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Ilya [14]
3 years ago
7

Static Electrictiy __________?​

Physics
1 answer:
rosijanka [135]3 years ago
5 0
Is an imbalance of electric charges within or on the surface of a material.
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A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down
Len [333]

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

4 0
3 years ago
How fast does a 500 Hz wave travel if its wavelength is 0.5 m?
olga_2 [115]

Answer:

250 m/s

Explanation:

4 0
3 years ago
Read 2 more answers
Explain why liquids solidify when they are cooled.
Setler [38]
Because the molecules that move freely begin to compact closer together, with less heat, means less molecular activity. 
8 0
3 years ago
Read 2 more answers
SOMEONE PLEASE HELP!!
Arisa [49]

Hello!

\large\boxed{\text{C. 7,350,000 J}}

Use the equation:

PE = mgh

Where:

m = mass of the object (kg)

g = acceleration due to gravity (≈9.8 m/s)

h = height above ground (m)

Plug the given values into the equation:

PE = 7500 · 9.8 · 100

PE = 7,350,000 Joules.

7 0
3 years ago
A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t
Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

T = \frac{2u Sin\theta }{g}

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}

v_{y}=u Sin 30 - 2u Sin 30

vy = - u Sin 30

The magnitude of final velocity is given by

v = \sqrt{v_{x}^{2}+v_{y}^{2}}

v = \sqrt{\left (uCos 30  \right )^{2}+\left (uSin 30  \right )^{2}}

v = u

Thus, the velocity is same as it just reaches the ground.

6 0
4 years ago
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