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Akimi4 [234]
3 years ago
8

You and your friend peter are putting new shingles on a roof pitched at 25 degrees. you are sitting on the very top of the roof

when Peter who is at the edge of the roof directly below you, 5.0m away, asks you for a box of nails. Rather than carry the 2.5kg box of nails down to peter, you decide to give the box a push and have it slide down to him. If the coefficient of the kinetic friction between the box and the roof is 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof?
Physics
1 answer:
Anna11 [10]3 years ago
6 0

Answer:2.72 m/s

Explanation:

Given

Inclination of shingles\theta =25^{\circ}

Length of inclined Plane s=5 m

coefficient of kinetic friction \mu _k=0.55

Let u be the initial velocity

F_{net}=mg\sin \theta -f_r

F_{net}=mg\sin \theta -\mu mg\cos \theta

a_{net}=\frac{F_{net}}{m}

a_{net}=g\sin \theta -\mu g\cos \theta

a_{net}=4.141-4.884

a_{net}=-0.744 m/s^2 i.e. deceleration

v^2-u^2=2a_{net}s

0-u^2=2\times (-0.744)\cdot 5

u=\sqrt{7.439}

u=2.72 m/s

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Find the volume of the bottom and top separately and then add them.
Cylinder volume is the area of the bottom times the height
(22/7)(5^2)•175=13750 ft^3

The volume of a sphere is
V=(4/3)(22/7)r^3
where r is the radius. Here that's also 5 since it fits on the cylinder.
Also we only want half the sphere so use
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6 0
3 years ago
Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC
Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

q_1=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 1)

q_2=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N

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