Answer:
Many drivers follow the “three-second rule.” In other words, you should keep three seconds worth of space between your car and the car in front of you in order to maintain a safe following distance.
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The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
A
Answer:
Explanation:
Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s
An impulse results in a change of momentum
The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s
The remaining momentum is 4.5e6 - 0.12e6 = 4.38e6 kg•m/s
The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s
The tug applies 0.012e6 N•s of impulse each second.
The initial barge momentum will be zero in
t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds
To stop the barge in one minute(60 s), the tug would have to apply
4.5e6 / 60 = 75000 N•s /s or 75 000 N
Answer:
6.58m
Explanation:
The kinetic energy = Workdone on the roller
Workdone = Force * distance
Given
KE = Workdone = 362J
Force = 55N
Required
Distance
Substitute into the formula;
Workdone = Force * distance
362 = 55d
d = 362/55
d = 6.58m
Hence the student must push at a distance of 6.58m
Answer:
Compression is the answer
Explanation:
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