Ask question

Ask question

All categories

2 years ago

6

A well-greased, essentially frictionless, metal bowl has the shape of a hemisphere of radius 0.150 m. You place a pat of butter

of mass 5.00×10−3kg at the rim of the bowl and let it slide to the bottom of the bowl. What is the speed of the pat of butter when it reaches the bottom of the bowl?

1 answer:

shepuryov [24]2 years ago

8 0

Answer:

$v = 1.71 m/s$

Explanation:

As we know that there is no friction in the bowl so total mechanical energy is conserved here

So we can say that

initial potential energy of the butter = final kinetic energy at the bottom

so we can say

$mgH = \frac{1}{2}mv^2$

here we can solve it for speed v

$v^2 = 2gH$

now plug in all values in it

$v^2 = (2 \times 9.8\times 0.150)$

$v = 1.71 m/s$

You might be interested in

In the system shown above, the pulley is a uniform disk with a mass of .75 kg and a radius of 6.5 cm. The coefficient of frictio

lord [1]

Answer:

i am answering the same question 3rd time

please find the answer in the images attached.

5 0

1 year ago

The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp

olganol [36]

Answer:

Ft

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

Acceleration, $a=\frac{F}{m}$

Final velocity , $v=a+ut$

Substitute the values

$v=0+\frac{F}{m}t=\frac{F}{m}t$

We know that

Momentum, p=mv

Using the formula

$p=m\times \frac{F}{m}t=Ft$

6 0

2 years ago

Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but

valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

$K=\frac{1}{2}I \omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

$K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2$

For disk 2:

$K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2$

so the total energy is

$K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2$

So, none of the options is correct.

5 0

2 years ago

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

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

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

2 years ago