Given that,
Mass of trackler, m₁ = 100 kg
Speed of trackler, u₁ = 2.6 m/s
Mass of halfback, m₂ = 92 kg
Speed of halfback, u₂ = -5 m/s (direction is opposite)
To find,
Mutual speed immediately after the collision.
Solution,
The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Answer:
• 36.4 kg of coal.
• 80 pounds of coal.
Explanation:
Using proportionality constant,
Mass of coal = 1,000,000/27,500,000 btus/metric ton
= 0.0364 metric tons of coal
Mass of coal = 1,000,000/25,000,000 btus/ton
= 0.04 tons of coal.
Converting metric tons to kilogram,
1 metric ton = 1000kg,
0.0364 metric ton;
= 36.4 kg of coal.
Converting tons to pounds,
1 ton = 2000 pounds,
0.04 metric ton;
= 80 pounds of coal.
Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut +
at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) +
x a x 5.2
53 = 2.6a
a =
= 20.4m/s²