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olga2289 [7]
3 years ago
10

In longitudinal waves the places where the coils are bunched together are called *

Physics
1 answer:
Firdavs [7]3 years ago
3 0
In longitudinal waves the places where the coils are bunched together are called *
Compressions
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Which statement about moons is true?
kotegsom [21]

Answer:

A

Explanation:

All of the other answers don't make much sense

8 0
3 years ago
Newton's first law of motion states that an object will keep a constant speed and direction unless acted upon by an unbalanced f
trasher [3.6K]

Answer:

The unbalanced force that caused the ball to stop was friction

Explanation:

As Newton's second law states, the acceleration of an object is proportional to the net force applied on the object:

F=ma

therefore, in order to move at constant speed, an object should have a net force of zero (balanced forces) acting on it.

In this case, the ball slows down and eventually comes to a stop: it means that the ball is decelerating, so there are unbalanced forces (net force different from zero) acting on it. The unbalanced force acting on the ball is the friction: friction is a force against the motion of the object, which is due to the contact between the surface of the ball and the surface of the street, and this force is responsible for slowing down the ball.

3 0
3 years ago
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Which property increases as an electromagnetic wave's energy decreases?
aleksklad [387]
The answer is B frequency. When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases
4 0
3 years ago
A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
Artist 52 [7]

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f =  μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

6 0
3 years ago
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