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DiKsa [7]
3 years ago
5

What keeps both the cars pressed down on the road? ​

Physics
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

Gravity

Explanation:

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Which of these would be an example of unbalanced forces?
mylen [45]

A snowball picks up speed as it rolls down the mountain.<em> (D)</em>

Since the description includes acceleration ("picks up speed"), we know that the forces on the snowball must be unbalanced.

3 0
3 years ago
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In order for an object to have kinetic energy it must have a mass and a ?
almond37 [142]

Answer:

Velocity

Explanation:

  • The mechanical energy of the body is defined as the sum of the potential energy and kinetic energy.

                                   E = P.E + K.E

  • The potential energy of a body is due to the height from the surface of the earth.

                                  P.E = mgh

  • The kinetic energy of the is possessed by the body due to the virtue of its motion,

                                  K.E = ½ mv²

  • If there is no velocity associated with the body, there is no K.E in the body.
8 0
3 years ago
Which of the following shows the units of angular motion?
Whitepunk [10]

Answer:

  • The units are <em>(</em><em>rad</em><em>/</em><em>s</em><em>^</em><em>2</em><em>)</em><em> </em>
4 0
3 years ago
1) Displacement is to velocity as ____________ is to acceleration. A) direction B) speed C) time D) velocity
vivado [14]

Answer: C) time

Explanation:

7 0
3 years ago
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
4 0
3 years ago
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