Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?
Answer:
g₂ = 11 m/s²
Explanation:
The value of free-fall acceleration on the surface of a planet is given by the following formula:
[tex]g = \frac{Gm}{r^2}[/tex]
where,
g = free-fall acceleration
G = Universal Gravitational Constant
m = mass of the planet
r = radius of planet
FOR PLANET 1:
[tex]g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2[/tex] --------------------- equation (1)
FOR PLANET 2:
[tex]g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\[/tex]
using equation (1):
[tex]g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}[/tex]
g₂ = 11 m/s²
A box-shaped metal can has dimensions 8 in. by 4 in. by 10 in. high. All of the air inside the can is removed with a vacuum pump. Assuming normal atmospheric pressure outside the can, find the total force on one of the 8-by-10-in. sides
Answer:
The force on the side is 5252 N.
Explanation:
Area, A = 8 in x 10 in = 80 in^2 = 0.052 m^2
height, h = 10 in
The force on the area is
F = P x A
where, P is the atmospheric pressure and A is the area.
P = 1.01 x 10^5 Pa
Force = 1.01 x10^5 x 0.052 = 5252 N
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?
Answer:
7.9 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:
[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]
Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:
[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]
Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]
Plug everything in:
7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]
(1590kg the initial weight plus the weight of the added passenger)
* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)
Explanation:
Given
Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]
time taken to reach ground is given by
[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]
(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]
From the figure, it can be written
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]
Considering horizontal motion
[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]
(c) The vertical velocity with which it strikes the ground is given by
[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]
Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]
Explanation:
Given
Ball is projected horizontally from a building of height
time taken to reach ground is given by
(b) Line joining the point of projection and the point where it hits the ground makes an angle of
From the figure, it can be written
Considering horizontal motion
(c) The vertical velocity with which it strikes the ground is given by
Thus, the ball strikes with a vertical velocity of
the Period T of oscillation of a Single Pendulum depends on the length l, and acceleration g. Determine the exact form of the dependence.
Answer:
[tex]{ \tt{check \: in \: the \: pic}}[/tex]
An object is moving from north to south what is the direction of the force of friction of the object
Answer:
North
Explanation:
Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.
Answer:
South To North
Explanation:
Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north
The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resistor that makes up part of the parallel circuit?
Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at m/s. Find the speed of the golf ball just after impact.
The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity = [tex]$\frac{4}{0.050}$[/tex]
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.
Every object around you is attracted to you. In fact, every object in the galaxy is attracted to every other object in the galaxy.
a. True
b. False
Answer:
True
Explanation:
With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!
1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (Hint: At what angle do field lines intersect equipotential lines?) Draw sufficient field lines that you can "see" the electric field.
Answer:
The angle between the electric field lines and the equipotential surface is 90 degree.
Explanation:
The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.
The electric field lines are always perpendicular to the equipotential surface.
As
[tex]dV = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
For equipotential surface, dV = 0 so
[tex]0 = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
The dot product of two non zero vectors is zero, if they are perpendicular to each other.
A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Answer:
2.80N/m
Explanation:
Given data
mass m= 56kg
perios T= 11.2s
The expression for the period is given as
T=2π√m/k
Substitute
11.2= 2*3.142*√56/k
square both sides
11.2^2= 2*3.142*56/k
125.44= 351.904/k
k=351.904/125.44
k= 2.80N/m
Hence the spring constant is 2.80N/m
Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.
Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
Explanation:
Let us assume that
[tex]q_{1} = q_{2} = +q[/tex]
[tex]q_{3} = -q[/tex]
As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).
Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r
Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]
where,
k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]
Substitute the values into above formula as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)
Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex] ... (2)
As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.
Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec
A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m along the axis and above the center of the loop
Answer:
[tex]B=2.91\ \mu T[/tex]
Explanation:
Given that,
The current in the loop, I = 2 A
The radius of the loop, r = 0.4 m
We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :
[tex]B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}[/tex]
Put all the values,
[tex]B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T[/tex]
So, the required magnetic field is equal to [tex]2.91\ \mu T[/tex].
