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AysviL [449]
3 years ago
10

EMERGENCY: What does the voltmeter read if X is a resistor??

Physics
1 answer:
frosja888 [35]3 years ago
5 0

Here's link to the answer:

tinyurl.com/wtjfavyw

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All steroids are illegal.<br> A. True<br> B. False
TiliK225 [7]
False noth all steroids are legal
7 0
3 years ago
Read 2 more answers
Column A is in the x-axis, and column B is on the y-axis. Which titles should replace A and B
pogonyaev

Column A: x-axis, input, domain

Column B: y-axis, output, range

Those are other ways to describe them

hope i helped:)

8 0
3 years ago
A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find
uranmaximum [27]

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

E_{k1}=\frac{1}{2}*m*v1^{2}

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

6 0
3 years ago
Predict the deformation or elongation of a spring that has a constant of elasticity of 400 N/m when a force of 75 N is applied i
morpeh [17]

Answer:

Explanation:

Give that,

Spring constant (k)=40N/m

Force applied =75N

Since the force is applied to the right, we don't know if it is compressing or stretching the spring

So let assume it compress

Using hooke's law

F=-ke

e=-F/k

Then, e=-75/40

e=-1.875m

The deformation is 1.875m.

Let assume it stretch

Using hooke's law

-F=-ke

e=F/k

Then, e=75/40

e=1.875m

The elongation is 1.875m

3 0
3 years ago
a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec
MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.  

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

\frac{1}{2}  mu_1^2+0=\frac{1}{2}  mv_1^2+\frac{1}{2}  mv_2^2

∴  \frac{1}{2}  m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2

∴ \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2}  mv_1^2+\frac{1}{2} mv_2^2

∴ mv₁v₂ = 0

  1. It is impossible for the mass to be zero.
  2. Because the second ball moves, velocity v2 cannot be zero.
  3. As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0
2 years ago
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