Answer:
The mass of the ice added = 16.71 g
Explanation:
The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.
But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.
Heat gained by the ice = Heat lost by the 326 g of water.
Let the mass of ice be m
The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)
Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT
m = unknown mass of ice
C = Specific Heat capacity of ice = 2.108 J/g°C
ΔT = change in temperature = 0 - (-5.5) = 5.5°C
Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J
Heat used by the ice to melt at 0°C = mL
m = unknown mass of ice
L = Latent Heat of fusion of ice to water = 334 J/g
Heat used by the ice to melt at 0°C = m×334 = (334m) J
Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT
m = unknown mass of water (which was ice)
C = Specific Heat capacity of water = 4.186 J/g°C
ΔT = change in temperature = 15 - 0 = 15°C
Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J
Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J
Heat lost by the water in the calorimeter cup = MCΔT
M = mass of water in the calorimeter cup = 326 g
C = specific heat capacity of water = 4.186 J/g°C
ΔT = change in temperature = 20 - 15 = 5°C
Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J
Heat gained by the ice = Heat lost by the 326 g of water.
408.384m = 6,823.18
m = (6,823.18/408.384)
m = 16.71 g
Hope this Helps!!!
Name and draw the devices that can convert digital signal to analog.
Answer:
modem, digital music players, optical communication
Explanation:
Photos of devices are attached. An example of R2R logic circuit is also attached. DAC conversion is also done via pulse code modulation
Consider five charged particles: A,B,C,D,and E.
(A attracts B),(C attracts D ), (B repels C), and (D repels E). If C is Positive,what is the charge of the other particles?
Answer:
A_negative
B_positive
C_positive
D_positive
E_negative
Explanation:
according to the law of electrostatic which is
like charge repels and unlike charge attract.
Answer:A=negative, B=positive, C=positive, D=negative, E=negative
Explanation:
Like poles or charges repels and unlike poles or charges attract each other
Dispositivo que muestra tanto la dirección como la magnitud de la corriente eléctrica.
Answer:
Galvanómetro
Explanation:
Un galvanómetro es un dispositivo eléctrico utilizado para detectar la presencia de corriente y voltaje pequeños o para medir su magnitud. Los galvanómetros se utilizan principalmente en puentes y potenciómetros.
A neutron star has about one and a half times the mass of our Sun but has collapsed to a radius of 10 kmkm . Part A What is the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface?
Answer:
gₓ = 1.36 x 10¹³ g
Explanation:
The value of acceleration due to gravity at a certain place is given by the following formula:
gₓ = GM/R²
where,
gₓ = acceleration due to gravity on the surface of neutron star
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the star = 10 * Mass of sun = (10)(2 x 10³⁰ kg) = 2 x 10³¹ kg
R = 10 km = 10⁴ m
Therefore,
gₓ = (6.67 x 10⁻¹¹ N.m²/kg²)(2 x 10³¹)/(10⁴)²
gₓ = 1.334 x 10¹⁴ m/s²
Hence, comparing it with the free-fall acceleration at Earth's Surface:
gₓ/g = (1.334 x 10¹⁴)/9.8
gₓ = 1.36 x 10¹³ g
The acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].
The given parameters:
Mass of the neutron star, m = 1.5 MRadius of the neutron star, R = 10 kmkmThe acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is calculated as follows;
[tex]F = mg = \frac{GM_sm}{R^2} \\\\(1.5 M_s)g = \frac{GM_s(1.5 M_s)}{R^2} \\\\g = \frac{GM_s}{R^2} \\\\[/tex]
where;
[tex]M_s[/tex] is the mass of the Sun = 1.989 x 10³⁰ kg.
[tex]g = \frac{6.67 \times 10^{-11} \times 1.989 \times 10^{30} }{(10,000,000)^2} \\\\g = 1.326 \times 10^{6} \ m/s^2[/tex]
In terms of gravity of Earth [tex](g_E)[/tex];
[tex]= \frac{1.326 \times 10^6}{9.81} = 1.35 \times 10^5 \\\\= 1.35 \times 10^5 \ g_E[/tex]
Thus, the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].
Learn more about acceleration due to gravity here: https://brainly.com/question/88039
A 0,2 kg ball is whirled at constant angular velocity in a vertical circle at the end of a 1.25m long string. If the maximum tension the string can stand is 6N, what is the maximum angular velocity a ball can have in radians per second?
