Answer:
Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)
Explanation:
Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.
or by rearanging the drivers equation.
Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.
To solve this equation we use the following formulas
Where a=1; b=-28.75; c=154
So we get:
At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.
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This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).
Best of luck
Answer: Smaller than ; larger than
Explanation:
When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of
N= mg +ma, (g is gravity and a is acceleration)
here ma is negative so the N= mg-ma
Hence, it feels smaller than its original weight.
When the elevator is moving downward , then the force acting will be positive in nature
N= mg+ma,
here ma will be positive so it feels larger the original weight of passenger.
Yes. A roulette ball circulating in a spinning wheel, a car going around a curved
road at 30 mph, and a planet in a circular orbit are all being accelerated.
"Acceleration" does NOT mean "speeding up". It means any change in the
speed or DIRECTION of motion.
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
A.a sign that breaks loose from the ground when a force is applied to it
