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zaharov [31]
3 years ago
5

A spacecraft is traveling in interplanetary space at a constant velocity. What is the estimated distance traveled by the spacecr

aft in 9.50 seconds?
Physics
2 answers:
aniked [119]3 years ago
7 0
You haven't said what the constant velocity is, so the best answer that
I can give you is

    Distance = (9.5) times (magnitude of the velocity, in meters per second) .
zhuklara [117]3 years ago
5 0
What is the constant velocity? 
<span>We need that number in order to solve the problem
</span>Or
<span>It has a graph in which it describes the different acceleration</span>
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An alloy that contains mainly copper and tin is bronze

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If a high jumper needs to make his center of gravity rise 1.50 m, how fast must he be able to sprint? Assume all of his kinetic
sergiy2304 [10]

Answer:

v = 5.42 m/s

Explanation:

given,

height of the jumper = 1.5 m

velocity of sprinter = ?

kinetic energy can be transformed into potential energy

m g h = \dfrac{1}{2}mv^2

g h = \dfrac{1}{2}v^2

v =\sqrt{2gh}

v =\sqrt{2\times 9.8 \times 1.5}

v = 5.42 m/s

Speed of the sprinter is equal to v = 5.42 m/s

7 0
4 years ago
Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran
alina1380 [7]

Answer:

Datos:

q1 = -50 μC = 50*10^{-6}

q2 = +30 μC = 30*10^{-6}

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = \sqrt\frac{k0 * q1 *q2}{F} = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10} = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em>\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}<em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50} = 0,520m

3 0
3 years ago
1. An astronaut in a spacesuit has a mass of 80 kilograms. What is the weight of this astronaut on the surface of the Moon where
Andrews [41]
Weight = 80 x (9.8/6) = .... N
8 0
3 years ago
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