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2 years ago

A 75N force is applied to a 15kg box. If the box

Physics

1 answer:

icang [17]2 years ago

6 0

Answer:

The coefficient of kinetic friction is 0.51.

Explanation:

Given that,

Applied force, F = 75 N

Mass of a box, m = 15 kg

The acceleration of the box is 3 m/s²

We need to find the coefficient of kinetic friction. It is given by the formula as follows :

$F=\mu_k F_N$

Where,

$F_N=mg$ , is the normal force

$\mu_k=\dfrac{F}{mg}\\\\\mu_k=\dfrac{75}{15\times 9.8}\\\\\mu_k=0.51$

So, the coefficient of kinetic friction is 0.51.

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Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript

Pavel [41]

Answer:

Part a)

$F_A = 4.59 N$

Part B)

$F_B = 1.28 N$

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

$F_A + F_B = ma$

as we know that

$a = 0.554 m/s^2$

m = 10.6 kg

now we will have

$F_A + F_B = 10.6(0.554)$

$F_A + F_B = 5.87 N$

Now two forces are in opposite direction then we have

$F_A - F_B = 10.6(0.313)$

$F_A - F_B = 3.32 N$

Part A)

Now we will have from above two equation

$F_A = 4.59 N$

Part B)

Similarly for other force we have

$F_B = 1.28 N$

5 0

2 years ago

A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th

N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0

2 years ago

Neon has 10 protons and 10 electrons. How many additional electrons would it take to fill the outer (second) electron shell?

Yakvenalex [24]

None because Neon is already stable

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3 years ago

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

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Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

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$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

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3 years ago

A 70-kg boy is sitting 3 m from the ground in a tree. What is his gravitational potential energy

steposvetlana [31]

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7 0

3 years ago