A white dwarf star has a density of about 1.0 x 10^9 kg/m3. If the earth were to suddenly become as dense as a white dwarf star,
1 answer:
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Answer:
a. 2.668 m/s
b. 0.00494
Explanation:
The computation is shown below:
a. As we know that
As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.
F = 3.70 cos 45 = 2.62 N
We know that
KE1 = Initial kinetic energy
KE2 = kinetic energy following 100 m
The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.
So, the equation is
KE2 = KE1 - W
Now solve for v2
= 2.668 m/s
b. Now the minimum value of Ug is
As we know that
Ff = force of friction
Us = coefficient of static friction
N = Normal force = weight of skater
So,
Now solve for Us
= 0.00494
Answer:
the final velocity of the wagon is 6 m/s.
Explanation:
Given;
initial velocity of the wagon, u = 4 m/s
mass of the wagon, m = 35 kg
energy applied to the wagon, E = 350 J
The final velocity of the wagon is calculated as;
E = ¹/₂m(v² - u²)
Therefore, the final velocity of the wagon is 6 m/s.