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3 years ago

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Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice w

as so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00×108 m/s) .

1 answer:

abruzzese [7]3 years ago

8 0

Answer:

3.85*10^8m

Explanation:

We know that

Speed v = distance/time t

So distance = v x t

So = 3*10^8 x 2.56s

= distatance = 7.7*10^8m

However this is double the distance from the earth to the moon, so to get distance from earth to the moon we divide by 2

7.7*10^8/2= 3.85*10^8m

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A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

2 years ago

A small airplane takes on 302 l of fuel. If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?

ella [17]

To calculate the mass of the fuel, we use the formula

$m = V \times \rho$

Here, m is the mass of fuel, V is the volume of the fuel and its value is $V =302 \ L = 302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL$ and $\rho$ is the density and its value of 0.821 g/mL.

Substituting these values in above relation, we get

$m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg$

Thus, the mass of the fuel 247 .94 kg.

8 0

3 years ago

A ball hits a wall horizontally at 6m/s and rebounces at 4.4m/s the ball is in contact with wall for 0.04 sec. what is the accel

Svetlanka [38]

Answer:

Acceleration (a) = 40 m/s²

Explanation:

Given:

Initial velocity (u) = 6 m/s

Final velocity (v) = 4.4 m/s

Time taken (t) = 0.04sec

Find:

Acceleration (a) = ?

Computation:

We know that,

⇒ v = u + at

⇒ a = (v - u) / t

⇒ Acceleration (a) = (4.4 - 6) / 0.04

⇒ Acceleration (a) = (-1.6) / 0.04

Acceleration (a) = 40 m/s²

8 0

3 years ago

Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t

LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

$F_{c} = \frac{mv^{2}}{R}$

$f_{s} = \mu_{s}mg$

where

$\mu_{s}$ = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

$F_{net} = f_{s} - F_{c}$

$0 = f_{s} - F_{c}$

$\mu_{s}mg = \frac{mv^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{gR}$

$\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544$

3 0

3 years ago

A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

6 0

3 years ago