Answer:
24 rad/s
Explanation:
.................
A certain radio wave has a wavelength of 6.0 × 10-2m. What is its frequency in hertz?
Answer:
The frequency of the wave is 5 x 10⁹ Hz
Explanation:
Given;
wavelength of the radio wave, λ = 6.0 × 10⁻²m
radio wave is an example of electromagnetic wave, and electromagnetic waves travel with speed of light, which is equal to 3 x 10⁸ m/s².
Applying wave equation;
V = F λ
where;
V is the speed of the wave
F is the frequency of the wave
λ is the wavelength
Make F the subject of the formula
F = V / λ
F = (3 x 10⁸) / (6.0 × 10⁻²)
F = 5 x 10⁹ Hz
Therefore, the frequency of the wave is 5 x 10⁹ Hz
A long metallic wire is stretched along the x direction. The applied potential difference along
the wire is 12 volts. The resistance of the wire is estimated to be 8 Ohm.
a. Calculate the current in the wire and the passing charge in 2 seconds.
b. Calculate the electric power dissipated in the wire.
This wire has a length of 100 m and the material of the wire has a resistivity of 1.6 x 10 -8 Ωm.
c. Calculate the cross-section area of the wire.
d. Calculate the conductivity of the wire.
e. What is the capacitance of the capacitor, that is connected to a series resistor of 8 MΩ in an RC
circuit that has a time constant of one second.
Answer:
a) I = 1.5 A , b) P = 18 W , c) A = 8 10⁻⁷ m², d) σ = 0.625 10⁸ (Ω m)⁻¹,
e) C = 0.125 10⁻⁶ F
Explanation:
a) for this exercise let's use Ohm's Law
V = I R
I = V / R
I = 12/8
I = 1.5 A
b) the power is given by the expression
P = V I
P = V V / R = V² / R
P = 12²/8
P = 18 W
c) the resistance of the wire is given by
R = ρ l / A
A = ρ l / R
A = 1.6 10⁻⁸ 100/8
A = 8 10⁻⁷ m²
d) conductivity is the inverse of resistivity
σ = 1 / ρ
σ = 1 / 1,610⁻⁸
σ = 0.625 10⁸ (Ω m)⁻¹
e) In an RC circuit the response time is
τ = RC
C = τ / R
C = 1/8 10⁶
C = 0.125 10⁻⁶ F
A solenoid is 2.50 cm in diameter and 30.0 cm long. It has 300 turns and
carries 12 A.
(a) Calculate the magnetic field inside the solenoid.
(b) Calculate the magnetic flux through the surface of a circle of radius 1.00 cm, which is
positioned perpendicular to and centered on the axis of the solenoid.
(c) The perimeter of this circle is made of conducting material. The current in the solenoid
uniformly goes from 12 A to 10 A in 0.001 seconds. What e.m.f. is generated in the
conducting material?
(d) Now, inside the solenoid we put a bar of steel, a ferromagnetic material. Steel has got a
magnetic permeability μm = 4000μ0 . If the current in the solenoid is 12 A, what’s the
total magnetic field in the steel bar?
