A plane monochromatic radio wave (λ = 0.3 m) travels in vacuum along the positive x-axis, with a time-averaged intensity I = 45

Answer:

1350 m/s

Clarification:

Bullet speed

The bullet travels 450 m

Sound travels a distance of 450 m

Using the equation S= V × t

==> t= S/V

Thus, the time for the bullet t1=450/vb

and the sound's travel time t2=450/vs

Given that there's a 1/2 sec interval from when the shot is fired to the moment the shooter hears the sound

==>  t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec  ---- (1)

Firing sound duration

==> T1 = x/vs

During this time period, the bullet covers 450 m and the sound of impact travels a distance 'x'

The time taken for sound = 450/vb

The time it takes sound to travel distance 'x'= x/vs

therefore let T2= 450/vb + x/vs

However, all this occurs within 3 seconds, i.e., T = 3 sec

because firing takes place before hitting the target, implying the strike sound is heard in time T = T2-T1= 450/vb + x/vs -x/vs

Making T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From equations 1 and 2

applying the same principle, in 3 seconds the observer sees the bullet travel 450 m and perceives the sound

Answer:

The kinetic energy is higher for the first cart.

Explanation:

For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to

resulting in

.

Consequently, the kinetic energy for the 3kg cart ends up as

indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.

Response:

Clarification:

Impulse is equal to change in momentum

mv - mu, where v and u represent the final and initial velocities during the surface impact

For the downward motion of the baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

This becomes the initial velocity upon impact.

For the upward movement

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes the final velocity post-impact

change in momentum is

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

=.7056 N.s.

Impulse exerted by the floor in the upward direction is

=.7056 N.s

Answer:

0.130

Explanation:

The coefficients of static friction recorded for each trial are listed as follows:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

Adding these coefficients together results in: 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

Consequently;

the mean coefficient of static friction =

                                              =

                                              = 0.12983

The mean coefficient of static friction is 0.130