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3 years ago

Calculate the coefficient of friction of a 12kg object that has a force of friction of 3N

Physics

1 answer:

postnew [5]3 years ago

3 0

Answer:

0.026

Explanation:

The force of friction acts in the direction perpendicular to the norm force of the surface on which the object rests, induced by gravity. The magnitude of the friction force is

(Friction) = (mass) x (gravitational acceleration g) x (coefficient of friction)

from which the coefficient of friction can be determined:

(coefficient of friction) = (Friction) / ((mass)x(g)) = 3 N / (12 kg * 9.8 m/s^2) = 0.026

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Uranus is unique among the planets in that it

Anvisha [2.4K]

The answer is A)

Here are some informations you should know:

Uranus is the seventh planet from the sun and the third giant gas planet. It rotates like all planets. But Uranus tilts far to one side. Instead of rotating in an upright position, Uranus rolls along like a bowling ball. Uranus also travels very slowly. Uranus has 27 moons. It takes 84 Earth days to complete one orbit.

3 0

4 years ago

Read 2 more answers

What is the distance from the crest to the equilibrium of a wave called?

diamong [38]

The answer would be amplitude. Amplitude is the largest displacement of a wave particle from its equilibrium position.

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3 years ago

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A charge −1.3 × 10−5 C is fixed on the x-axis at 7 m, and a charge 1 × 10−5 C is fixed on the y-axis at 4 m. Calculate the magni

wariber [46]

Answer:

6104 N/C.

Explanation:

Given:

k = 8.99 × 10^9 Nm2/C^2

Qx = 1.3 × 10^-5 C

rx = 7 m

Qy = 1 × 10−5 C

ry = 4 m

E = F/Q

= kQ/r^2

Ex = (8.99 × 10^9 × 1.3 × 10^−5) ÷ 7^2

= 2385.1 N/C.

Ey = (8.99 × 10^9 × 1.0 × 10^−5) ÷ 4^2

= 5618.75 N/C

Eo = sqrt(Ex^2 + Ey^2)

= sqrt(3.157 × 10^7 + 5.69 × 10^6)

= 6104 N/C.

5 0

3 years ago

It’s your birthday, and to celebrate you’re going to make your first bungee jump. You stand on a bridge 100 m above a raging riv

amm1812

Answer:

$X'=50.4\,m$

Explanation:

Given that:

Height of jump, $h=100\,m$

length of elastic cord, $l=30\,m$

spring constant of the cord, $k=40\,N.m^{-1}$

mass of the body that jumps, $m=80\,kg$

Force on the bungee elastic cord:

$F=m.g$

$F=80\times 9.8$

Now this force F will be responsible for the elongation in the elastic cord, so:

$F=k.x$ ............................(1)

where :

k = spring constant

x = extension in the elastic cord

using eq. (1)

$80\times 9.8=40\times x$

$x=19.6\,m$

So the cord stretches 19.6 meters more beyond its original length of 30 meters.

Hence, the remaining distance from the river surface at the bottom is:

$X'=100-(30+19.6)$

$X'=50.4\,m$

3 0

3 years ago

An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8

Aloiza [94]

Answer:

<em>The mass of the apple is 0.172 kg (172 g)</em>

Explanation:

<u>The Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

$m_2v_2-m_2v'=m_1v'-m_1v_1$