Answer:
the answer is b. b velocity. remember v=v
Which is a belief held by sociologists who work from a social-conflict
perspective?
O A. The best approach for a study is from a micro-level orientation.
O B. Personal background has little impact on how individuals react
with one another.
C. Some social patterns are helpful, while others are harmful.
D. Data are irrelevant to the study of sociology.
SUBMIT
Answer:
C. Some social patterns are helpful, while others are harmful.
Explanation:
Hope this was helpful, Have an amazing,spooky Halloween!!
Time it takes stone to fall from the height of 80 m is approximately equal to *
A. 1 s
B. 2 s
C. 4 s
D. 8 s
Answer:
D
Explanation:
Answer:
c.4s
Explanation:
1 Which of the following is an example of a physical change?
A. Water freezing into ice.
B. A piece of wood burning.
C. A toy car rusting.
D. Zinc producing hydrogen gas when mixed with water.
Answer:
it's A trust me ok
Explanation:
i can't explain
A plane travelling at 100 m/s accelerates at 5 m/s² for a distance of 125 m. What is the final velocity of the plane?
Analyzing the question:
We are given:
initial velocity (u) = 100 m/s
final velocity (v) = v m/s
distance (s) = 125 m
acceleration (a) = 5 m/s²
Solving for Final Velocity (v):
from the third equation of motion:
v² - u² = 2as
v² - (100)² = 2(5)(125)
v² - 10000 = 1250
v² = 1250 + 10000
v² = 11250
v = 106.06 m/s
An element with 7 valence electrons will most likely-
be found in group 17 and be highly reactive
be found in group 7 and be highly reactive
be found in group 17 and be inert
be found in group 17 and be somewhat reactive
Answer: be found in group 17 and be highly reactive
Explanation:
Elements are distributed in groups and periods in a periodic table.
Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.
Flourine, chlorine, bromine and iodine are elements which belong to Group 17. All of them contain 7 valence electrons each and need one electron to complete their octet.
The chemical reactivity of elements is governed by the valence electrons present in the element and thus all of them are highly reactive.
A thin, horizontal copper rod is 1.43 m long and has a mass of 236 g. What is the minimum current in the rod that can cause it to float in a horizontal magnetic field of 0.546 T
Answer:
The minimum current in the rod that can cause it to float 2.96 A.
Explanation:
Given;
length of the copper rod, L = 1.43 m
mass of the rod, m = 236 g = 0.236 kg
magnetic field, B = 0.546 T
The magnetic of the rod is given by;
Fm = BIL
The downward force on the rod is given by;
F = mg
For the rod to float, the difference in the two force will be zero;
BIL - mg = 0
BIL = mg
I = mg / BL
I = (0.236 x 9.8) / (0.546 x 1.43)
I = 2.96 A
Therefore, the minimum current in the rod that can cause it to float 2.96 A.
A man walking at 0.5 m/s increases his pace to 0.8 m/s in 3 seconds. What is his acceleration?
Acceleration = (change in speed) / (time for the change)
Change in speed = (New speed) - (old speed)
Change in speed = (0.8 m/s - 0.5 m/s) = 0.3 m/s
Time for the change = 3 seconds
Acceleration = (0.3 m/s) / (3 s)
Acceleration = 0.1 m/s²
(a hair more than 1 percent of 1 G.)
What are arrecture pill muscles
Answer:
Arrector Pili Muscle - This is a tiny muscle that attaches to the base of a hair follicle at one end and to dermal tissue on the other end. In order to generate heat when the body is cold, the arrector pili muscles contract all at once, causing the hair to "stand up straight" on the skin.
Explanation:
Water flows from one reservoir to another a height, 41 m below. A turbine (η=0.77) generates power from this flow. 1 m3/s passes through the turbine. If 12 m head loss occurs between the two reservoirs, determine the actual (i.e. useable) power generated by the turbine (in kW).
