Ask question

Ask question

All categories

2 years ago

14

A sphere with a radius of 15 cm rolls on a level surface with a constant angular speed of 10 rad/s. To what height on a 30° incl

ined plane will the sphere roll before coming to rest? (Neglect frictional losses.) [ 0.16 m]

1 answer:

Scrat [10]2 years ago

8 0

Explanation:

Below is an attachment containing the solution.

You might be interested in

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0

3 years ago

An old-fashioned LP record rotates at 33 1/3 RPM

Ksenya-84 [330]

Answer:

Part a) $\frac{5}{9}\ \frac{rev}{sec}$

Part b) $\frac{9}{5}\ \frac{sec}{rev}$

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

$33\frac{1}{3}\ \frac{rev}{min}$

Convert mixed number to an improper fraction

$33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}$

Remember that

$1\ min=60\ sec$

Convert rev/min to rev/sec

$\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}$

Simplify

$\frac{5}{9}\ \frac{rev}{sec}$

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

$\frac{9}{5}\ \frac{sec}{rev}$

4 0

2 years ago

A roller coaster speeds up with constant acceleration for 2.3 s until it reaches a velocity of 35 m/s. During this time, the rol

Arada [10]

The initial velocity is 0.65 m/s

Explanation:

The motion of the roller coaster is a uniformly accelerated motion, so we can solve the problem by using the following suvat equation:

$s=(\frac{u+v}{2})t$

where

s is the displacement

u is the initial velocity

v is the final velocity

t is the time

For the roller coaster in this problem:

v = 35 m/s

t = 2.3 s

s = 41 m

Solving the equation for u, we find the initial velocity:

$u=\frac{2s}{t}-v=\frac{2(41)}{2.3}-35=0.65 m/s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

5 0

3 years ago

A fisherman moves from one end of a boat to another

Naily [24]

Answer:

That isnt a question so no one will know the answer to what you are talking about. I suggest adding a sceenshot or picture of the question.

5 0

2 years ago

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)

where

m is the mass of the block

$v_{max}=1.25 m/s$ is the maximum speed

We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

$m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg$

b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

$k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m$

c)

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

#LearnwithBrainly

3 0

2 years ago