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tamaranim1 [39]
3 years ago
7

An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?

Physics
1 answer:
Norma-Jean [14]3 years ago
6 0
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s
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What characteristics of EM waves did you discover?
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4 0
3 years ago
Read 2 more answers
A soccer ball was kicked off the roof of a building going
WINSTONCH [101]

Answer:

Final vertical velocity = -29m/s

Horizontal distance = 100m

Height = 20.41m

Explanation:

1. The vertical final velocity can be calculated thus: vy = vyo - gt

Where;

vy = vertical velocity (m/s)

vyo = vertical initial velocity (20m/s)

g = acceleration due to gravity (9.8m/s²)

t = time (5s)

Hence, vy = vyo - gt

vy = 20 - (9.8 × 5)

vy = 20 - 49

vy = -29m/s

2. x = V0 x t

Where;

x = horizontal distance (m)

Vo = initial velocity

t = time (s)

x = 20 × 5

x = 100m

3. Maximum height = (voy)²/2g

= 20²/ 2 × 9.8

= 400/19.6

= 20.41m

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3 years ago
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nikdorinn [45]

Answer:

1/4 times your earth's weight

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assuming the Mass of earth = M

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∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = \frac{GM}{R^{2} }

where G is the gravitational constant

Gravitational force of the planet = \frac{G4M}{(4R)^{2} }

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∴Gravitational force of planet = 1/4 times the gravitational force of the earth

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