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GREYUIT [131]
3 years ago
9

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

ring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.
Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

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I think it is c density and temperature
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3 years ago
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A cold front moves through and the temperature drops by 20 degrees. in which temperature scale would this 20 degree change repre
nalin [4]
In the Celsius scale each degree is one part of 100 degrees. This is because in this scale the difference between boiling and freezing temperatures of water is 100 ° - 0 ° = 100 °, so one degree Celsius is one part of 100.

In the Farenheit scale, each degree is one part of 180 degrees. This is because in this scale the difference between the boiling and freezind temperatures are 212 ° - 32 ° = 180°, so one degree Farenheti is one part of 180.

That means that 1 °C is a larger amount than 1 °C, so 20°C is a larger amount than 20°F.

Conclusion: 20 degree change represents a larger change in Celsius scale.

4 0
3 years ago
If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni
Lelu [443]

Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

(F₁/F) = (0.5446K/0.309K) = 1.7625

F₁ = 1.7625 F

Hope this Helps!!!

5 0
3 years ago
A 110 kg quarterback is running the ball downfield at 4.5 m/s in the positive direction when he is tackled head-on by a 150 kg l
melisa1 [442]

Answer:

v_f=-0.29\frac{m}{s}

Explanation:

The principle of conservation of momentum, states that if the sum of the forces acting on a system is null, the initial total momentum of the system before a collision equals the final total momentum of the system after the collision. The collision is completely inelastic, which means that the players remain stick to each other after the collision:

p_i=p_f\\m_1v_1+m_2v_2=(m_1+m_2)v_f\\v_f=\frac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\v_f=\frac{(110kg)4.5\frac{m}{s}+150kg(-3.8\frac{m}{s})}{(110kg+150kg)}\\v_f=-0.29\frac{m}{s}

5 0
3 years ago
If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo
Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0
3 years ago
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