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3 years ago

Please re-explain the following phrases in terms of momentum

1 answer:

8 0

A. An object in motion will remain in motion unless acted upon by an outside force : The momentum of an object is constant unless an outside force acts on the object.

B. Force is defined as mass times acceleration : the rate of change of the momentum of a particle is proportional to the force F acting on it, hence the force is equal to <span>mass times acceleration.

C. </span>For every action there is an equal and opposite reaction : <span>to every action force there is an equal and opposite reaction force. </span>

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What component of a longitudinal sound wave is analogous to a trough of a transverse wave?

anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.

6 0

2 years ago

If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

3 years ago

A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the

Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

$W_{r} = |F_{r}|*|d|*cos(\theta_{1})$

Where:

$F_{r}$ : is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

$W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J$

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

$W_{g} = |F_{g}|*|d|*cos(\theta_{2})$

The force of gravity is given by the weight of the bike.

$F_{g} = -mgsin(24)$

And the angle between the force of gravity and the direction of motion is 180°.

$W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})$

$W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J$

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.

I hope it helps you!

3 0

2 years ago

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of LaTe

RideAnS [48]

Answer:

5.5x 10^-11 C

Explanation:

Pls see attached file

7 0

3 years ago

A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperat

liberstina [14]

Answer:

45000 K .

Explanation:

Given :

A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure

We need to find the temperature in which 1 litre of the same gas weigh 1 gram

in pressure 75 atm.

We know, by ideal gas equation :

$PV=nRT$

Here , n is no of moles , $n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}$

Putting initial and final values and dividing them :

$\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}$

$\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.$

Hence , this is the required solution.

7 0

3 years ago