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2 years ago

A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?

Physics

2 answers:

S_A_V [24]2 years ago

7 0

Answer:

Force, F = 57.134 N

Explanation:

It is given that,

Mass of the sled, m = 53 kg

Coefficient of kinetic friction, $\mu_k=0.11$

We need to find the force of friction felt by sled. The force of friction is given by :

$F=\mu mg$

$F=0.11\times 53\ kg\times 9.8\ m/s^2$

F = 57.134 N

So, the force of friction felt by the sled is 57.134 N. Hence, this is the required solution.

Anon25 [30]2 years ago

6 0

Answer:

57N

Explanation:

$F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }$

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At what position or positions on the x-axis is the electric field zero?

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Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

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$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

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Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

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If x = ∞, then the equation is be satisfied.

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