Answer:
a) B = 0.015 T
b) ФB = 4.71*10⁻6 W
c) emf = 7.89*10^-4 V
d) B = 60.31 T
Explanation:
(a) To find the magnitude of the magnetic field inside the solenoid you use the following formula:
[tex]B=\frac{\mu_oNI}{L}[/tex] (1)
B: magnitude of the magnetic field
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
N: turns of the solenoid = 300
L: length = 2.50cm = 0.025 m
I: current = 12 A
You replace the values of the variables in the equation (1):
[tex]B=\frac{(4\pi *10^{-7}T/A)(300)(12A)}{0.3m}=0.015T[/tex]
(b) The magnetic flux is given by:
[tex]\Phi_B=BA[/tex]
A: area = π(0.01m)^2 = 3.1415*10^-4 m^2
[tex]\Phi_B=(0.015T)(3.1415*10^{-4}m^2)=4.71*10^{-6}W[/tex]
(c) The induced emf is given by the following formula:
[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=\frac{A(B_f-B_i)}{\Delta t}[/tex] (2)
Bf and Bi are the final and initial magnetic field. You use the equation (1) into the equation (2):
[tex]emf=-A\mu_o\frac{N}{L}\frac{(I_2-I_1)}{\Delta t}\\\\emf=-\pi(0.01m)^2(4\pi*10^{-7}T/A)\frac{300}{0.3m}\frac{(10A-12A)}{0.001s}\\\\emf=7.89*10^{-4}V[/tex]
(d) With the ferromagnetic material inside the solenoid the magnetic field inside is modified as following:
[tex]B=\frac{\mu NI}{L}\\\\\mu=4000\mu_o=4000(4\pi*10^{-7}T/A)=5.026*10^{-3}T/A\\\\B=\frac{(5.026*10^{-3}T/A)(300)(12A)}{0.3m}=60.31T[/tex]
Following are the calculation to the given points:
Given:
[tex]r = 1.25\ cm \\\\l=30\ cm\\\\ N= 300 \\\\I=12\ A\\\\[/tex]
To find:
[tex]B=?\\\\\Phi_B=?\\\\[/tex]
Solution:
For point a:
Using formula:
[tex]\to \mu_o= 4 \pi 10^{-7} \ \frac{T}{A}[/tex]
[tex]\to N= 300\\\\ \to L= 2.50\ cm = 0.025\ m\\\\ \to I= 12 \ A\\\\[/tex]
[tex]\to B = \frac{\mu_o \ N\ I}{L}\\[/tex]
[tex]=\frac{4 \pi \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{4 \times 3.14 \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{48 \times 3.14\ T \times 300 }{0.3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 3000 }{3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 1000 }{10^{7}}\\\\= 0.015\ T\\\\[/tex]
For point b:
[tex]\to A : area = \pi (0.01m)^2 = 3.14 \times 10^{-4}\ m^2\\\\[/tex]
Using formula:
[tex]\to \Phi_B = BA\\\\[/tex]
[tex]= (0.015\ T)(3.14 \times 10^{-4} \ m^2)\\\\ = 4.71 \times 10^{-6}\ W[/tex]
For point c:
Using formula:
[tex]\to emf = -A \mu_o \frac{N}{L} \frac{(I_2-I_1)}{\Delta t}[/tex]
[tex]= -\pi(0.01 \ m)^2(4\pi \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3\ m} \frac{(10A-12A)}{0.001\ s} \\\\= -3.14 (0.01 \times 0.01 \ m)(4\times 3.14 \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3} \frac{(-2A)}{0.001\ s} \\\\= -3.14 (1 \ m)(12.56\times 10^{-7}\ \frac{T}{A}) 10^6 \frac{(-2A)}{ 10^3\ s} \\\\= -3.14 (1 \ m)(12.56 \frac{T}{A}) \frac{(-2A)}{ 10^4\ s} \\\\= 7.89 \times 10^{-4}\ V\\\\[/tex]
For point d:
[tex]\to B = \frac{\mu NI}{L}\\\\\to \mu = 4000 \mu_o = 4000(4\pi \times 10^{-7}\ \frac{T}{A}) = 5.026 \times 10^{-3} \frac{T}{A}\\\\ \to B = \frac{(5.026\times 10^{-3} \ \frac{T}{A})(300)(12\ A)}{0.3\ m} = 60.31\ T\\\\[/tex]
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A gas is collected from a radioactive material; upon inspection, the gas is identified as helium. The presence of the helium indicates the radioactive sample is most likely decaying by: A). alpha B). beta+ C). beta- D). gamma
Answer:
option (a) alpha I have doublt
3. Calculate the work done to move an object of 50g in a circular path of radius 10cm.
Answer:
Work done in moving an object in circular path is always 0 J
Explanation:
Work doen on any moving object, in general, is given by the following formula:
W = F d Cos θ
where,
W = Work Done
F = Force applied
d = Distance covered
θ = Angle between the direction of motion and the force applied
When the object is moving in a circular path the direction of its motion is always tangential to the circular path. While, the force applied on the object is the centripetal force. And centripetal force is always directed towards the center of the circular path. Therefore, the force and direction of motion are always perpendicular to each other. This means θ = 90°
Therefore,
W = F d Cos 90° = F d(0)
W = 0 J
Consider a system of an 85.0 kg man, his 14.5-kg dog, and the earth. The gravitational potential energy of the system increases by 1.85 103 J when the man climbs a spiral staircase from the first to the second floor of an apartment building. If his dog climbs a normal staircase from the same first floor to the second floor, by how much does the potential energy of the system increase (in J)
Answer:
Explanation:
Increase in gravitational potential energy = m x g x h
where m is mass , g is gravitational acceleration and h is height
In the first case when man climbs
increase in potential = 85 x g x h = 1.85 x 10³ J
gh = 21.7647
when dog climbs
increase in potential = 14.5 x g x h J
= 14.5 x 21.7647
= 315.6 J
A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.11 m and a mass of 0.010 kg. A uniform magnetic field of magnitude 0.055 T is directed from the ceiling to the floor. When a current of I = 29 A exists in the wire, the wire swings upward and, at equilibrium, makes an angle φ with respect to the vertical, as the drawing shows. Find (a) the angle and (b) the tension in each of the two strings.
Answer:
Explanation:
The magnetic force acting horizontally will deflect the wire by angle φ from the vertical
Let T be the tension
T cosφ = mg
Tsinφ = Magnetic force
Tsinφ = BiL , where B is magnetic field , i is current and L is length of wire
Dividing
Tanφ = BiL / mg
= .055 x 29 x .11 / .010 x 9.8
= 1.79
φ = 61° .
Tension T = mg / cosφ
= .01 x 9.8 / cos61
= .2 N .
The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.95 × 103 N with an effective perpendicular lever arm of 3.1 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia of the boxer's forearm?
Answer:
I = 0.483 kgm^2
Explanation:
To know what is the moment of inertia I of the boxer's forearm you use the following formula:
[tex]\tau=I\alpha[/tex] (1)
τ: torque exerted by the forearm
I: moment of inertia
α: angular acceleration = 125 rad/s^2
You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)
[tex]\tau=Fr=(1.95*10^3N)(0.031m)=60.45J[/tex]
Next, you replace this value of τ in the equation (1) and solve for I:
[tex]I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2[/tex]
hence, the moment of inertia of the forearm is 0.483 kgm^2
A transverse sinusoidal wave is traveling rightward along a string stretched along the X-direction. The wavelength is 70.0 cm and 8.00 wave peaks per second pass any fixed point on the string. The maximum speed of the string particles in the direction perpendicular to the string is 3.20 m/s. At time t 0, the element of string at x = 20.0 cm is 4.00 cm above its resting position. (a) Construct two distinct wave functions yfx,t) consistent with the given information. (If the only difference between two wavefunctions is that their phase constants differ by an integer multiple of 2n, then they are not distinct.) (b) Sketch both of these wavefunctions for time t 0 between x 0 and x A har 2 20 /s A= 0.0Ch 560 Ch a)yo) Acos CRxt wt +o) w- 2TF : 50 265 S A Vwa 3 20 5a.2655 =0.0656n 0.7854
Answer:
1111
1Explanation:
111111
1
A 2020 kg car traveling at 14.2 m/s collides with a 2940 kg car that is initally at rest at a stoplight. The cars stick together and move 2.12 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Answer:
The answer is 3,064x[tex]e^{-4}[/tex]
Explanation:
When the collision happens, the momentum of the first car is applied to the both of them.
So we can calculate the force that acts on both cars as:
The momentum of the first car is P = 2020 kg x 14.2 m/s = 28,684 kg.m/sThe acceleration of both cars after the crash is going to be a = P / mtotal which will give us a = 28,684 / (2020+2940) = 5.78 m/s Since the second car was initially not moving, the final acceleration was calculated with the momentum of the first car.Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:
F = (2020+2940) x 5.78 = 28,684The friction force acts in the opposite direction and if they stop after moving 2.12 meters;
Friction force is Ff = μ x N where μ is the friction coefficient and the N is the normal force which is (2020+2940) x 10 if we take gravitational force as 10, equals to 49,600.F - Ffriction = m x V 28,684 - μ x 49,600 = 4960 x 5.78μ = 3,064x[tex]e^{-4}[/tex]How much heat is necessary to increase temperature of 4 kg water from 20 degree Celsius to 80 degree Celsius?
Answer:
2.4 × 10⁵ cal
Explanation:
Step 1: Given data
Mass of water (m): 4 kgInitial temperature: 20°CFinal temperature: 80°CSpecific heat capacity of water (c): 1 cal/g.°CStep 2: Convert the mass to grams
We will use the relationship 1 kg = 1,000 g.