Complete Question
A diagram representing this question is shown on the first uploaded image
Answer:
The value is [tex]P = 294594.3 \ W[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 41 \ m[/tex]
The efficiency of the turbine is [tex]\eta = 0.77[/tex]
The flow rate is [tex]\r V = 1 m^3 / s[/tex]
The head loss is g = 2 m
Generally the head gain by the turbine is mathematically represented as
[tex]H = h - d[/tex]
=> [tex]H = 41 - 2[/tex]
=> [tex]H = 39 \ m [/tex]
Generally the actual power generated by the turbine is mathematically represented as
[tex]P = \eta *\gamma * \r V * H[/tex]
Here [tex]\gamma[/tex] is the specific density of water with value
[tex]\gamma = 9810 N/m^3 [/tex]
So
[tex]P =0.77 *9810 * 1 * 39[/tex]
[tex]P = 294594.3 \ W[/tex]
I need help with this answer
Choose the best description of a magnet
O A. Something with magnetic domains
O B. Something that attracts iron
O C. Iron, cobalt, and nickel
O D. Something that becomes magnetic with the application of a
current
Answer:
Explanation:
Something that attracts iron...duuhhh
Which is the simplest?
Organ
Organism
Tissue
Cell
Which is most complex?
Tissue
Organ
Organ system
Whole organism
Answer:
1. Cell2. OrganHopes it helps you....Thank you ☺️
When working with vectors, you will often see right triangles. What are the consistent properties of these triangles?
a. The hypotenuse of the triangle is always the vector itself.
b. The trangles are always equilateral.
c. The sides of the triangles always ine up with the x- and y-axes.
d. The triangles always have the same base and height
Answer:
a) and c)
Explanation:
Any vector can be expressed as a vector sum of its x- and -y components, as follows:v = vₓ* x + vy* y
where vₓ = x- component, x= unit vector in the x direction, vy = y-component, y = unit vector in the y direction.If we add the components graphically, using the head-to-tail method, it will be defined a right triangle, being the vector sum of both components (the vector itself) the hypotenuse of the triangle.As both components are perpendicular, they will be always lined up with the x- and y- axes.1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal. The coefficient of friction between the crate and the floor is 0.12.
a) Calculate the net force and acceleration of the crate.
b) If the crate was moving at 1.0 m/s when it was pulled, what would be its velocity after pulling it for 5.0 s?
Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]
Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]
Apply Newton's second law of motion;
F = ma
[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]
(b) the velocity of the crate after 5.0 s
[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]
Resistors of 10 and 30 ohms are connected in series to a 120-V source. What is the current l owing in the 30 ohm resistor
Answer:
the current through 30 ohms resistor is 30 A.
Explanation:
Given;
resistor, R₁ = 10 ohms
resistor, R₂ = 30 ohms
voltage of the source = 120 V
The effective resistance, Rt = 30 ohms + 10 ohms = 40 ohms
The current through 30 ohms resistor is the same as the current through 10 ohms resistor because they are connected in series.
Appy ohms law to determine the currecnt through 30 ohms resistor;
I = V / Rt
I = 120 / 40
I = 3 A
Therefore, the current through 30 ohms resistor is 30 A.
Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
Learn more about "Circuit":
https://brainly.com/question/15767094?referrer=searchResults
Enterprising students set an enormous slip-n-slide (a plastic sheet covered in water to reduce friction) on flat ground. If the slip-n-slide is 250 meters long.
a. How small does the average acceleration have to be for a student starting at 5 meters/second to slide to the end?
b. If the acceleration is 0.34 m/s^2, what is the minimum initial speed a student would have to run to make it to the end? Is this speed possible?
Answer:
Explanation:
a) Using the equation of motion
v² = u²+2as
v is the final velocity = 0m/s
u is the initial velocity = 5m/s
S is the distance = 250m
a is the acceleration
Substitute given values:
0² = 5²+2a(250)
-25 = 500a
a = -25/500
a = -0.05m/s²
b) If a = 0.34m/s²
We need to get the initial speed u
v² = u²+2as
0 = u²+2(0.34)(250)
-u² = 170
u² = -170
u = √-170
The square root of a negative number will give a complex number, hence the speed is not possible
A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sphere about an axis through its center
Answer:
2I/7
Explanation:
Formula for moment of inertia through the centre of mass of a solid sphere is given as; I_sc = (2/5) mR²
Now, from parallel axis theorem, moment of inertia of solid sphere from tangent is given as;
I = (2/5) m R² + m R²
I = (7/5) mR²
Thus,mR² = 5I/7
Putting 5I/7 for mR² in first equation, we have;
I_sc = (2/5) × 5I/7
I_sc = 2I/7
Listening to the radio, you can hear two stations at once. Describe this wave interaction.
