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Pachacha [2.7K]
3 years ago
9

A spring stretches from 10cm to 22cm when a force of 4n is applied, if it obeys Hooke’s law,its total length in cm when a force

of 6N is applied is?
Physics
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

Explanation:

Hooke's Law is a linear representation that, in equation form, looks like this:

F = -kΔx where F is force, k is the spring constant, and Δx is the displacement of the spring when a Force is applied to it in the form of a mass hanging on the end of it. We need to use the equation to solve for the spring constant, k. Doing this by filling in the values we were given:

4 = -k(-12) the 12 is negative because the mass hangs below the point of equilibrium. This gives us that

k = 1/3. Now we'll use that in the next equation where we need to solve for displacement:

6 = -1/3(Δx) and

Δx = -18 cm

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In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the
MrRa [10]

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v

where:

m_1 = 2.5 kg is the mass of the ball

u_1 = 7.5 m/s is the initial velocity of the ball

m_2 = 70 kg is the mass of the man

u_2 = 0 is the initial velocity of the man

v is the final velocity of the man and the ball after the collision

Re-arranging the equation and substituting the values, we find the final velocity:

v=\frac{m_1 u_1}{m_1+m_2}=\frac{(2.5)(7.5)}{2.5+70}=0.259 m/s

So, the man and the ball slides on the ice at 0.259 m/s.

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3 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

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