[tex]4kg \times \frac{1,000g}{1kg} = 4,000 g[/tex]
Step 3: Calculate the change in the temperature
[tex]\Delta T = T_f - T_i = 80 \° C - 20 \° C = 60 \° C[/tex]
Step 4: Calculate the heat required (Q)
We will use the following expression.
[tex]Q = c \times m \times \Delta T = \frac{1cal}{g. \° C} \times 4,000 g \times 60 \° C = 2.4 \times 10^{5} cal[/tex]
A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it is released and explain how you know. b. If work is done on the particle after its release, is the work positive or negative
Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.
Linear charge density 4.00×10−12 C/m surrounds an infinitely long line charge. A positively charged elementary particle (mass 1.67×10−27 kg, charge +1.60×10−19 C) is 15.0 cm from this line charge. Consider that this elementary particle is moving at speed 3.20×103 m/s directly toward the line charge.
Part A- Find the initial kinetic energy of this elementary particle.
Part B- Find the closest distance that the elementary particle get to the line charge?
Answer:
A)Kopya
B)YASAK
Explanation:
kopya yasak dostum adın da belli. Başın belaya girmesin
a body of 2.0kg mass makes an elastic collision with another at rest and continues to move in the original direction but with 1/4 of it's original speed.what is the mass of the struck body?
2.0 kg into the bifactor of 1/4 so the mass struck would be 32
An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the negative x direction and the magnetic field is pointing in the negative y direction. In which direction does the released electromagnetic wave travel
Answer: the magnetic wave will travel out of the screen.
Explanation:
Electric field direction is perpendicular to the magnetic field direction. Both are also perpendicular to the direction of the particles.
Using right hand rule to solve this problem,
This pointed finger depicts the electric field direction which the curly fingers depict the direction of the magnetic field. The pointed thumb will depict the direction in which the wave travel. Which is out of the screen.
A baton twirler in a marching band competition grabs one end of her 1.2 kg, 1.0 meter long baton. She throws her baton into the air such that it rises to a height of 5.0 meters while spinning end over end at a rate of 3.5 revolutions per second. How much work did she do on the baton?
Answer:
349 J
Explanation:
Length L of baton = 1.0 m
Mass m of baton = 1.2 kg
Weight W of baton = 1.2 kg x 9.81 m/[tex]s^{2}[/tex] = 11.772 N
Height h reached = 5.0 m
Angular speed ω = 3.5 rev/s = 2π x 3.5 (rad/s) = 21.99 rad/s
Total work done on baton will be the work done in taking it to a height of 5.0 m and the kinetic energy with which the baton rolls.
Work done to bringing it to the height of 5.0 m = weight x height above ground
W x h = 11.772 x 5 = 58.86 J
Velocity v of spinning baton = ω x L = 21.99 x 1 = 21.99 m/s
Kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] =
Total work done on baton = 58.86 + 290.14 = 349 J
A human expedition lands on an alien moon. One of the explorers is able to jump a maximum distance of 16.0 m with an initial speed of 2.90 m/s. Find the gravitational acceleration on the surface of the alien moon. Assume the planet has a negligible atmosphere. (Enter the magnitude in m/s2.)
Answer:
Gravitational acceleration (g) = 0.4205 m/s²
Explanation:
Given:
Distance (R) = 20 m
Initial speed (u) = 2.90 m/s
Find:
Gravitational acceleration (g)
Computation:
⇒ Distance (R) = [Initial speed (u)]²/ Gravitational acceleration (g)
⇒ Gravitational acceleration (g) = [Initial speed (u)]² / Distance (R)
⇒ Gravitational acceleration (g) = 2.90 ² / 20
⇒ Gravitational acceleration (g) = 8.41 / 20
⇒ Gravitational acceleration (g) = 0.4205 m/s²
A 12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s2. What is the magnitude of the contact force exerted on the block by the surface?
Answer:
157.36 N
Explanation:
Contact force is the force which is created due to contact and it is applied on the contact point . The force applied by body on the surface is its weight .