A)
diffraction
B)
interference. C)
reflection
D)
refraction
Answer:
Answer is C or B
Explanation:
A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from the center of the bead? (mass of an electron= 9.1x10^-31 kg)
Answer:
The acceleration is 3.16x10¹⁷ m/s².
Explanation:
First, we need to find the magnitude of the Coulombs force (F):
[tex] |F| = \frac{Kq_{1}q_{2}}{d^{2}} [/tex]
Where:
K is the Coulomb constant = 9x10⁹ Nm²/C²
q₁ is the charge = 20x10⁻⁹ C
q₂ is the electron's charge = -1.6x10⁻¹⁹ C
d is the distance = 1.0 cm = 1.0x10⁻² m
[tex]|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N[/tex]
Now, we can find the acceleration:
[tex] a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2} [/tex]
Therefore, the acceleration is 3.16x10¹⁷ m/s².
I hope it helps you!
A Labrador retriever runs 50 m in 7.2 s to retrieve a toy bird. The dog then runs half way
back in 3.85 s. Determine the average speed and velocity of the dog
Answer:
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
Explanation:
From Physics we must remember the definitions of average speed and average velocity, both measured in meters per second. Velocity is a vectorial quantity, that is, it has both magnitude and direction, whereas speed is an scalar quantity, which is a quantity that is represented solely by its magnitude. We assume that dog moves at constant speed.
For the case of the dog, we get that average speed and average velocity of the animal are, respectively:
Average velocity:
[tex]\vec v_{avg} = \frac{1}{\Delta t}\cdot (\vec r_{B}-\vec r_{A})[/tex] (Eq. 1)
Where:
[tex]\Delta t[/tex] - Travelling time of the dog, measured in seconds.
[tex]\vec r_{A}[/tex] - Initial vector position of the dog, measured in meters.
[tex]\vec r_{B}[/tex] - Final vector position of the dog, measured in meters.
Average speed:
[tex]v_{avg} = \frac{1}{\Delta t} \cdot (s_{A}+s_{B})[/tex] (Eq. 2)
Where [tex]s_{A}[/tex] and [tex]s_{B}[/tex] are the travelled distances of each stage, measured in meters.
If we know that [tex]\Delta t = 11.05\,s[/tex], [tex]\vec r_{A} = 0\,\hat{i}\,\,\,[m][/tex] and [tex]\vec r_{B} = 25\,\hat{i}\,\,\,[m][/tex], [tex]s_{A} = 50\,m[/tex] and [tex]s_{B} = 25\,m[/tex], average velocity and average speed are, respectively:
[tex]\vec v_{avg} = \frac{1}{11.05\,s}\cdot (25\,\hat{i})\,\,\,[m][/tex]
[tex]\vec v_{avg} = 2.262\,\hat{i}\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]v_{avg} = \frac{75\,m}{11.05\,s}[/tex]
[tex]v_{avg} = 6.787\,\frac{m}{s}[/tex]
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
WHAT IS ACCURACY, PRECISION, AND REPRODUCIBILITY? AND WHY ARE THEY SO NECESSARY IN CONDUCTING/DESIGNING EXPERIMENTS? 30 POINTS AND WILL MARK BRAINLIEST
Answer:
Explanation:
Accuracy can be said to mean the degree to which the particular result of a measurement, or calculation, and even possibly specification agrees or is the same with respect to the correct value or an established standard. Succinctly put, it's is how close a value is to the actual value it ought to be.
Precision on the other hand, is a change in a measurement, or calculation, and even as far as specification, much especially as represented by the number of digits that has been established. In other words, it is the proximity of two or more measurements with respect to one another.
Reproducibility occurs when a measurement(for example) is made by another person, or a different instrument is used. Yet, the same values are obtained.
They are very important in design because they account for very important part of an experiment. Neglecting these quantities means exposing an instrument to unknown danger to the factory and even the personnels.
Also, neglect in taking note of accuracy, precision and reproducibility can lead to poor data processing and even human errors.
Now, vote brainliest, will you? :)
How would you measure and compare the size of the two balls ?
