PHYSICS COLLEGE

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.

Answers

Answer 1

Answer:

Answer:

42.67N

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

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Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 56 m long that spin at 13 rpm .A. At the tip of a blade, what is the speed?
B. At the tip of a blade, what is the centripetal acceleration?
C. A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .
D. What force is required to keep a 10 mg water droplet moving in this circular arc?
E. What is the ratio of this force to the weight of a droplet?

Answers

A) The speed of the tip of the blade is 76.2 m/s

B) The centripetal acceleration of the tip of the blade is $103.7 m/s^2$

D) The force required to keep the droplet moving in circular motion is 0.39 N

E) The ratio of the force to the weight of the droplet is 4.0

Explanation:

We know that the blade of the turbine is rotating at an angular speed of

$\omega = 13 rpm$

First, we have to convert this angular speed into radians per second. Keeping in mind that

$1 rev = 2 \pi$

$1 min = 60 s$

We get

$\omega = 13 rpm \cdot (2\pi rad/rev)/(60 s/min)=1.36 rad/s$

The linear speed of a point on the blade is given by:

$v=\omega r$

where

$\omega=1.36 rad/s$ is the angular speed

r is the distance of the point from the axis of rotation

For a point at the tip of the blade,

r = 56 m

Therefore, its speed is

$v=(1.36)(56)=76.2 m/s$

The centripetal acceleration of a point in uniform circular motion is given by

$a=(v^2)/(r)$

where

v is the linear speed

r is the distance of the point from the axis of rotation

In this problem, for the tip of the blade we have:

v = 76. 2 m/s is the speed

r = 56 m is the distance from the axis of rotation

Substituting, we find the centripetal acceleration:

$a=((76.2)^2)/(56)=103.7 m/s^2$

The force required to keep the 10 mg water droplet in circular motion on the dog's fur is equal to the centripetal force experienced by the droplet, therefore:

$F=m(v^2)/(r)$

where

m is the mass of the droplet

v is the linear speed

r is the distance from the centre of rotation

The data in this problem are

m = 10 mg = 0.010 kg is the mass of the droplet

v = 2.5 m/s is the linear speed

r = 16 cm = 0.16 m is the radius of the circular path

Substituting,

$F=(0.010)(2.5^2)/(0.16)=0.39 N$

The weight of the droplet is given by

$F_g = mg$

where

m = 10 mg = 0.010 kg is the mass of the droplet

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

Substituting,

$F_g = (0.010)(9.8)=0.098 N$

The force that keeps the droplet in circular motion instead is

F = 0.39 N

Therefore, the ratio between the two forces is

$(F)/(F_g)=(0.39)/(0.098)=4.0$

Learn more about circular motion:

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Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?

Answers

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608* 10^(-9)\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=(mD\lambda)/(a)$

where

a = width of the slit

$a=(mD\lambda)/(y)$

$a=(5* 0.885\ m* 608* 10^(-9)\ m)/(0.0161\ m)$

a = 0.000167 m

$a=1.67* 10^(-4)\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

Clouds form from _____ temperature change, which occurs when an expanding gas cools.

Answers

The answer is Adiabatic

While testing at 30 feet below the surface in Lake Minnetonka, with the sub stopped and in equilibrium, one of the students aboard the sub drops a hammer that goes through the hull of the submarine, and sticks out of the submarine handle first. When this happens, a seal forms immediately around the handle, so that no water enters the sub. What is the new equilibrium position for the sub?

Answers

Answer:

Explanation:

The equilibrium position of the sub is at the surface of the lake

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

Answers

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) $h*(1 - 1/2 g * h/v_0^2)$

b) $h = v_0^2/ g$

c) $h = v_0^2/ g$

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

$y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t$

where:

y = height at time t

y0 = initial height

v0 = initial velocity

a = acceleration

t = time

v = velocity

a) When the balls collide, h1 = h2. Then,

$h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0$

Replacing in the equation of the height of the first ball:

$h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)$

b) that the balls collide at t = h/v0. Then:

$h/ v_0 = -v_0/-g\nh = v_0^2/ g$

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

$h = v_0^2/ g$

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An unladen swallow that weighs 0.03 kg flies straight northeast a distance of 125 km in 4.0 hours. With the x x direction due east and the y y direction due north, what is the average momentum of the bird (in unit vector notation)?

Answers

The average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Total displacement

Since the unladen swallow that weighs 0.03 kg flies straight northeast (that is at a bearing of 45°) a distance of 125 km in 4.0 hours.

Its position vector after 4.0 hours is d = (125kmcos45)i + (125kmsin45)j = (125000 × 1/√2)i + (125000 × 1/√2)j

= (62500√2)i + (62500√2)j.

If the initial position of the swallow is d' = 0i + 0j, then its total displacement after 4 hours is, D = d - d'

= (62500√2)i + (625000√2)j - (0i + 0j)

= (62500√2)i + (62500√2)j m

Average velocity

The unladen swallow's average velocity, v = D/t where

D = total displacement = (62500√2)i + (62500√2)j m and
t = time = 4.0 hours = 4 × 60 min/hr × 60 s/min = 14400 s

So, v = [(62500√2)i + (62500√2)j m]/14400 s = (88388.35)i/14400 + (88388.35)j /1440

= 6.14i + 6.14j m/s

Average momentum

The average momentum of the unladen swallow is p = mv where

m = mass of unladen swallow = 0.03 kg and
v = average velocity = 6.14i + 6.14j m/s

So, p = mv

p = 0.03 kg × (6.14i + 6.14j m/s)

p = (0.1842i + 0.1842j) kgm/s

So, the average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Learn more about average momentum here:

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Answer:

The average momentum of the bird is 0.26 kgm/s

Explanation:

The formula to be used here is that of momentum which is

momentum (in kgm/s) = mass (in kg) × velocity (in m/s)

The velocity of the bird is

velocity (in m/s) = distance (in meter) ÷ time (in seconds)

distance in meters = 125km × 1000 = 125,000 m

time in seconds = 4 hrs × 60 × 60 = 14,400 secs

velocity = 125000/14400

velocity = 8.68 m/s

momentum (p) = 0.03 × 8.68

p = 0.26 kgm/s

The average momentum of the bird is 0.26 kgm/s

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