If R be the reaction force of the ground
R = mg + F son30
= 12 x 9.8 + 75 sin 30
= 117.6 + 37.5
= 155.10 N .
friction force = f
Net force in forward direction = F cos 30 - f = ma
75cos 30 - f = 12 x 3.2
f = 65 - 38.4
= 26.6 N
Total force on the surface =√( f² + R² )
√ (26.6² + 155.1²)
= √707.56 + 24056²
=√ 24763.57
= 157.36 N.
contact force = 157.36 N .
list and discuss how the nature of a rural settlements affect the type and expanse of agricultural activities
Answer:
Availability of land for Agricultural activities- The rural areas are known for a lesser degree of development which means lesser factories and other work buildings. The area has undeveloped lands which are usually used for a commercial type of agricultural activities.
Bad road networks: Bad road networks are mainly associated with rural settlements. This hinders to an extent the agricultural activities of planting and harvesting of crops due to difficulties in moving of the crops.
An isotope has 46 electrons, 60 neutrons, and 46 protons. Name the isotope.
Answer:
Palladium
Explanation:
Answer:
palladium-106
Explanation:
46 protons -46 electrons=no charge
46 electrons +60neutron = 106
Thus this is called palladium -106
Which action is due to field forces?
A. an apple falling from a tree
B. a moving car stopping when the brakes are applied
C. the rowing of a boat
D. pushing a chair against the wall
Answer:
a
an apple falling from a tree
Answer an apple falling from a tree
Explanation:
A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45 degrees. With what speed was it thrown?
Answer:
Explanation:
This is the case of horizontal projection from a height:
Time, t = sqrt ( 2h / g )
= sqrt ( 2 * 20 / 9.8 )
= 2.02 s
Vfx = V
Vfy = g* t = 2.02 g
theta (θ)= 45 deg
tan theta (tan θ) = Vfy / Vfx
tan 45 = 2.02 g / V
V = 2.02 * 9.8
= 19.8 m/s
≅ 20m/s
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 52.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.936 m/s2. Calculate her mass.
Answer:
55.56kg
Explanation:
Given:
F= 52N
a=0.936m/s²
Applyinc Newton's second law, that states: force is equal to mass times acceleration.
F = ma
m=F/a =>52 / 0.936
m=55.56kg
Two identical metal balls of radii 2.50
cm are at a center to center distance of
1.00 m from each other. Each ball is
charged so that a point at the surface of
the first ball has an electric potential of
+1.20 x 103 V and a point at the surface
of the other ball has an electric
potential of -1.20 x 103 V. What is the total charge on each ball?
Answer:
+1.33 × [tex]10^{-7}[/tex] C and -1.33 × [tex]10^{-7}[/tex] C respectively.
Explanation:
Electric potential (V) is the work done in moving a unit positive charge from infinity to a reference point within an electric field. It is measured in volts.
V = [tex]\frac{kq}{r}[/tex] ............. 1
where: k is a constant = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex], q is the charge and r is the distance between the charges.
From equation 1,
q = [tex]\frac{Vr}{k}[/tex] ............... 2
The charge on each ball can be determined as;
given that; V = 1.2 × [tex]10^{3}[/tex], k = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex] and r = 1.00 m.
From equation 2,
q = [tex]\frac{1.2*10^{3} * 1.0}{9*10^{9} }[/tex]
= 1.33 × [tex]10^{-7}[/tex] C
Thus, the charge on the first ball is +1.33 × [tex]10^{-7}[/tex] C, while the charge on the second ball is -1.33 × [tex]10^{-7}[/tex] C.
An AC voltage is applied to a purely capacitive circuit. Just as the applied voltage is crossing the zero axis going negative, what is the value of capacitor current? Ic is at its positive peak Ic is zero Ic is at its negative peak
Answer:
The Ic will be zero.
Explanation:
Capacitors have a working principal as follows:
As the current flows through the circuit, they store the electrical energy according to certain attributes they have such as the area of the plates and the material's capacitence in between the plates.An AC voltage increases and decreases between certain maximum and minimum points periodically. So while the AC voltage is on the positive side, the capacitor charges up and when the AC voltage crosses to the negative side, the capacitor takes over and it's current starts increasing as the current coming from the AC source decreases.
So in this case, as the AC voltage crosses zero, the capacitor current was decreasing because the AC voltage was on the positive side and it was charging. The capacitor current will be zero as well and it will start to increase when AC voltage is on the negative.
I hope this answer helps.
