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Wittaler [7]
2 years ago
8

A student catches a can of soda dropped from some unknown height by Mr. Fineman. If

Physics
1 answer:
nikdorinn [45]2 years ago
3 0

Answer:

v² = u² + 2gh

8.9² = 0² + 2(9.8)h

h = 4.0 m

v = at

t = 8.9/9.8

t = 0.91 s

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A force in the negative x-direction is applied for 27 ms to a 0.4 kg mass initially moving at 14 m/s in the x-direction. The for
noname [10]

Answer:

-67 m/s

Explanation:

We are given that

Mass of ball,m=0.4 kg

Initial speed,u=14 m/s

Impulse,I=-32.4 N-s

Time,t=27 ms=27\times 10^{-3} s

We have to find the mass's velocity in the x- direction.

We know that

Impulse=mv-mu

Substitute the values

-32.4=0.4v-0.4(14)

-32.4+0.4(14)=0.4 v

-26.8=0.4v

v=\frac{-26.8}{0.4}=-67m/s

6 0
3 years ago
Isn’t it funny how the people who delete my questions are the ones with my brainliests? P.S Free Points! Your Welcome!
Nuetrik [128]

Answer:

yessss, hilarious.

Explanation:

thank you, btw

6 0
3 years ago
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How does our body control blood pressure?
k0ka [10]

Short-term regulation of blood pressure is controlled by the autonomic nervous system . Changes in blood pressure are detected by baroreceptors. These are located in the arch of the aorta and the carotid sinus. Increased arterial pressure stretches the wall of the blood vessel, triggering the baroreceptors.

6 0
2 years ago
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A similarity between radio waves and microwaves is that both are used for _____.
Cloud [144]

Answer:

(B) Sending Messages

Explanation:

  • A similarity between radio waves and microwaves is that both are used for sending messages and communication

Radio Waves and Microwaves have longer wavelengths and lower frequencies. Longer wavelengths than an x-ray which is in the electromagnetic spectrum and lower frequencies than gamma rays, which are also in the electromagnetic spectrum. They are also both electromagnetic radiation.

5 0
3 years ago
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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
Read 2 more answers
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