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1 year ago

The component of the normal force on the object parallel to the plane balances the friction force on the object.True/False

1 answer:

3 0

True. The component of the normal force on the object parallel to the plane balances the friction force on the object at rest.

<h3>What is component of normal force parallel to the plane? </h3>

An object placed on an inclined plane is subject to three forces, namely;

Normal force acting downwards
Component of the normal force acting parallel to the plane
Frictional force, parallel to the plane

<h3>Net force acting on the object at rest</h3>

∑F = 0

mg sinθ - μFₙ cosθ = 0

mg sinθ = μFₙ cosθ

where;

mg sinθ is component of the normal force acting parallel to the plane
μFₙ cosθ is frictional force, parallel to the plane

Thus, the component of the normal force on the object parallel to the plane balances the friction force on the object at rest.

Learn more about normal force here: brainly.com/question/14486416

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How many kids under the age of 21 are drinking and smoking in the us and California combined

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2 years ago

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A railroad tank car contains milk and rolls at a constant speed along a level track. The milk begins to leak out the bottom. The

kifflom [539]

Answer:

option C

Explanation:

The correct answer is option C

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3 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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3 years ago

Charlie Brown kicks a football at 24.5 m/s at 35.0. What is the maximum height of the ball?

lozanna [386]

Answer:

d = 10.076 m

Explanation:

We need to obtain the velocity of the ball in the y direction

Vy = 24.5m/s * sin(35) = 14.053 m/s

To obtain the distance, we use the formula

vf^2 = v0^2 -2*g*d

but vf = 0

d = -vo^2/2g

d = (14.053)^2/2*(9.8) = 10.076 m

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4 years ago