Answer:
ΔE = 37.8 x 10^9 J
Explanation:
The energy required will increased the potential energy and increase the kinetic energy.
As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative
Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²
G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²
ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J
The centripetal force at orbit must be equal to the gravity force
mv²/R = mg'
v²/8.0e6 = 6.272
v² = (6.272(8.0e6)) = 50.2e6 m²/s²
The maximum velocity when resting on earth at the equator is about 460 m/s.
The change in kinetic energy is
ΔKE = ½m(vf² - vi²)(1000)
ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J
Total energy increase is
25e9 + 12.857e9 = 37.8e9 J
For B, it is because water is a really good conductor of electricity, so the electrician will get shocked
Answer:
given -
initial velocity = 4.09 m/s
acceleration = 1.01 m/s²
distance = 23.4 m
time = ?
using second formula of motion,
s = ut + 1/2 at².
where, s = distance
u = initial velocity
t = time
a = acceleration
23.4 = 4.09(t) + 1/2(1.01)(t) ²
23.4 = 4.09t + 2.02t²
2.02t² + 4.09t - 23.4 = 0
solving the equation by using quadratic formula
Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :
a = 2.02
b = 4.09
c = -23.4
we get t=2.539 or t= -4.563
time cannot be negative so
t= 2.539 sec = 2.6 Sec is the answer
Answer:
141.42 km/h
Explanation:
Since this is a 100km/h right angle crosswind, the speed vector with respect to the ground is the vector speed of the airplane with respect to air + the vector speed of the cross-wind with respect to the ground
<100,0> + <0,100> = <100,100>
This vector has a magnitude of
km/h
Answer:
-9.8 m/s²
0 m/s
Explanation:
Average acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (-9.8 m/s − 9.8 m/s) / 2.0 s
a = -9.8 m/s²
Average velocity is the change in position over change in time.
v = Δx / Δt
v = (0 m − 0 m) / 2.0 s
v = 0 m/s