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3 years ago

Sound waves with frequency higher that 20,000 Hz are referred to as

Physics

1 answer:

Solnce55 [7]3 years ago

4 0

<span>Sound waves with frequency higher that 20,000 Hz are referred to as "Ultrasonic"

Hope this helps!</span>

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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0

3 years ago

How much gravitational potential energy does an object have if it is located 20 m above the point that we define as O height and

raketka [301]

Answer:

Explanation:

PE = mgh where m is the mass in kg, g is the pull of gravity which is 9.8, and h is the height of the object above the point to which it could possibly fall, measured in meters. Plugging in:

PE = 10(9.8)(20) so

PE = 1960 J

This should be rounded to 1 sig fig according to the rules of sig fig and your numbers here, but I imagine you're not following them all that much. It should be 2000 J

8 0

3 years ago

A cyclical motion occurs because of density differences in the mantle. Heated, less dense lower regions of the fluid mantle rise

Marta_Voda [28]

The answer for this is B

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3 years ago

While at a party, you pull up a sound intensity level app on your phone (everyone does stuff like that, right?), and it reads 83

allochka39001 [22]

To solve this problem it is necessary to apply the concepts related to Sound Intensity. The unit most used in the logarithmic scale is the decibel and mathematically this is expressed as

$\beta_{dB} = 10log_{10}\frac{I}{I_0}$