distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years
Answer:
the weight of the wire + 25kg
Explanation:
Answer:
a) W = 10995.6 J
b) W = - 9996 J
c) Kf = 999.6 J
d) v = 5.77 m/s
Explanation:
Given
m = 60 Kg
h = 17 m
a = g/10
g = 9.8 m/s²
a) We can apply Newton's 2nd Law as follows
∑Fy = m*a ⇒ T - m*g = m*a ⇒ T = (g + a)*m
where T is the force exerted by the cable
⇒ T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)
⇒ T = 646.8 N
then we use the equation
W = F*d = T*h = (646.8 N)*(17 m)
W = 10995.6 J
b) We use the formula
W = m*g*h ⇒ W = (60 Kg)(9.8 m/s²)(-17 m)
⇒ W = - 9996 J
c) We have to obtain Wnet as follows
Wnet = W₁ + W₂ = 10995.6 J - 9996 J
⇒ Wnet = 999.6 J
then we apply the equation
Wnet = ΔK = Kf - Ki = Kf - 0 = Kf
⇒ Kf = 999.6 J
d) Knowing that
K = 0.5*m*v² ⇒ v = √(2*Kf / m)
⇒ v = √(2*999.6 J / 60 Kg)
⇒ v = 5.77 m/s
Answer:
Radio waves heading upward from Earth often bounce off the thermosphere-D.
