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mixas84 [53]
3 years ago
11

Two masses of size m and 4m are connected by a massless thread and are strung over a frictionless pulley of radius R and moment-

of-inertia I. The two masses are held motionless at the same height and then released from rest. After the mass 4m has fallen a distance L and the mass m has risen the same distance L, what is the total kinetic energy of the entire system
Physics
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

Total kinetic energy of entire system is 3 mgl

Explanation:

Given two masses: m and 4m.

Since the pulley is frictionless and the thread is massless, the energy here is linked to the two masses.

Total kinetic energy of entire system = decrease in gravitational potential energy of the system.

Therefore, we have :

ΔKE = Δp

ΔKE = 4mgl - mgl

= 3 mgl

Total kinetic energy of entire system is 3 mgl

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What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a
Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660  N

==>T× 0.7660 = 29.4

==> T= 38.38 N

8 0
3 years ago
The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
3 years ago
1. A 1,000-kg car has 50,000 joules of kinetic energy. What is its speed?
bulgar [2K]
Use KE= 1/2mv^2
So... 
50,000=(.5)(1,000)v^2
50,000=500 x v^2
Divide 500 on both sides 
100 = v^2 
Square root both sides to get rid of v^2 
Therefore v = 10 m/s
4 0
3 years ago
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A sample contains 110 g of a radioactive isotope. how much radioactive isotope will remain in the sample after 1 half-life?
Lerok [7]

Answer:

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Thus, if we start with A grams of a given radioactive isotope, after a 1 half-life, we will have A/2 grams of the radioactive isotope.

In this case, we know that the sample has 110g of a radioactive isotope.

Then, after 1 half-life, we should have half of 110g, which is:

110g/2 = 55g

Then we should have 55 g of a radioactive isotope.

The answer that is closer to this result is option d (50 g), so that is the correct one.

5 0
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If we see a galaxy that is 100 light years from us and it appears red, that means that the galaxy is actually red in color
Sliva [168]
Hello!

I believe the answer is false. Please let me know if you have any other questions. Have a bless day!

-Rose


5 0
2 years ago
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