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3 years ago

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Physics

1 answer:

coldgirl [10]3 years ago

4 0

Answer:

An independent variable is the variable in an experiment that is changed. A dependent variable is the resulting variable that "depends" on the changes made to the independent variable. Controlled variables are all of the other variables that are kept the same between all changes made to the experiment. You would graph the independent variable and dependent variable. The independent variable will be found on the x-axis while the dependent variable can be found on the y-axis.

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A 42.2 kg sled is pulled forward

zaharov [31]

The net force on the sledge is 31.64N.

Frictional force = µkR

= 0.269 x 42.2 x 9.81 = 111.36

net force = 143N - 111.36N

= 31.64N

refer brainly.com/question/24557767

#SPJ2

7 0

2 years ago

Plz answer this very soon

tester [92]

Answer:

Im gonna say it is answer A:) Hope this helps!

Explanation:

6 0

3 years ago

Read 2 more answers

The escape speed from the moon is much smaller than from earth. True or False

Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

$v_{e} = \sqrt{\frac{2GM}{r}}$ (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is $5.972x10^{24}kg$ and its radius is $6371000m$

Then, replacing those values in equation 1 it is gotten.

$v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}$

$v_{e} = 11.18m/s$

For the case of the Moon:

$v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}$

$v_{e} = 2.38m/s$

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

4 0

3 years ago

An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s

DENIUS [597]

A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.

6 0

3 years ago

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe

dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is 3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F $_{hand$

using equation for impulse in momentum

F $_{hand$ × t = m( v - v₀ )

we substitute

F $_{hand$ × 0.00265 = 1.75( 0 - 5.25 )

F $_{hand$ × 0.00265 = 1.75( - 5.25 )

F $_{hand$ × 0.00265 = -9.1875

F $_{hand$ = -9.1875 / 0.00265

F $_{hand$ = -3466.98 N

Next we determine force on the leg F $_{leg$

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F $_{leg$ = - F $_{hand$

we substitute

F $_{leg$ = - ( -3466.98 N )

F $_{leg$ = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N

5 0

3 years ago

HELP ASAP THIS IS WORTH ALOT ​

HELP ASAP THIS IS WORTH ALOT