Left block: f k=0.462(2.45N)=1.13N. The maximum static friction force would be larger, so the spring force would produce no motion of this block or the right-hand block, which could feel even more friction force. For both, a=0.
<h3>What is kinetic energy?</h3>
- A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
- Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
- The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
- Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.
(c) 0.462:
Frictional force is the one which opposes the motion of an object. Here, the force of kinetic friction on the heavier block is greater than the force exerted by the spring on the block. Hence, the block will not move and the acceleration is zero.
F k=0.462(2.45N)=1.13N for the left block. Because the maximum static friction force would be higher, neither this block nor the right-hand block, which might experience even greater static friction force, would move in response to the spring force. A=0 for both.
To learn more about kinetic energy, refer to:
brainly.com/question/25959744
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Answer:
36,000j
Explanation:
W= P•t
120•300= 36000
I checked my answer with P= W/t and got the right answer
hope this helps
The resultant of the acceleration can be found using:
a = √(ax² + ay²)
a = √(1.5² + 3.7²)
a = 3.99 m/s²
v = u + at, u = 0
v = 3.99 x 6.4
v = 25.5 m/s
D, melting point and reactivity/solubility in water!
Answer:
induced emf = 28.65 mV
Explanation:
given data
diameter = 7.3 cm
magnetic field = 0.61
time period = 0.13 s
to find out
magnitude of the induced emf
solution
we know radius is diameter / 2
radius = 7.3 / 2
radius = 3.65 m
so induced emf is dπ/dt = Adb/dt
induced emf = A × ΔB / Δt
induced emf = πr² × ΔB / Δt
induced emf = π (0..65)² × ( 0.61 - (-0.28)) / 0.13
induced emf = 0.0286538 V
so induced emf = 28.65 mV