both velocity and acceleration point down).
4. TRUE
Since, there is no air friction. Therefore, the acceleration for both upward an downward motion will be constant, without any opposing force. Hence, the time taken by the stone to go up will be equal to time taken by the stone to come down.
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.540 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision. (a) What is the speed of the 20.0 gram block after the collision
Answer:
0.180 m/s
Explanation:
Solving the equations for conservation of momentum and energy for an elastic collision gives ...
v₁' = ((m₁ -m2)v₁ +2m₂v₂)/(m₁ +m₂) . . . . v₁' is the velocity of m₁ after collision
__
Here, we have (m₁, m₂, v₁, v₂) = (20 g, 40 g, 0.540 m/s, 0 m/s).
Substituting these values in to the equation for v₁', we have ...
v₁' = ((20 -40)(0.540) +2(40)(0))/(20 +40) = (-20/60)(0.540)
v₁' = -0.180 . . . m/s
The speed of the 20 g block after the collision is 0.180 m/s to the left.
Mike ran ten laps around his school's 1/4 mile track in 15 minutes. He
finishes his run at the exact same point that he started. What was his average
velocity in miles per hour?
Answer:
v = 0
Explanation:
It is given that,
Mike ran ten laps around his school's 1/4 mile track in 15 minutes. He finishes his run at the exact same point that he started. We need to find his average velocity. Average velocity is equal to net displacement divided by total time taken.
As he returns exactly at the same point that he started, it means his displacement or the shortest distance covered is equal to 0. As a result, its average velocity is equal to 0.
