Choice-C is a correct statement.
Answer:
Explanation:
from the question we have the following:
distance between Sacramento and los angles = 400 miles
speed of car A = 60 mph
start time of car A = 11 am
speed of car B = 75 mph
start time of car B = 12 pm
distance of Fresno from Los Angeles = 150 miles
- To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
- Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
- From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
- Distance covered by car A = speed x time(t) = 60 x t = 60t
- Distance covered by car B = speed x time(t) = 75 x t = 75t
- 60t + 75t = 340 miles
- 135t = 340
- t = 2.51 hours
- Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
- Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
- 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.
Answer:
7.04 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement on Earth = 1.2 m
a = Acceleration due to gravity on Moon = 1.67 m/s²
a = Acceleration due to gravity Earth= 9.81 m/s²
Accelration going up is considered as negetive
Initial Velocity of the ball
Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is
The displacement of the ball on the moon is 7.04 m
Explanation:
There are three forces on the bicycle:
Reaction force Rp pushing up at P,
Reaction force Rq pushing up at Q,
Weight force mg pulling down at O.
There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.
Sum of the forces in the y direction:
Rp + Rq − (15)(9.8) = 0
Rp + Rq − 147 = 0
Sum of the moments at P:
(15)(9.8)(0.30) − Rq(1) = 0
44.1 − Rq = 0
Sum of the moments at Q:
Rp(1) − (15)(9.8)(0.70) = 0
Rp − 102.9 = 0
Sum of the moments at O:
Rp(0.30) − Rq(0.70) = 0
0.3 Rp − 0.7 Rq = 0
Any combination of these equations will work.
After impact velocity = 14.968 ft/s
Weight and mass of Bullet and wooden block:
Bullet: w = 1oz = 1/16 lb m = 0.001941 lb
wooden block : W = 5lb M = 0.15528 lb
velocity of block and bullet immediately after impact:
Σmv1 + ΣImp = mv2
Resolving vertical component
( m× v₀cos30⁰) + 0 = ( m+M) v'
v' = ( m× v₀cos30⁰)/ (m+M)
v' = 14.968 ft/s
Horizontal and vertical component of the impulse exerted by block on the bullet:
Here we will apply the principle of impulse and momentum.
Horizontal component:
-mv₀ cos30⁰ + RxΔt =0
RxΔt = mv₀sin30⁰
= 0.001941 × 1400sin30⁰
RxΔt = 1.3587 lb.s
Vertical component:
-mv₀cos30⁰ + RyΔt = -mv'
RyΔt = m( v₀cos30⁰-v')
RyΔt = 0.001941(1400cos30⁰ - 14.968)
= 2.32 lb.s
Learn more about impact here:
brainly.com/question/15008937
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