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What is the main difference between the gas water vapor and the liquid water?

Vinil7 [7]

Water vapour has more kinetic energy.

Hope this helps :)

6 0

3 years ago

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M

Katena32 [7]

The net force needed to accelerate a 200 kg piano is 50N

According to Newton's law of motion, the formula for the net force is calculated as:

∑F = ma

Given the following parameters

Mass = 200kg

acceleration = 0.25m/s²

Substitute the given parameters into the formula

∑F = 200 * 0.25

∑F = 50N

Hence the net force needed to accelerate a 200 kg piano is 50N

Learn more here: brainly.com/question/19030143

7 0

3 years ago

The triceps muscle in the back of the upper arm extends the forearm.

iren [92.7K]

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

$\tau = F \times r$

Where,

F = Force

r = Distance

Replacing,

$\tau = 2*10^3*0.03$

$\tau = 60N\cdot m$

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

$\tau = I \alpha$

I = Moment of inertia

$\alpha$ = Angular acceleration

Replacing with our values we have that

$I = \frac{\tau}{\alpha}$

$I = \frac{60}{120}$

$I = 0.5kg\cdot m^2$

Therefore the value of moment of inertia is $0.5kg\cdot m^2$

8 0

3 years ago

If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d

Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

$v_{i}=-44.5m/s\\v_{f}=55.5m/s$

Now we need to find the initial momentum

So

$P_{1}=m*v_{i}$

Substitute the given values

$P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s$

Now for final momentum

$P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s$

So the change in momentum is given as:

ΔP=P₂-P₁

$=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s$

ΔP=20 kg.m/s

3 0

4 years ago

A car moves at a constant velocity of 30 m/s and has 3.6 × 105 J of kinetic energy. The driver applies the brakes and the car st

LekaFEV [45]

The force needed to the stop the car is -3.79 N.

Explanation:

The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.

So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be

$K.E = \frac{1}{2} * m*v^{2} \\\\m = \frac{2*KE}{v^{2} } = \frac{2*3.6*10^{5} }{30*30}=800 kg$

Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,

Work done = Final kinetic energy - Initial kinetic energy

Work done = Force × Displacement

So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.

$Force * 95 = 0-3.6*10^{5}\\ \\Force =\frac{-3.6*10^{5} }{95}=-3.79 N$

Thus, the force needed to the stop the car is -3.79 N.

5 0

3 years ago