Answer:
b) 4781 N
Explanation:
Because there is a redius do this question is talking about the acceleration force which= mv^2/r
so a=15^2/80=2.8125 m^2/s
so the force will be = m.a
F =1700×2.8125=4781.25 N
Answer:
I_v = 2,700 W / m^2
I_m = 610 W / m^2
I_s = 16 W / m^2
Explanation:
Given:
- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W
- Radius of Venus r_v = 1.08 * 10^11 m
- Radius of Mars r_m = 2.28 * 10^11 m
- Radius of Saturn r_s = 1.43 * 10^12 m
Find:
Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.
Solution:
- We know that Power is related to intensity and surface area of an object follows:
I = P / 4*pi*r^2
Where, A is the surface area of a sphere models the atmosphere around the planets.
a)
- The intensity at the surface of Venus is calculated as:
I_v = P_s / 4*pi*r^2_v
I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2
I_v = 2,700 W / m^2
b)
- The intensity at the surface of Mars is calculated as:
I_m = P_s / 4*pi*r^2_m
I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2
I_m = 610 W / m^2
c)
- The intensity at the surface of Saturn is calculated as:
I_s = P_s / 4*pi*r^2_s
I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2
I_s = 16 W / m^2
Answer:
-6327.45 Joules
650.375 Joules
378.47166 N
Explanation:
h = Height the bear slides from = 15 m
m = Mass of bear = 43 kg
g = Acceleration due to gravity = 9.81 m/s²
v = Velocity of bear = 5.5 m/s
f = Frictional force
Potential energy is given by

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules
Kinetic energy is given by

Kinetic energy of the bear just before hitting the ground is 650.375 Joules
Change in total energy is given by

The frictional force that acts on the sliding bear is 378.47166 N