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torisob [31]
2 years ago
6

A student pulls horizontally on a block with a spring scale. The block reads 24.5 Newtons before the block starts to move. It re

ads 23.7 Newtons as she pulls the block at a constant velocity. How should she explain this?
Physics
1 answer:
Ahat [919]2 years ago
7 0

Answer:

The coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.

Explanation:

Since the block reads 24.5 N before the block starts to move, this is its weight. Now, when the block starts to move at a constant velocity, it experiences a frictional force which is equal to the force with which the student pulls.

Now, since the velocity is constant so, there is no acceleration and thus, the net force is zero.

Let F = force applied and f = frictional force = μN = μW where μ = coefficient of friction and N = normal force. The normal force also equals the weight of the object W.

Now, since F - f = ma and a = 0 where a = acceleration and m = mass of block,

F - f = m(0) = 0

F - f = 0

F = f

Since the force applied equals the frictional force, we have that

F =  μW and F = 23.7 N and W = 24.5 N

So, 23.7 N = μ(24.5 N)

μ = 23.7 N/24.5 N

μ = 0.97

Since μ = 0.97 < 1, the coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.

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Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

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Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

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January 1st 2000

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A ball rolls up a hill, going from a speed of 8.09 m/s to a speed of 5.87 m/s in 9.17 seconds. What is its acceleration?
PilotLPTM [1.2K]

Answer:

Explanation:

Comment

It's actually a deceleration. You should get a negative answer because the ball is slowing down.

Givens

vi = 8.09 m/s

vf = 5.87 m/s

t = 9.17

Formula

a = (vf  - vi) /t                     Substitute the givens into the formula

Solution

a = (5.87 - 8.09)/9.17

Answer

a = - 2.22 m/s^2

4 0
1 year ago
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