Answer: E = 941738.537J
Explanation:
to begin,
given that the mass = 2320 pound = 1052.334 kg
Δh = 110 ft = 33.528 m
given that distance (d) = 1283 ft = 391.058 m
also the speed (v) is 65 mph = 29.058 m/s
force (F) = 87 pounds = 386.995 N
from our knowledge in work energy theory;
E = Fd + 1/2mv² + mgh
E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)
E = 151337.491 + 444278.2 + 346122.84
E = 941738.537J
i hope this helps, cheers.
Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.
Answer:
20 m
Explanation:
From the equation of motion,
S = ut+1/2gt²................................. Equation 1
Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.
Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative
Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)
Substituting into equation 1
S = 0(2) + 1/2(10)(2)²
S = 5(4)
S = 20 m
Hence the height of the the cliff above the pool is 20 m
Answer: Pretty sure the answer is B but this kind of looks like a test question.
Explanation:
