We can find the acceleration as follows:
Answer:
D. 2 m/s^2
A rocket achieves a lift-off velocity of 500.0 m/s from rest in 30.0 seconds. Calculate the average acceleration of the rocket.
1 answer:
Answer:
The average acceleration is 16.6 m/s² ⇒ 1st answer
Explanation:
A rocket achieves a lift-off velocity of 500.0 m/s from rest in
30.0 seconds
The given is:
→ The initial velocity = 0
→ The final velocity = 500 meters per seconds
→ The time is 30 seconds
Acceleration is the rate of change of velocity of the rocket
→
where a is the acceleration, v is the final velocity, u is the initial velocity
and t is the time
→ u = 0 , v = 500 m/s , t = 30 s
Substitute these values in the rule
→ m/s²
<em>The average acceleration is 16.6 m/s²</em>
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Answer:
687.5 A
Explanation:
We are given that
Magnetic field,B=0.11 T
Diameter,d=2.5mm
Radius,r=
We know that current in coil
Substitute the values
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα=
Senα=
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s