Answer:
you would use a measuring tape and compare your answers
Part D
Next, we'll examine magnetic force. Bring the ends of your two magnets together. Explore the three
possible combinations. In two of the combinations, the two ends are the same. In one combination, the
two ends are different. Describe the force you feel in each combination
Answer:
i. The magnetic force of repulsion.
ii. The magnetic force of attraction.
Explanation:
A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.
i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.
ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.
A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s. At this moment, the truck's momentum (in kg · m/s) is how many times as large as the truck's speed (in m/s)?IncorrectThis means that as the truck travels, the truck's momentum (in kg · m/s) is always Incorrect times as large as the truck's speed (in m/s).If the truck is traveling at 16 m/s, what is the truck's momentum?
Answer:
At the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
If the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
Explanation:
From the question,
The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).
Let the truck's momentum be P and the truck's speed be v,
Then we can write that
P∝v
Then,
P = kv
Where k is the proportionality constant
From the question,
At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s,
To determine how many times the truck's speed is as large as the truck's momentum at this moment, we will divide the truck's momentum by the speed, that is
50600 ÷ 23 = 2200
Hence, at the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
Since, dividing the truck's momentum by the truck's speed gives the proportionality constant k (that is, P/v = k), then
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
From
P = kv
Then, k = P/v
At a moment, P = 50600 kg · m/s and v = 23 m/s
∴ k = 50600 kg · m/s ÷ 23 m/s = 2200 kg
k = 2200 kg
To determine the truck's momentum if the truck is traveling at 16 m/s
From
P = kv
k = 2200 kg
v = 16 m/s
∴ P = 2200 kg × 16 m/s
P = 35200 kg · m/s
Hence, if the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
In the figure below the two blocks are connected by a string of negligible mass passing over a frictionless pulley. m1 = 10.0 kg and m2 = 3.80 kg and the angle of the incline is = 37.0°. Assume that the incline is smooth. (Assume the +x direction is down the incline of the plane.)
For what value of m1 (in kg) will the system be in equilibrium?
Answer:
For equilibrium, the mass m1 must be about 2.29 kg
Explanation:
The forces acting on m1 are : the weight (m1 * g) and the tension (T1) of the string.
The forces acting on m2 are:
1) along the ramp: the component of m2's weight (m2 g * sin(37)) and the tension (T2) of the string. Which by the way, must be equal in magnitude to T1 since the string is inextensible.
2) perpendicular to the ramp; the component of m2's weight (m2 g * cos(37)) and the normal from the contact with the ramp. These two compensate each other.
We therefore want the net force on each block to be zero for the system to be in equilibrium. This means:
T1 = m1 g
T2 = T1 = m2 g sin(37)
Then we have that if m2 = 3.8 kg, then:
m1 g = m2 g sin(37)
cancelling "g" on both sides:
m1 = m2 * sin(37) = 3.8 * sin(37) = 2.28689 kg
which may be rounded to about 2.29 kg.
In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Answer in units of J.
Answer:
The energy is [tex]U = 18.98 \ J [/tex]
Explanation:
From the question we are told that
The inductor is [tex]L = 4.74 \ H[/tex]
The resistance of the resistor is [tex]R = 9.33 \ \Omega[/tex]
The voltage of the battery is [tex]V = 26.4 \ V[/tex]
Generally the current flowing in the circuit is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
=> [tex]I = \frac{26.4}{9.33 }[/tex]
=> [tex]I = 2.83 \ A[/tex]
Generally the corresponding energy stored in the circuit is
[tex]U = \frac{1}{2} * L * I^2[/tex]
[tex]U = \frac{1}{2} * 4.74 * 2.83 ^2[/tex]
[tex]U = 18.98 \ J [/tex]
Watching the World Series (only as an example of physics in action), you wonder about the ability of the catcher to throw out a base runner trying to steal second. Suppose a catcher is crouched down behind the plate when he observes the runner breaking for second. After he gets the ball from the pitcher, he throws as hard as necessary to second base without standing up. If the catcher throws the ball at an angle of 30 degrees from the horizontal so that it is caught at second base at about the same height as that catcher threw it, how much time does it take for the ball to travel the 120 feet from the catcher to second base
