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3 years ago

A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?

Physics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

<em>v=40 m/s south</em>

Explanation:

<u>Momentum </u>

It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

$p=m.v$

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

$\displaystyle v=\frac{p}{m}$

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

$\displaystyle v=\frac{6}{0.15}=40\ m/s$

The velocity is directed to the south

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

sdas [7]

Answer:

The vertical distance is $d = \frac{2}{k} *[mg + f]$

Explanation:

From the question we are told that

The mass of the cylinder is m

The kinetic frictional force is f

Generally from the work energy theorem

$E = P + W_f$

Here E the the energy of the spring which is increasing and this is mathematically represented as

$E = \frac{1}{2} * k * d^2$

Here k is the spring constant

P is the potential energy of the cylinder which is mathematically represented as

$P = mgd$

And

$W_f$ is the workdone by friction which is mathematically represented as

$W_f = f * d$

$\frac{1}{2} * k * d^2 = mgd + f * d$

=> $\frac{1}{2} * k * d^2 = d[mg + f ]$

=> $\frac{1}{2} * k * d = [mg + f ]$

=> $d = \frac{2}{k} *[mg + f]$

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3 years ago

Which of these describes the part of the water cycle happening at D

tino4ka555 [31]

The answer is Runoff.

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3 years ago

Read 2 more answers

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.

marishachu [46]

Answer:

a) the magnitude of the force is

F= Q( $\frac{kqs}{r^3}$ ) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b) the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = $\frac{kq}{r^{2} }$

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = $\frac{kp}{r^{3} }$ , where p = q × s

E(r) = $\frac{kqs}{r^{3} }$ where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q( $\frac{kqs}{r^3}$ )

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

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What important factor does the ozone layer serve in earth's atmosphere?

Dmitry_Shevchenko [17]

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