268.6567 mph is its velocity when it crosses the finish line
d=(v1+v2 /2) x t
.25=(0+v2 /2) x 6.7/3600 hours
900=v2/2 x 6.7
v2=268.6567 mph as the speed with which the dragster crosses the finish
<h3>When acceleration is not zero, can speed remain constant?</h3>
The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.
Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.
Learn more about velocity refer
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Answer:
156.8 Watts
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 10 kg
Height (h) = 8 m
Time (t) = 5 s
Power (P) =?
Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:
Mass (m) = 10 kg
Height (h) = 8 m
Acceleration due to gravity (g) = 9.8 m/s²
Energy (E) =?
E = mgh
E = 10 × 9. 8 × 8
E = 784 J
Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:
Time (t) = 5 s
Energy (E) = 784 J
Power (P) =?
P = E/t
P = 784 / 5
P = 156.8 Watts
Therefore, the power output of the motor is 156.8 Watts
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
Answer:
A 100 N force acting on a lever 2 m from the fulcrum balances an object 0.5 m from the fulcrum on. ... What is the weight of the object(in newtons)? What is its mass (in kg)? ... mass at the one end and effort arm is the distance between pivot and effort applied at the other end.
Explanation:
hpoe this helps you.
