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2 years ago

A block of mass m=3.0 kg is pushed down a 50° incline surface with a force of magnitude F=40 N.

Physics

1 answer:

Sauron [17]2 years ago

7 0

Answer:

I really don't know but please if you find the correct answer tell me because I'm eager to know

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An elevator was a full mass of 2400 kg has a power rating on 16.5 horsepower how much time will it take the elevator to lift a f

ehidna [41]

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6 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

5 0

3 years ago

Take a look at a bicycle with gears. Using what you have just learned, answer the following questions:

Harlamova29_29 [7]

Answer:

answer a: a large front gear with a small back gear

answer b: a small front gear with a large back gear

Explanation:

just simple gearing ratios

4 0

2 years ago

2. A 1.30-m long gas column that is open at one end and closed at the other end has a fundamental resonant frequency 80.0 Hz. Wh

Elina [12.6K]

To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.

This is mathematically given as

$f_n (2n+1)(\frac{v}{4L})$

For fundamental frequency n is 0, then,

$f_0 = \frac{v}{4L}$

When,

v = Velocity of sound

L = Length,

Rearranging to find the velocity,

$v = f_0 (4L)$

$v = (80Hz)(4)(1.3m)$

$v = 416m/s$

Therefore the speed of sound in this gas is 416m/s

7 0

3 years ago

Im back and need help

const2013 [10]

Always here to help. Bring it!!!

7 0

3 years ago