Answer:
3234.2 W
Explanation:
Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.
So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².
Now, the power radiated on the patch of area 0.570 m² at the equator is
P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W
Around 3.70 seconds unless traveling through media
M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).
The momentum before colision is equal with the momentum after colision:
m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem
=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
Answer:
Explanation:
Given:
- mass of the object on a horizontal surface,
- coefficient of static friction,
- coefficient of kinetic friction,
- horizontal force on the object,
<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>
where:
normal force of reaction acting on the body= weight of the body
As we know that the frictional force acting on the body is always in the opposite direction:
So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.
so, the frictional force will be:
Gas "floats" so if there are examples or pictures it would be the one with the most evenly spread out "dots".
