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4 years ago

Which one of the following situations is not possible?

Physics

1 answer:

Tems11 [23]4 years ago

3 0

Answer:

(d)

Explanation: because an object cannot go in the same speed and accelerate at the same time

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Suppose a moving car has 2000 J of kinetic energy. If the carʹs speed doubles, how much kinetic energy would it then have?

Bond [772]

Answer:

option A

Explanation:

given,

Kinetic energy of the car = 2000 J

speed of the car is doubled

we know,

$KE_1 = \dfrac{1}{2}mv^2$

$2000= \dfrac{1}{2}mv^2$ ........(1)

now, speed of the car is doubled

v' = 2 v

$KE_2 = \dfrac{1}{2}mv'^2$

$KE_2 = \dfrac{1}{2}m(2v)^2$

from equation (1)

$KE_2 = 4\times \dfrac{1}{2}m(v)^2$

$KE_2 = 4\times 2000$

$KE_2 = 8000\ J$

Hence, the Kinetic energy would be equal to 8000 J.

The correct answer is option A.

8 0

3 years ago

A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat

agasfer [191]

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{2.8\times 9.8}{0.018}$

k = 1524.44 N/m

Since, $\omega=\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2 f^2}$

$m=\dfrac{1524.44}{4\pi^2 \times 3^2}$

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.

4 0

3 years ago

Suppose two waves collide and the temporary combined wave that results is smaller than the original waves. What term best descri

babymother [125]

The correct answer is B

7 0

3 years ago

Read 2 more answers

A boy kicks a football from ground level. The ball takes 3 seconds to reach its maximum height. What is the angle of the initial

ruslelena [56]

<h2>The angle of the initial velocity with respect to the horizontal is 85.14°</h2>

Explanation:

Given that the ball takes 3 seconds to reach its maximum height.

Consider the vertical motion of ball till maximum height.

We have equation of motion v = u + at

Acceleration, a = -9.81 m/s²

Final velocity, v = 0 m/s

Time, t = 3 s

Substituting

v = u + at

0 = u + -9.81 x 3

u = 29.43 m/s

Initial vertical velocity is 29.43 m/s.

Now consider horizontal motion of ball.

Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s

Displacement = 15 m

We have equation of motion s = ut + 0.5 at²

Displacement, s = 15 m

Acceleration, a = 0 m/s²

Time, t = 6 s

Substituting

s = ut + 0.5 at²

15 = u x 6 + 0.5 x 0 x 6²

u = 2.5 m/s

Initial horizontal velocity is 2.5 m/s

Let r be the initial velocity and θ be the angle with horizontal

Initial vertical velocity = rsinθ = 29.43 m/s

Initial horizontal velocity = rcosθ = 2.5 m/s

Dividing

$\frac{rsin\theta }{rcos\theta }=\frac{29.43}{2.5}\\\\tan\theta=11.77\\\\\theta=85.14^0$

The angle of the initial velocity with respect to the horizontal is 85.14°

4 0

3 years ago

Select all that apply.

bonufazy [111]

Mechanical energy of the ball will remain conserved

so here we have

$E = mgh + \frac{1}{2}mv^2$

$E = 0.200(9.8)(2) + 0$

$E = 3.92 J$

so mechanical energy will remain same at all positions

Now when ball comes to position of 1 m height then potential energy is given as

$U = mgh$

$U = (0.200kg)(9.8 m/s^2)(1 m)$

$U = 1.96 J$

Now since total mechanical energy is conserved so we will say

$KE + U = E$

$KE + 1.96 = 3.92$

$KE = 1.96 J$

so we have

its mechanical energy = 3.92 J

its potential energy = 1.96 J

its kinetic energy = 1.96 J

4 0

3 years ago