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2 years ago

what is the acceleration of a softball if it has a mass of 0.50 kg and hits the catcher’s glove with a force of 25 newtons

Physics

1 answer:

forsale [732]2 years ago

3 0

Answer:

a = 50 [m/s²]

Explanation:

This type of problem can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force =25 [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

$a=F/m\\a=25/0.5\\a= 50 [m/s^{2}]$

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If the mass of a moving object were quartered, it’s inertia would be

AVprozaik [17]

Answer:

Quartered

Explanation:

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2 years ago

the gas in a balloon has P=100000 pa and v=0.0279 m^3. if the pressure increases to 120000 pa at constant temperature, what is t

mash [69]

Answer:

New volume of the baloon is 0.02325m^3

Explanation:

To answer this question we need to know the ideal gas law, which says:

p•V = n•R•T

p is pressure, V is volume, n is amount of substance (in moles), R is constant value and T is temperature.

Since it's stated that n and T are constant, and we know that R is a constant too, that means that p•V = constant value. Basically, that means that p1•V1 (pressure and volume before the pressure increase) equals to p2•V2 (pressure and volume after the pressure increase).

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6 0

3 years ago

Read 2 more answers

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$