Answer:
1311900 m far away from the earth
Explanation:
Given data
astronauts = 1.5 m
sound waves to travel = 343 m/s
to find out
How far away from the earth
solution
we know 2 astronauts are at distance 1.5 m from each other
and sound travel 343 m/s
so time period = distance / sound speed
time period = 1.5 / 343 = 4.373 × sec
and time for sound reach at earth
speed of sound = 3 × m/s
so
time = distnace / 3 ×
4.373 × = distnace / 3 ×
distance = 4.373 × × 3 ×
distance = 1311900 m
so 1311900 m far away from the earth
Answer:
(D) 3
Explanation:
The angular momentum is given by:
Thus, the magnitude of the angular momenta of both solar systems are given by:
where we have taken that both systems has the same radius.
By taking into account that T1=3T2, we have
but L1=L2=L:
Hence, the answer is (D) 3
HOPE THIS HELPS!!
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
v = 2 m/s
Hence, their speed after collision is 2 m/s.
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11