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Ber [7]
3 years ago
9

Give an example of hypothesis for an experiment and then identify its dependent and independent variables. Write all the steps o

f the scientific method. Explain why it is good to limit an experiment to test only one variable at a time whenever possible ?
Please somebody !!!!
Physics
1 answer:
e-lub [12.9K]3 years ago
6 0
An example of a hypothesis for an experiment might be: “A basketball will bounce higher if there is more air it”

Step one would be to make an observation... “hey, my b-ball doesn’t have much air in it, and it isn’t bouncing ver high”

Step two is to form your hypothesis: “A basketball will bounce higher if there is more air it”

Step three is to test your hypothesis: maybe you want to drop the ball from a certain height, deflate it by some amount and then drop it from that same height again, and record how high the ball bounced each time.


Here the independent variable is how much air is in the basketball (what you want to change) and the dependent variable is how high the b-ball will bounce (what will change as a result of the independent variable)

Step four is to record all of your results and step five is to analyze that data. Does your data support your hypothesis? Why or why not?

You should only test one variable at a time because it is easier to tell why the results are how they are; you only have one cause.

Hope this helps!
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A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti
Zigmanuir [339]

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: Vf^{2} = Vo^{2} - 2gh and h=Vo*t - \frac{g*t^{2} }{2}, First we need to find the height to time iqual to 3.04 s using the formula: h=Vo*t - \frac{g*t^{2} }{2} and remember that golf ball was released from the rest (Vo= 0 (m/s)) so h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}, we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using Vf^{2} = Vo^{2} - 2gh solving for Vf, we get: Vf = \sqrt{Vo^{2}-2*g*h } replacing the values given Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }, so we get: Vf = 29.79 (m/s).

5 0
4 years ago
Read 2 more answers
?????????????????????
NeTakaya

Answer:

me too

Explanation:

5 0
3 years ago
What is an example of a stable system
oksian1 [2.3K]
Some examples of stable system are:

1) functions of sine

2) DC

3) signum

4) unit step 

5) cosine.

Happy Studying! ^^


6 0
3 years ago
Read 2 more answers
While taking the stairs it takes you 10 seconds to reach the top. The next time you take the same stairs, it takes you 5 seconds
Ksju [112]

Answer:

  P₂ = 2 P₁

we conclude that in the second time the power used is double that in the first rise

Explanation:

In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.

Now we can analyze the required power,

         P = W / t

From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,

therefore the first time the power is

           P₁ = E / 10

           P₁ = 0.1 E

for the second time the power is

          P₂ = E / 5

          P₂ = 0.2 E

we see that the power in the second case is

         P₂ = 2 P₁

Therefore, we conclude that in the second time the power used is double that in the first rise.

6 0
3 years ago
A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What m
Goryan [66]

Answer:

ΔE = GMm/24R

Explanation:

centripetal acceleration a = V^2 / R = 2T/mr

T= kinetic energy

m= mass of satellite, r= radius of earth

= gravitational acceleration = GM / r^2

Now, solving for the kinetic energy:

T = GMm / 2r = -1/2 U,

where U is the potential energy

So the total energy is:

E = T+U = -GMm / 2r

Now we want to find the energy difference as r goes from one orbital radius to another:

ΔE = GMm/2 (1/R_1 - 1/R_2)

So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R

ΔE = GMm/2R (1/3 - 1/4)

ΔE = GMm/24R

6 0
3 years ago
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