A wave has a wavelength of 13 mm and a frequency of 18 hertz. What is its speed?

If a satellite weighs 321 lb. on the earth's surface (R = 4,000 miles), how much does it weigh 12,000 miles above the surface? (

Sever21 [200]

The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.

On the surface, their centers are separated by 1 Earth radius.

12,000 miles above the surface, they're separated by 4 Earth radiii.

(4/1) = 4

So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.

(321 lb) / (16) = (20 and a hair) lb

The correct choice from the given list is " <em>>20 lb "</em> .

3 0

3 years ago

1. A student demonstrates electromagnetic induction using a

Katen [24]

answer is :D it would be a great answer

4 0

3 years ago

Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10

USPshnik [31]

Answer:

D.

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

$y = \frac{1}{2} * g * t^{2}$

where y = height = 40 ft.

As all the parameters are given in SI units, it is advisable to convert this value to m, as follows:

$y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m$

Now, we can solve for t, as follows:

$t = \sqrt{\frac{2*y}{g}} = \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s$

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

$y = v0*t +\frac{1}{2} * g * t^{2}$

This leaves us a quadratic equation on t, as follows:

$t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s$

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

5 0

4 years ago

Do frictional forces act in the same direction or in the opposite direction to the applied force?

Evgesh-ka [11]

Frictional forces act in the direction opposite to the MOTION. That direction could be the same OR opposite to applied force.

-- If you push a loaded heavy wagon from behind, trying to get it going faster, friction is acting against you, opposite to your force.

-- If you push a loaded rolling heavy wagon from in front, trying to make it slow down, friction is acting with you, in the same direction as your force.

-- Opposite to the motion both times.

3 0

3 years ago

A stone is tossed horizontally from the highest point of a 95 m building and lands 105 m from the base of the building. Ignore a

aalyn [17]

Answer:

A) t = 4.40 s , B) v = 23.86 m / s , c) v_y = - 43.12 m / s , D) v = 49.28 m/s

Explanation:

This is a projectile throwing exercise,

A) To know the time of the stone in the air, let's find the time it takes to reach the floor

y = y₀ + $v_{oy}$ t - ½ g t²

as the stone is thrown horizontally v_{oy} = 0

y = y₀ - ½ g t²

0 = y₀ - ½ g t²

t = √ (2 y₀ / g)

t = √ (2 95 / 9.8)

t = 4.40 s

B) what is the horizontal velocity of the body

v = x / t

v = 105 / 4.40

v = 23.86 m / s

C) The vertical speed when it touches the ground

v_y = $v_{oy}$ - g t

v_y = 0 - 9.8 4.40

v_y = - 43.12 m / s

the negative sign indicates that the speed is down

D) total velocity just hitting the ground

v = vₓ i ^ + v_y j ^

v = 23.86 i ^ - 43.12 j ^

Let's use Pythagoras' theorem to find the modulus

v = √ (vₓ² + v_y²)

v = √ (23.86² + 43.12²)

v = 49.28 m / s

we use trigonometry for the angle

tan θ = v_y / vₓ

θ = tan⁻¹ (-43.12 / 23.86)

θ = -61

7 0

3 years ago