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exis [7]
3 years ago
8

Do you ever have a sensation of loneliness?

Physics
1 answer:
Reptile [31]3 years ago
8 0

Answer:

nope

Explanation:

sige bigyan kitang Happy pills

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A young man heaves a 6.5 kg rock at a velocity of 6.9 m/s. What is the kinetic energy of the rock?
earnstyle [38]

Answer:

Hope it will help you a lot.

7 0
2 years ago
Children are told to avoid standing too close to a rapidly moving train because they might get sucked under it. Is this possible
storchak [24]

Answer:

no its not like the undertow in the ocean

Explanation:

4 0
3 years ago
1)Calculate the mass of a girl that has a weight of 113 Newtons.
Blababa [14]

Answer:

Explanation:We should know that weight = mass * gravity.

That is weight equals mass times gravity.

Gravity is a force of attraction between any two bodies in the universe. It is directly proportional to product of their masses and inversely proportional to the square of the distance between them.

Gravity is generally measured in terms of acceleration due to gravity, denoted as g. For Earth it is, 9.8 m/s². And for moon, it is about 1.62 m/s².

On Earth, your weight is 70 kg = W

W = mass x 9.8

70 = mass x 9.8

Your mass is 70/ 9.8

i.e approximately 7.14

Weight at the Moon, W' = 7.14 x 1.62

Hence, your weight on the surface of the moon is just 11.56 kg.

Congratulations, you've lost about 58.14 kilograms without any hard exercise. And you're as light as a Sweedish Vallhund! Cheers!

3 0
3 years ago
98 POINTS, 5 simple questions!! HELP
lozanna [386]

25,000 Feet = 7620m

PE = mgh where m is mass, g is gravity accel: 9.8 n h is height

= 90 x 9.8 x 7620

= 6720840J

= 6.72MJ

F = ma where m is mass, a is accel = gravity = 9.8

= 90 x 9.8

= 882N

Accel = gravity = 9.8m/s^2

KE = 1/2mv^2 where m is mass n v is vel

if no wind resistance, PE leaving airplane = KE at net

6720840 = 1/2 x 90 x v^2

v^2 = 149352

v = 386.5m/s


3 0
3 years ago
Read 2 more answers
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
3 years ago
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