Answer: I hope this helps you
Explanation:
Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = ( ) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K =
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth
Answer:
34 m/s
Explanation:
Centripetal acceleration is:
a = v² / r
Given a = 6.55 × 9.8 m/s² = 64.2 m/s², and r = 18 m:
64.2 m/s² = v² / (18 m)
v = 34 m/s
The speed of the centrifuge is 34 m/s.
Answer:
Force A=-−2,697.75 N
Force B=13, 488.75 N
Explanation:
Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.
25 mg-20Fb=0
25*1100g=20Fb
Fb=25*1100g/20=1375g
Taking g as 9.81 then Fb=1375*9.81=13,488.75 N
The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g
-275*9.81=−2,697.75. Therefore, force A pulls downwards
Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle
Principles of conservation of linear momentum
Kinetic Energy before collision 60kJ
Kinetic Energy after collision 56.76kJ
Explanation:
1. Given data
mass m1= 1000kg
velocity u1= 10m/s
mass m2= 800kg
velocity u2= 5m/s
2. We want to find the common velocity after the collision
Applying the principle of conservation of linear momentum for inelastic collision we have
m1u1+m2u2= (m1+m2)V
substituting and solving for V
1000*10+800*5= (1000+800)V
10000+4000= 1800V
14000=1800V
3. Divide both sides by 1800
V= 14000/1800
V= 7.8m/s
4. Now let us find the kinetic energy of the two cars before the collision
For the first car
KE= 1/2m1u1^2
KE= 1/2*1000*10^2
KE=(1000*100)/2
KE= 100000/2
KE= 50000 Joules
KE= 50 kJ
For the second car
KE= 1/2m2u2^2
KE= 1/2*800*5^2
KE=(800*25)/2
KE= 20000/2
KE= 10000 Joules
KE= 10 kJ
Hence the total kinetic energy before the collision is
=50+10
=60kJ
After collision
KE= 1/2m1V^2
KE= 1/2*1000*7.8^2
KE=(1000*60.84)/2
KE= 60840/2
KE= 32420 Joules
KE= 32.42kJ
For the second car
KE= 1/2m2u^2
KE= 1/2*800*7.8^2
KE=(800*60.84)/2
KE= 48.672/2
KE= 24336 Joules
KE= 24.34 kJ
Hence the total Kinetic energy after collision is
=32.42+24.34
=56.76kJ
See the link below for more information on linear momemtum
https://brainly.in/question/17792506