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3 years ago

15

Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x directi

on. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's x-component of velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s to the baseball?

1 answer:

Readme [11.4K]3 years ago

7 0

Answer:

25.9 m/s

Explanation:

mass of ball, m = 0.145 kg

initial velocity, u = + 32 m/s

It bounce back with the velocity but in opposite direction so final velocity,

v = - v

Impulse, I = - 8.4 Ns

Impulse is defined as the change in momentum

I = m v - mu = m (v - u)

- 8.4 = 0.145 x (- v - 3 2)

- 57.9 = - v - 32

v = 57.9 - 32 = 25.9 m/s

Thus, the final speed of the ball is 25.9 m/s

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The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

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Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

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3 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

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