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2 years ago

8

A mover pushes a 30.0 kg crate across a wooden floor at a constant speed of 0.75 m/s. If the coefficient of static friction for

wood-on-wood is 0.20, what is the normal force exerted by the floor on the crate?

1 answer:

liraira [26]2 years ago

3 0

Answer:

294.3 N

Explanation:

In this situation, we are told that the crate is not accelerating in the horizontal plane. But also it is not accelerating in the vertical plane. Meaning that the sum of all vertical forces add up to zero.

Fnet = ma

Weight + Normal force = mass * acceleration

-(30 kg * 9.81 m/s²) + Normal force = 30.0 kg * 0 m/s²

Normal force = 294.3 N

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Explanation:

Given that,

Initial speed of the electron, $u=4\times 10^5\ m/s$

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, $a=6\times 10^{12}\ m/s^2$

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

$v^2=u^2+2as$

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(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

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$t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}$

$t=7.85\times 10^{-8}\ s$

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Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

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The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
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An object is moving in the plane according to these parametric equations:

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d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
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i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)

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A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

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The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$

Where,

$k = 8.99 * 10^9 N m^2 / C^2.$

q = charges of the objects

r= distance/radius

Our values are previously given, so

$q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59$

Replacing,

$F=\frac{kqQ}{r^2}$

$F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}$

$F= 4.8423N$

The force acting on the block are given by,

$F-mgsin\theta = ma$

$a = \frac{F-mgsin\theta}{m}$

$a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}$ $a = 10.31m/s^2$

Therefore the box is accelerated upward.

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A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly abov

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Answer:

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Refractive index of water (n2) = 1.33

So,

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Apparent depth (Da) = 80 (1/1.33)

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