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alexandr402 [8]

3 years ago

13

If you push an object horizontally on a surface, (a) how many forces will act on the object? (b) Name the forces, and (c) will g

ravity be acting on the object?

1 answer:

Whitepunk [10]3 years ago

5 0

Answer:

a. 5

b. gravity, weight, pull, velocity, and energy.

c. gravity will be acting on the object by pulling it to wherever the object headed.

Explanation:

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A scuba tank, when fully submerged, displaces 16.7 L of seawater. The tank itself has a mass of 13.4 kg and, when "full," contai

meriva

Answer:

$m = 17.87 kg$

Explanation:

As we know that tank itself has mass given as 13.4 kg

so it is given as

$m_1 = 13.4 kg$

also the tank contains air in it and the mass of air inside the tank is also given as 4.47 kg

so it is

$m_2 = 4.47 kg$

so total mass of the tank is the mass of the empty tank and mass of air in the tank

$m = m_1 + m_2$

$m = 13.4 + 4.47$

$m = 17.87 kg$

3 0

3 years ago

Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the ed

I am Lyosha [343]

Answer:

c. remains the same, but the RPMs decrease.

Explanation:

Because there aren't external torques on the system composed by the person and the turntable it follows that total angular momentum (I) is conserved, that means the total angular momentum is a constant:

$\overrightarrow{L}=constant$

The total angular momentum is the sum of the individual angular momenta, in our case we should sum the angular momentum of the turntable and the angular momentum of a point mass respect the center of the turntable (the person)

$\overrightarrow{L_{turnatble}}+\overrightarrow{L_{person}}=constant$ (1)

The angular momentum of the turntable is:

$\overrightarrow{L_{turnatble}}=I\overrightarrow{\omega}$ (2)

with I the moment of inertia and ω the angular velocity.

The angular momentum of the person respects the center of the turntable is:

$\overrightarrow{L_{person}}=\overrightarrow{r}\times m\overrightarrow{v}$ (3)

with r the position of the person respects the center of the turntable, m the mass of the person and v the linear velocity

Using the fact $v=\omega r$ :

$\overrightarrow{L_{person}}=\overrightarrow{r}\times rm\overrightarrow{\omega}$ (3)

By (3) and (2) on (1) and working only the magnitudes (it's all that we need for this problem):

$I\omega+r^{2}m\omega=constant$

$\omega(I+r^{2}m)=constant$

Because the equality should be maintained, if we increase the distance between the person and the center of the turntable (r), the angular velocity should decrease to maintain the same constant value because I and m are constants, so the RPM's (unit of angular velocity) are going to decrease.

4 0

4 years ago

Water in central Texas comes from the Edwards

kodGreya [7K]

The answer is d my homie

8 0

3 years ago

Read 2 more answers

A student says that the two terms speed and frequency of the wave refer to the same thing. What is your response?

rjkz [21]

Answer:

He is wrong!

Explanation:

Frequency refers to how many wave lengths pass in a second and speed is how fast the wave is traveling for example the speed of light goes really fast but has a mid-level frequency.

Hope this helps! ;-)

3 0

3 years ago

A ball of mass m = 4.6 kg, at one end of a string of length L= 6.6 m, rotates in a vertical circle just fast enough to prevent t

VashaNatasha [74]

Answer:

v₂ = 17.98 m/s

Explanation:

given,

mass of ball = m = 4.6 Kg

length of string = L = 6.6 m

force acting toward the center is equal to the force exerted by centripetal acceleration

$m g = \dfrac{mv_1^2}{r}$

$v_1 = \sqrt{gr}$

now, calculating the speed of ball at the bottom of the circlr

work done by the gravity = change in kinetic energy

$- m g (2R) = \dfrac{1}{2}m(v_1^2-v_2^2)$

$-4 gR =v_1^2-v_2^2$

$-4 gR =g R-v_2^2$

$v_2^2 = 5 g R$

$v_2=\sqrt{5\times 9.8 \times 6.6}$

v₂ = 17.98 m/s

8 0

3 years ago