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S_A_V [24]
2 years ago
5

What are Radio/Tv waves?

Physics
1 answer:
Lady_Fox [76]2 years ago
6 0
An electromagnetic wave of a frequency between about 104<span> and 10</span>11<span> or 10</span>12<span> Hz, as used for long-distance communication</span>
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Which formula represents final velocity of an object with average acceleration?
zheka24 [161]

Answer:

The equation v – = v 0 + v 2 v – = v 0 + v 2 is reflects the fact that when acceleration is constant, v – is just the simple average of the initial and final velocities.

Explanation:

hope this is it

5 0
2 years ago
Read 2 more answers
The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he
VashaNatasha [74]

Answer:

dV/dt  = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

V = \pi r^{2}h

put r = h/2 so,

V = \pi \frac{h^{3}}{4}

Differentiate both sides with respect to t.

\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3

dV/dt  = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0
2 years ago
What do we get from the area between the speed-time graph of a body and the time axis?​
maxonik [38]

Explanation:

The area between the speed-time graph f a body and time axis measures the distance travelled by the body

7 0
2 years ago
A 1180kg car is moving at a speed of 85.5 km/h. Find the force needed to bring the car to rest in a distance of 300m
yarga [219]

Answer:

1110 N

Explanation:

First, find the acceleration.

Given:

Δx = 300 m

v₀ = 85.5 km/h = 23.75 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (23.75 m/s)² + 2a (300 m)

a = -0.94 m/s²

Find the force:

F = ma

F = (1180 kg) (-0.94 m/s²)

F = -1110 N

The magnitude of the force is 1110 N.

3 0
3 years ago
A large 10.kg medicine ball is caught by a 70.kg student on the track team. If the ball was moving at 4.0 m/s, how fast will the
exis [7]

v2 = ?

m1 = 10kg

m2 = 70kg

v1 = 4m/s

E1 = E2

E1 = 1/2 * m1 * v1^2 = 1/2 * 10kg * 4m/s^2 = 80J

E2 = 1/2 * m2 * v2^2 = 80 J

v2 = √(E2/(2 * m2)) = √(80J/(2 * 70kg)) = about 0.76m/s

7 0
3 years ago
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