The following contribute to accidents when a teen driver has other teens as passengers
Answer:
When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
What are the factors contribute to accidents when a teen driver has other teens as passengers?When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.
Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.
Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
To learn more about factors contribute to accidents refer to:
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A horizontal, mass spring system undergoes simple harmonic motion. which of the following statements is correct reguarding the mass in the system when it is located at its maximum distance from the equilibrium position?
a. The acceleration of the mass is zero.
b. The potential energy of the spring attached to the mass is at a minimum.
c. The total mechanical energy of the mass is zero.
d. The kinetic energy of the mass is at a maximum.
e. The speed of the mass is zero.
Answer:
Option (e)
Explanation:
A body executing SHM moves to and fro or back and forth about its mean position.
When the particle is at mean position, its velocity is maximum and when it is at extreme position its velocity is zero.
So, when it is at maximum distance:
a.
The acceleration is maximum.
b.
The potential energy is maximum.
c.
The total mechanical energy is non zero.
d.
The kinetic energy is zero.
e. The speed is zero. Correct
A water-balloon launcher with mass 2 kg fires a 0.75 kg balloon with a
velocity of 14 m/s to the west. What is the recoil velocity of the launcher?
What is the answer
Answer:
5.25 m/s to the east
Explanation:
Applying,
MV = mv.............. Equation 1
Where M = mass of the launcher, V = recoil velocity of the launcher, m = mass of the balloon, v = velocity of the balloon
make V the subject of the equation
V = mv/M............ Equation 2
From the question,
M = 2 kg, m = 0.75 kg, v = 14 m/s
Substitute these values into equation 2
V = (0.75×14)/2
V = 5.25 m/s to the east
Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2.
(a) When a constant force is applied to object 1, it accelerates through a distance d. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (Select all that apply.)
K1 < K2 p1 = p2 p1 < p2 p1 > p2 K1 > K2 K1 = K2
(b) When a force is applied to object 1, it accelerates for a time interval ?t. The force is removed from object 1 and is applied to object 2. Which statements are true after object 2 has accelerated for the same time interval ?t? (Select all that apply.)
K1 > K2 K1 = K2 p1 = p2 p1 > p2 K1 < K2 p1 < p2
Answer:
Look at explanation
Explanation:
a) Kinetic energy= ΔW. W=Fd, and since in both scenarios the same force and same distance is travelled. K1=K2. I am assuming that the objects are at non zero height so by P=mgh, P1>P2
b. Again I am assuming that the objects are at non zero height so by P=mgh, P1>P2. A heavier mass, a constant force means a smaller acceleration. So a1<a2. We can then use x-x0=v0t+1/2at² and since v0=0, x-x0(d)=1/2at². Solve for t²=2d/a. Since t is the same for both but a1<a2, d1<d2. And since Kinetic Energy=ΔW, W=Fd and F is constant while d1<d2, K1<K2.
According to the question,
Potential energy be "P".Kinetic energy be "K".(a)
Word done towards both the block will be similar.
So,
→ [tex]P1 = P2[/tex]
→ [tex]K1= K2[/tex]
(b)
We know,
→ [tex]a = \frac{F}{M}[/tex]
or,
→ [tex]V = a\times t[/tex]
Now,
→ [tex]K = \frac{1}{2} MV^2[/tex]
[tex]= 0.5\times M\times V^2[/tex]
[tex]=0.5\times M\times (\frac{F^2}{M^2} )\times t^2[/tex]
[tex]= 0.5\times F^2\times \frac{t^2}{M}[/tex]
The force and t will be same. So K of the smaller mass will be greater than the larger mass.
hence,
→ [tex]K1<K2[/tex]
Thus the above responses are correct.
Learn more about friction here:
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Pure water is an example of alan
A. Insulator
B. Metalloid
C. Conductor
D. Nonmetal
Answer: I think its A or C I'm not sure though sorry.
A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?
A) 3.57 m.
B) 4.28 m.
C) 3.14 m.
D) 2.68 m.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×10^3 m/s, given the collision lasts 6.00×10^8s.
Answer:
F = 6666.7 N
Explanation:
Given that,
Mass of a chip, m = 0.1 mg
Initial speed, u = 0
Final speed,[tex]v=4\times 10^{3}\ m/s[/tex]
Time of collision,[tex]t=6\times 10^{-8}\ s[/tex]
We know that,
Force, F = ma
Put all the values,
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N[/tex]
So, the required force is 6666.7 N.
I need help with this physics question.
Answer:
5.04 m
Explanation:
You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.
Answer:
5.0384m
Explanation:
% increase = 100 x (Final - Initial / | initial | )
( |~~| Bars indicate absolute value since you can't have a negative height)
the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic field for that wave at P
Answer:
[tex]6.63\times 10^8\ N/C[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 2.21
We need to find the magnitude of the electric field. Let it is E. So,
[tex]\dfrac{E}{B}=c\\\\E=Bc[/tex]
Put all the values,
[tex]E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C[/tex]
So, the magnitude of the electric field is equal to [tex]6.63\times 10^8\ N/C[/tex].
While an object near the earths surface is in free fall, its
A) velocity increases
B) acceleration increases
Answer:
a
Explanation:
The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-axis
Answer: -5×10-3
Explanation:
E=kq/r
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 10.6 ft/s at point A and 15.6 ft/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.80 s to go from point B to point C. What is the cart's speed at point B
Answer:
a) [tex]a_{avg}=1.25ft/s^2[/tex]
b) [tex]v_b=13.35ft/s[/tex]
Explanation:
From the question we are told that:
Speed at point A [tex]v_A=10.6ft/s[/tex]
Speed at point C [tex]v_C=15.6ft/s[/tex]
Time from Point A to C [tex]T_{ac}=4.00s[/tex]
Time from Point B to C [tex]T_{bc}=1.80s[/tex]
Generally the equation for acceleration From A to B is mathematically given by
[tex]a_{avg}=\frac{v_c-v_a}{\triangle t}[/tex]
[tex]a_{avg}=\frac{15.6-10.6}{4.0 }[/tex]
[tex]a_{avg}=1.25ft/s^2[/tex]
Generally the equation for cart speed at B is mathematically given by
[tex]a_{avg}=\frac{v_c-v_a}{T_{bc}}[/tex]
[tex]v_b=v_c-a_{avg}*T_{bc}[/tex]
[tex]v_b=15.6ft/s-(1.25ft/s^2)(1.80)[/tex]
[tex]v_b=13.35ft/s[/tex]
TIME REMAINING
45:43
What are possible units for impulse? Check all that apply.
kg • m
kg • meters per second
N • s
N • m
StartFraction Newtons times meters per second EndFraction
Answer:
n.m maby
Explanation:
i think or its kg m/s
Answer:
answer (B) & (C)
Explanation:
kg • /N • s
~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)
a.) Determine the linear regression
b.) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
A proton is held at rest in a uniform electric field. When it is released, the proton will gain:_________
a) electrical potential energy.
b) kinetic energy.
c) both kinetic energy and electric potential energy.
d) either kinetic energy or electric potential energy.
the unit for
ΔL/L
is
Answer:
the unit for ΔL/L is "unitless".
Explanation:
Given;
ΔL/L
by physics convection, the above parameters can be defined as;
delta L (ΔL) is change in length, with SI unit as meters (m),
L is the original length of the material, with SI unit as meters (m)
The ratio of the change in length to the original length has no unit since both units cancel out during the division.
[tex]\frac{\Delta L }{L } = \frac{(m)}{(m)} = \ unitless[/tex]
This ratio (ΔL/L), is also called tensile strain and it has no unit.
Therefore, the unit for ΔL/L is "unitless".