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sukhopar [10]
3 years ago
5

In an experiment, the independent variable __________.

Physics
2 answers:
alexira [117]3 years ago
7 0

Answer:The correct answer is option D.

Explanation:

Independent variables are those variables which do not dependent on any other variable.These are those variables which can be controlled or varied by researchers to test their effect on the dependent variables.

Dependent variables are those variables which depends upon independent and other dependent variables.

Hence, from the given the option the correct answer is option D.

joja [24]3 years ago
4 0
D) is manipulated or varied
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If the 78.0 kg astronaut were in a spacecraft 6R from the center of the earth, what would the astronaut's weight be on earth? 76
den301095 [7]

(a) 764.4 N

The weight of the astronaut on Earth is given by:

F=mg

where

m is the astronaut's mass

g is the acceleration due to gravity

Here we have

m = 78.0 kg

g = 9.8 m/s^2 at the Earth's surface

So the weight of the astronaut is

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(b) 21.1 N

The spacecraft is located at a distance of

r=6R

from the center of Earth.

The acceleration due to gravity at a generic distance r from the Earth's center is

g=\frac{GM}{r^2}

where G is the gravitational constant and M is the Earth's mass.

We know that at a distance of r = R (at the Earth's surface) the value of g is 9.8 m/s^2, so we can write:

GM=9.8R^2 (1)

the acceleration due to gravity at r=6R instead will be

g'=\frac{GM}{(6R)^2}

And substituting (1) into this formula,

g'=\frac{9.8R^2}{36R^2}=0.27 m/s^2

So the weight of the astronaut at the spacecratf location is

F'=mg'=(78.0 kg)(0.27 m/s^2)=21.1 N

6 0
4 years ago
HELP I WILL GIVE BRAINLIEST!
Ann [662]

Answer:

Answer:B

Explanation:

Because it all stayed consistant

4 0
3 years ago
A golf club exerts an average horizontal force of 1000 n on a 0.045 -kg golf ball that is initially at rest on the tee. the club
OlgaM077 [116]
The impulse (the variation of momentum of the ball) is related to the force applied by
\Delta p = F \Delta t
where \Delta p is the variation of momentum, F is the intensity of the force and \Delta t is the time of application of the force. 
Using F=1000 N and \Delta t = 1.8 ms=1.8 \cdot 10^{-3}s, we can find the variation of momentum:
\Delta p = (1000 N)(1.8 \cdot 10^{-3} s)=1.8 kg m/s

This \Delta p can be rewritten as
\Delta p = p_f - p_i = mv_f - mv_i
where p_f and p_i are the final and initial momentum. But the ball is initially at rest, so the initial momentum is zero, and
\Delta p = mv_f
from which we find the final velocity of the ball:
v_f =  \frac{\Delta p }{m}= \frac{1.8 kg m/s}{0.045 kg}=  40 m/s
8 0
3 years ago
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