law of electromagnetic induction hope this helps
We have thatThe options that define Speed, Velocity,Acceleration.
- Speed is used to describe how fast the object is moving.
- Acceleration is the rate of change of velocity in time.
- Velocity is used to describe how fast the object is moving and tells us in which direction it is going.
Options
B,D,E
From the question we are told
Choose the correct definitions of speed, velocity, and acceleration. Check all that apply.
- Acceleration tells us in which direction the object is going. Speed is the rate of change of velocity in time.
- Acceleration is the rate of change of velocity in time.
- Velocity is used to describe changes in the movement direction of the object. Acceleration tells us how far the object will go in a certain amount of time.
- Velocity is used to describe how fast the object is moving and tells us in which direction it is going.
- Speed is used to describe how fast the object is moving.
- Velocity is the rate of change of speed in time.
Speed is used to describe how fast the object is moving and tells us in which direction it is going.
Generally
- Acceleration is the rate of change of velocity in time.
- Velocity is used to describe how fast the object is moving and tells us in which direction it is going.
- Speed is used to describe how fast the object is moving without its direction in consideration
Hence
The options that define Speed, Velocity,Acceleration.
- Speed is used to describe how fast the object is moving.
- Acceleration is the rate of change of velocity in time.
- Velocity is used to describe how fast the object is moving and tells us in which direction it is going.
Options
B,D,E
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Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
Normal force for the rock because that makes an object stable at its position.
static friction because micro-welts hold its particle on its position so it doesn't change in position by a potential energy. Gravity makes it stay on the ground because its force attraction between an object and the earth.
Hope this helps <span />
<h2>since weight is measured in newtons, convert the 6 kg to newtons</h2><h3>the formula to convert is kg x 9.807 = N</h3>
hope that helps :))
