Answer:
13.5
Explanation:
Mass: 5kg
Initial Velocity: -15
Final Velocity: 12
Force: 10
We can use the equation: Vf = Vi + at
We need to find acceleration, and we can use the equation, F=ma,
We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.
Now we have all the variables to find time.
Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)
Plugging them in into desmos gives 13.5 for time.
Answer:
5x10^-3
Explanation:
Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.
Hooke's Law can be represented as
<h3> F = kx, </h3>
<em>where F is the force </em>
<em> k is the spring constant</em>
<em> x is the extension of the material </em>
<em />
Plug values in the equation
Step 1 find the original extension
0.045 = (400)x
x = 1.125x 10^-4 m d
Step 2 find the new extension
0.045+2 = 400(x)
2.045 = 400x
x = 5.1125x10^-3
Step 3 subtract the new extension with original
Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3
Answer:
Explanation:
= 4190 J/kg.K
= 910 J/Kg. K
= 1.50 kg
= 1.80 kg
ΔT + 
ΔT
= (1.50)(910)(85.0-20)+(1.80)(4190)(85.0-20)
= 578,955 J
= 579 kJ
Answer:
if you slide a hockey puck on ice, it will eventually stop, because of friction on the ice
kite when the wind changes can be described by the first law
Explanation:
if you slide a hockey puck on ice, it will eventually stop, because of friction on the ice
kite when the wind changes can be described by the first law
<h2>
Answer: 7020.117 m/s</h2>
Explanation:
The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting, in this case, the Earth
.
the radius of the orbit (measured from the center of the planet to the satellite).
This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth 
and the altitude of the satellite above the Earth's surface 
.
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).
Now, rewriting equation (1) with the known values:
