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3 years ago

An object has a coefficient of kinetic friction of 0.2 and a normal force of 30N. Find the force of kinetic friction.

Physics

1 answer:

shutvik [7]3 years ago

6 0

Answer:

Force of kinetic friction, Fk = 6N

Explanation:

Given the following data;

Coefficient of kinetic friction, μ = 0.2

Normal force, Fn=30N

To find the Force of kinetic friction;

Mathematically, the force of kinetic friction is given by the formula;

Fk = μFn

Where;

Fk represents the force of kinetic friction.

μ represents the coefficient of friction.

Fn represents the normal force.

Substituting into the equation, we have;

Fk = 0.2*30

Fk = 6N

Therefore, the force of kinetic friction is 6N.

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First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
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Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
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$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
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Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

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