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zhenek [66]
3 years ago
14

What are the requirement for a valid hypothesis ?

Physics
1 answer:
kramer3 years ago
3 0
If you are talking about a scientific hypothesis, it essentially needs to be falsifiable, that is, that it can be proven wrong. Another requirement is for it to be testable, though under certain circumstances and definitions those two mean the same things. <span>
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You might be interested in
In the absence of a gravitational force, the weight of a body is?
azamat

Answer:

0

Explanation:

Weight = mass x gravity

if gravity = 0 then Weight =0 as well

5 0
2 years ago
An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete
Ksju [112]
 <span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:


R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km 

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s 

  the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h 

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu!  :)</span>
6 0
3 years ago
In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0
3 years ago
Read 2 more answers
C, N, Ne, Ar which contains a metal, nonmetal, noble gas , metalloid
postnew [5]

Answer:

c,carbono nonmetal

n, nitrogen nonmetal

ne, neón noble gas

ar,argon noble gas

4 0
3 years ago
Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1360 N. Assume that the play
Alona [7]

Answer:

7.59Ns

Explanation:

Given parameters:

Force  = 1360N

Time of contact  = 5.85 x 10⁻³s

Unknown:

Impulse  = ?

Solution:

The impulse of the ball is given as:

        Impulse  = Force x time

       Impulse  = 1360 x 5.85 x 10⁻³ = 7.59Ns

4 0
3 years ago
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