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Greeley [361]
4 years ago
10

Can someone please help me with this please❗️❗️❗️❗️❗️❗️❗️❗️

Physics
1 answer:
ICE Princess25 [194]4 years ago
4 0

Answer:

A the one that was thrown

B the one that was thrown

C the one that was just dropped

Explanation:

pl give brainliest

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If an object accelerates from rest, with a constant acceleration of 4.4 m/s2, what will its velocity be after 28s?
Norma-Jean [14]

Answer:

123.2 m/s after 28s

Explanation:

Vi= 0 m/s

a= 4.4 m/s^2

t=28s

Vf after 28s

To find Vf use your kinematics formula Vf=Vi+at

Vi is Zero so it gets removed and the equation becomes

Vf=at  

Simply Plug and Solve

Vf= 4.4(28)

Vf=123.2 m/s after 28s

6 0
3 years ago
15) What is the frequency of a pendulum that is moving at 30 m/s with a wavelength of .35 m?
____ [38]

A pendulum is not a wave.

-- A pendulum doesn't have a 'wavelength'.

-- There's no way to define how many of its "waves" pass a point
every second.

--  Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.

-- The frequency of a pendulum depends only on the length
of the string from which it hangs.


If you take the given information and try to apply wave motion to it:

             Wave speed = (wavelength) x (frequency)

             Frequency  =  (speed) / (wavelength) ,

you would end up with

             Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz

Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ?     That's pretty absurd. 

This math is not applicable to the pendulum.

6 0
3 years ago
Marcus wants to use two objects to compare the sizes of a proton and an electron. If he uses a basketball for a proton, which ob
Rudik [331]

Answer:

B

Explanation:

4 0
3 years ago
PLEASE HELP ME!!!!!!!!
Ray Of Light [21]

Answer:

I believe C

Explanation:

i never learned this sorry

5 0
3 years ago
Read 2 more answers
A 240 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg tr
lesya [120]

Answer:

The track's angular velocity is W2 = 4.15 in rpm

Explanation:

Momentum angular can be find

I = m*r^2

P = I*W

So to use the conservation

P1 + P2 = 0

I1*W1 + I2*W2 = 0

Solve to w2 to find the angular velocity

0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2

W2 = 0.435 rad/s

W2 = 4.15 rpm

8 0
3 years ago
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