The mass of the astronaut is the same on both, but weight is actually a force and it depends on the acceleration due to gravity. On the moon, the acceleration due to gravity is 1/6 of the Earth’s so the astronaut’s weight will be 1/6 lighter on the moon.
Answer:
Explanation:
We shall apply the formula for velocity in case of elastic collision which is given below
v₁ = (m₁ - m₂)u₁ / (m₁ + m₂) + 2m₂u₂ / (m₁ + m₂)
m₁ and u₁ is mass and velocity of first object , m₂ and u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.
Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm
v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 ) + 2 x 18 x - 14 / ( 7.7 + 18 )
= - 8.817 - 19.6
= - 28.4 cm / s
Answer:
θ = 33°
Explanation:
Here, we can use the formula for the total time of flight of a projectile to calculate the launch angle of frog:
where,
θ = launch angle = ?
T = Total time of flight = 0.6 s
g = acceleration due to gravity = 9.81 m/s²
u = launch speed = 5.4 m/s
Therefore,
<u>θ = 33°</u>
Answer:
20.6 cm
Explanation:
charge per cm = 0.14 μC
number of electrons (e) =
to get the length of tape pulled we can apply the formula below
length of tape =
therefore we need to find the magnitude of the charge of the electrons
- 1 electron = C
now that we have the magnitude of the charge, we can find the length of the tape
- length of the tape = = 20.6 cm
Answer:
<em>The rebound speed of the mass 2m is v/2</em>
Explanation:
I will designate the two masses as body A and body B.
mass of body A = m
mass of body B = 2m
velocity of body A = v
velocity of body B = -v since they both move in opposite direction
final speed of mass A = 2v
final speed of body B = ?
The equation of conservation of momentum for this system is
mv - 2mv = -2mv + x
where x is the final momentum of the mass B
x = mv - 2mv + 2mv
x = mv
to get the speed, we divide the momentum by the mass of mass B
x/2m = v = mv/2m
speed of mass B = <em>v/2</em>