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3 years ago

Baseball pitcher throws a fast ball with a 100 Ns impulse. If he applied the force in 0.15 seconds, What force did he apply?

Physics

2 answers:

VMariaS [17]3 years ago

5 0

Answer:

The velocity of the pitcher is 0.105 m/s in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

Therefore,

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

The pitcher has a velocity 0.105 m/s opposite to the direction of the velocity of the ball.

Explanation:

drek231 [11]3 years ago

5 0

Answer: The force applied on the ball is 666.7 N

Explanation:

Impulse is defined the change in momentum of an object when force is acted on the object for the given interval of time.

Mathematically,

$\Delta p=F\Delta t$

where,

$\Delta p$ = change in momentum or impulse = 100 Ns

F = force applied on the ball = ?

$\Delta t$ = time taken = 0.15 s

Putting values in above equation, we get:

$100Ns=F\times 0.15s\\\\F=\frac{100Ns}{0.15s}=666.7N$

Hence, the force applied on the ball is 666.7 N

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Which of the following statements about electric field lines due to static charges are true? (Select all that apply.)

anastassius [24]

Answer:

a. Electric field lines can never cross each other.

d. Wider spacing between electric field lines indicates a lower magnitude of electric field.

Explanation:

a. The electric field lines cannot be crossed, since this would mean that there would be more than one electric field vector for the same point at the place where the crossing occurs.

d. The space between the field lines is inversely proportional to the intensity of the electric field.

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3 years ago

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t

timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

Explanation: (a) Potential difference is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

$qvB=\frac{mv^{2}}{r}$

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

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