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Andrei [34K]
3 years ago
5

A block with mass 0.5kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2m. whe

n the spring is released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. the force constant k is 100n/m. what is the coefficient of kinetic friction μk between the block and the tabletop?
Physics
1 answer:
dangina [55]3 years ago
4 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al
irina [24]

Answer:

a = 15.1 g

Explanation:

The relation between mass and acceleration is given by :

m\propto \dfrac{1}{a}

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂

So,

\dfrac{m_1}{m_2}=\dfrac{a_2}{a_1}\\\\a_2=\dfrac{a_1m_1}{m_2}\\\\a_2=\dfrac{0.8g\times 1510}{80}\\\\a_2=15.1g

So, the car's acceleration would be 15.1 g.

6 0
2 years ago
27N-(u)(14kg)(9.8m/s^2)=0
atroni [7]
If u is about 0.2 (no unit) then the whole left side is zero.
4 0
3 years ago
A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.
Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

a=\frac{v-u}{t}

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2

And now we can calculate the force exerted on the running back, by using Newton's second law:

F=ma=(100 kg)(-10 m/s^2)=-1000 N

so, the magnitude of the force is 1000 N.

6 0
3 years ago
Read 2 more answers
A boy throws a ball up into the air with a speed of 8.2 m/s. The ball has a mass of 0.3 kg. How much gravitational potential ene
diamong [38]
We can use the law of conservation of energy to solve the problem.

The total mechanical energy of the system at any moment of the motion is:
E=U+K = mgh + \frac{1}{2}mv^2
where U is the potential energy and K the kinetic energy.

At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
E_i = K_i =  \frac{1}{2}mv^2 =  \frac{1}{2}(0.3 kg)(8.2 m/s)^2=10.09 J

When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
E_f = U_f
But the mechanical energy must be conserved, Ef=Ei, so we have
U_f = K_i
and so, the potential energy at the top of the flight is
U_f = K_i = 10.09 J
7 0
3 years ago
Read 2 more answers
Determine the elastic energy U stored in<br> the compressed spring.
hoa [83]

Answer:

Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.

Explanation:

7 0
2 years ago
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