Answer:
The time is [tex]t_t = 3.7583 \ s [/tex]
Explanation:
From the question we are told that
The angle is [tex]\theta = 30^o[/tex]
The horizontal distance is [tex]d = 120 \ ft[/tex]
Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as
[tex]v = u + at[/tex]
here a = -g the negative sign is because the direction of motion is against gravity
So
[tex]v = v_i + at[/tex]
Here[tex] v_i [/tex] is the vertical component of the initial velocity of the ball which is mathematically
represented as
[tex]v_i = usin(\theta )[/tex]
So
=> [tex]0 = usin(\theta ) -9.8t[/tex]
Generally the total time taken to travel the 120 ft is mathematically represented as
[tex]t_t = \frac{120}{v_h}[/tex]
Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as
[tex]v_h = u cos(\theta )[/tex]
So
[tex]t_t = \frac{120}{ u cos(\theta )}[/tex]
Generally the time taken to reach the maximum height is
[tex]t = \frac{t_t}{2}[/tex]
=> [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]
=> [tex]t = \frac{60}{ u cos(\theta )} [/tex]
So
[tex]0 = usin(\theta ) -9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) = 9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) * u cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(\theta ) cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(30 ) cos(30) =60* 9.8 [/tex]
[tex] u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2} =588.6 [/tex]
[tex] u^2 *\sqrt{3} =2354.4 [/tex]
[tex] u^2 = 1359.31 [/tex]
[tex] u = 36.87 \ ft/s [/tex]
Substituting this value into the equation for total time
[tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]
[tex]t_t = 3.7583 \ s [/tex]
A cannon fires a shell straight upward; 1.6 s after it is launched, the shell is moving upward with a speed of 19 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 5.5 s after the launch.
Answer:
-19.259m/s
Explanation:
Given;
Final velocity = 19m/s
time t = 1.6s
u is the initial velocity
g is the acceleration due to gravity = 9.81m/s²
Using the equation of motion to first get the initial velocity of the shell:
v = u-gt
19 = u - (9.81)(1.6)
19 = u - 15.696
u = 19+15.696
u = 34.696m/s
The initial velocity of the shell is 34.696m/s
Next is to find the speed of the shell 5.5s after the launch
Using the equation of motion:
v = u-gt
v = 34.696-9.81(5.5)
v = 34.696 - 53.955
v = -19.259m/s
The negative value of the velocity shows that the velocity is travelling in the downward direction
. A cathode ray tube (CRT) is a device that produces a focused beam of electrons in a vacuum. The electrons strike a phosphor-coated glass screen at the end of the tube, which produces a bright spot of light. The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both. Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display (LCD) or plasma screen. You still may come across a CRT in your study of science. Consider a CRT with an electron beam average current of 25.00μA25.00μA . How many electrons strike the screen every minute?
Answer:
The value is [tex]n= 9.375 *10^{15} \ electrons [/tex]
Explanation:
From the question we are told that
The average current is [tex]I = 25.0 \mu A = 25.0 *10^{-6} \ A[/tex]
Generally the quantity of charge (electron ) is mathematically represented as
[tex]Q = ne[/tex]
Here e is the charge on a single electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
Generally current is mathematically represented as
[tex]I = \frac{Q}{t}[/tex]
=> [tex]I = \frac{ne}{t}[/tex]
Here t is time which is given as 1 minutes = 60 seconds
and n is the number of electrons
So
[tex]25.0 *10^{-6} = \frac{ n* 1.60 *10^{-19}}{60}[/tex]
=> [tex] 60 * 25.0 *10^{-6} = n* 1.60 *10^{-19} [/tex]
=> [tex]n= \frac{60 * 25.0 *10^{-6} }{ 1.60 *10^{-19} }[/tex]
=> [tex]n= 9.375 *10^{15} \ electrons [/tex]
The number of electrons that strike the screen every minute is; n = 9.375 × 10¹⁵ electrons
What is the number of electrons?
We are given;
Average Current; I = 25 μA = 25 × 10⁻⁶ A
Formula for the current is;
I = Q/t = ne/t
where;
n is number of electrons
e is electron charge = 1.6 * 10⁻¹⁹ C
t is time = 1 minute = 60 seconds
Thus making n the subject gives;
n = It/e
n = (25 × 10⁻⁶ * 60)/(1.6 * 10⁻¹⁹)
n = 9.375 × 10¹⁵ electrons
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