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STALIN [3.7K]
3 years ago
10

Having trouble with a physics class problem. Help?

Physics
1 answer:
tekilochka [14]3 years ago
8 0
I just did this lol good luck
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You use a pulley system to lift a car engine. You apply a force of 120n and the pulley pulls on the engine with a force of 1050n
Savatey [412]

Given,

Effort force = 120 N

Load force= 1050 N

Mechanical advantage of a pulley is given by the ratio of load force to the effort force.

M.A=\frac{Load force}{Effort force}

=\frac{1050}{120}

=8.75

Therefore, the mechanical advantage of the given pulley is 8.75.

5 0
3 years ago
Read 2 more answers
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
Bas_tet [7]

Answer:

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

5 0
3 years ago
What problems can you imagine arising from a nation of mixed nationalities and faiths? Provide real world examples, if you can.
JulsSmile [24]
Many different problem such as racial slurring and racial profiling
4 0
3 years ago
What is the Weight of Earth?<br><br><br><br><br> Don't SpAm​
Lena [83]

5.972 × 10^24 kg

it is the weight of earth

hope it is helpful to you

6 0
3 years ago
Read 2 more answers
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi
steposvetlana [31]

Answer:

W=K_f-K_i

Explanation:

The work done on a particle by external forces is defined as:

W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,

According to Newton's second law F=ma. Thus:

W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\

Acceleration is defined as the derivative of the speed with respect to time:

W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,

Speed is defined as the derivative of the position with respect to time:

W=m\int\limits^{v_f}_{v_i} v \cdot dv \,

Kinetic energy is defined as K=\frac{mv^2}{2}:

W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i

3 0
3 years ago
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