A hydraulic system at the mechanic shop can perform up to 180,000 joules of work in a day. if throughout the day it can lift 10
1 answer:
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Given:
Distance = 50 yard = 45.72 meter
Speed = 40 km/hr = 11.11 m/s
To find:
Time required by ball to reach the receiver = ?
Formula used:
speed =
Solution:
The speed of the ball is given by,
speed =
Thus,
Time =
Distance = 50 yard = 45.72 meter
Speed = 40 km/hr = 11.11 m/s
Time = 4.12 second
Hence, ball reaches the receiver in 4.12 second.
You will have to use this formula:
Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs
Then:
-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2
Ps: It's value is negative because the she was in retrograde motion.
Answer: Her acceleration is -2 m/s^2.
Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs
Then:
-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2
Ps: It's value is negative because the she was in retrograde motion.
Answer: Her acceleration is -2 m/s^2.
Answer:
Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m
Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m
Explanation:
Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.
H = (v₀² Sin²θ)/2g
v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s
θ = 75°, g = 9.8 m/s²
H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m
Range of projectile
R = v₀² (sin2θ)/g
R = 250² (sin2×75)/9.8
R = 250² (sin 150)/9.8 = 3188.8 m
Height of mountain = 1.80 × 10³ = 1800 m
Maximum height of projectile = 2975.2 m
Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m
Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m
Range of projectile = 3188.8 m
Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m
