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1 year ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Physics

1 answer:

Rama09 [41]1 year ago

7 0

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

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What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

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1 year ago

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You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w

Artemon [7]

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .

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1500 x v₁ = ( 1500 + 2100 ) x 7.365

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1 year ago

What is the temperature when a solid begins to liquefy

MrRa [10]

Answer:

Explanation:

The temperature is at its Melting Point - temperature at which a solid begins to liquefy.

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1 year ago

In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north

Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

$\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}$ (1)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

Now by 1:

$F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}$

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