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pochemuha
2 years ago
13

At the same amplitude or volume, which sound would have the highest frequency?

Physics
1 answer:
weqwewe [10]2 years ago
5 0
Dog whistle ..................
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If you were to come back to our solar system in 6 billion years what might you expect to find
Mariulka [41]
Dead starts bursting new ones being born, maybe more dwarf planets
3 0
3 years ago
A 1200-kg car accelerates it’s speed from 4 m/s to 10 m/s in 3 seconds. Find the car average speed , the car acceleration , the
dsp73

Answer:

Given: mass 1200kg

initial velocity: 4m/s

finial velocity: 10 m/s

time 3 sec

then

speed; initial velocity + final velocity/2

4+10/3

: 4.66m/s2

8 0
2 years ago
Example: A wooden crate with mass 100kg is at rest on a stone floor. You know that the coefficients of kinetic and static fricti
alexgriva [62]

Answer

Any force greater 490N

Explanation

The force required just to make an object slide over a rough horizontal surface is any force greater that the static friction which given by;

F=\mu_s mg.............(1)

Given;

\mu_s=0.5\\m=100kg\\g=9.8m/s^2

Hence;

F = 0.5 x 100 x 9.8

F = 490N.

We will only need the coefficient of kinetic friction if we were asked to find the force required to keep the object moving uniformly. Usually, the force needed to keep an object moving uniformly over a rough surface is lesser that which is needed to start its motion.

In this problem, we were only asked to find the minimum force required to make the object move which we have done.

7 0
3 years ago
Can an object have both kinetic energy and gravitational potential energy? Explain.
marin [14]
Energy flows with kinetic energy
8 0
2 years ago
A 55-kg woman is wearing high heels.
Grace [21]

Answer:

Pressure, P=1.90\times 10^7\ Pa        

Explanation:

It is given that,

Mass of the woman, m = 55 kg

Diameter of the circular cross section, d = 6 mm

Radius, r = 3 mm = 0.003 m

Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{mg}{\pi r^2}

P=\dfrac{55\times 9.8}{\pi (0.003)^2}

P=1.90\times 10^7\ Pa

So, the pressure exerted on the floor is 1.90\times 10^7\ Pa. Hence, this is the required solution.

7 0
3 years ago
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