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3 years ago

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Object A of mass M is released from height H, whereas object B of mass 0.5M is released from height 2H. What is the ratio of the

velocity of object A to that of object B immediately before they hit the ground?
A. 1:2
B. 1.41 :1
C. 1: 1.41
D. 1:4
E. 1:1

1 answer:

skelet666 [1.2K]3 years ago

5 0

My solution. ......................

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Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

$D = 72 \ m$

b

Magnitude

$d = - 20 \ m$

Direction

West

Explanation:

From the question we are told that

The first distance covered westward is $d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2$

=> $D = 18 + 28 + 12 + 14$

=> $D = 72 \ m$

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

$d = -d_w1 -d_w2 + d_e1 + d_e2$

=> $d = -18-28 + 12 + 14$

=> $d = - 20 \ m$

The direction is west

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The Earth’s diameter is about 8,000 miles; our Moon’s diameter is about 2,000 miles; how

Vitek1552 [10]

Three moons can fit inside the volume of the sun.

<h3>What is the moon?</h3>

The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.

Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

Hence, three moons can fit inside the volume of the sun.

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3 years ago

Arocket with an initial velocity of 20 m/s another engine that gives it an acceleration of 4 m/s ^ 2 over 10 secondsHow far did

ExtremeBDS [4]

Answer:

50m

Explanation:

Given parameters:

Initial velocity = 20m/s

Acceleration = 4m/s²

Time = 10s

Unknown:

Distance traveled by the rocket = ?

Solution:

To solve this problem use the expression below;

v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

final velocity = 0

Insert the parameters and solve;

0² = 20² + 2 x 4 x s

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s = 50m

Disregard the negative sign because distance cannot be negative.

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It must be sliding friction, because the fish is already in motion.

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