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Alexus [3.1K]
3 years ago
7

Please could you guys help me. When an object falls from a height, the maximum energy transferred to its kinetic energy store is

Physics
1 answer:
lesya692 [45]3 years ago
3 0

Answer:

the energy transferred away from its gravitational potential energy store

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

P.E = mgh

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

When an object falls from a height, the maximum energy transferred to its kinetic energy store is equal to the energy transferred away from its gravitational potential energy store.

This ultimately implies that, while the object was at rest it possessed potential energy, which is then transformed into motion when it starts to fall.

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3 0
4 years ago
. Why are scientific models important? They help scientists prove scientific theories. They help scientists explain complex ques
sdas [7]
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3 0
4 years ago
A ball is thrown straight upward with a speed of 36 m/s. How long does it take to return to its starting point, assuming negligi
svetoff [14.1K]

Answer:

The time taken for the ball to return to the starting point is is 7.4 s

Explanation:

Given;

initial velocity of the ball, u = 36 m/s

the final vellocity at maximum height, v = 0

let time taken for the ball to reach maxmimum height = t

Time taken for the ball to return to the starting point is known as time of flight, calculated as;

t = \frac{v-u}{-g} \\\\T = 2t\\\\T = \frac{2(0-u)}{-g}\\\\T = \frac{-2u}{-g}\\\\ T = \frac{2u}{g}

T = (2 x 36) / 9.8

T = 7.4 s

Therefore, the time taken for the ball to return to the starting point is is 7.4 s

4 0
3 years ago
After t hours a freight train is s(t) = 18t2 − 2t3 miles due north of its starting point (for 0 ≤ t ≤ 9). (a) Find its velocity
BartSMP [9]

Answer:

Explanation:

Given the equation modelled by the height of the train given as:

s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9

a) Velocity is the rate of change of displacement.

Velocity = dS(t)/dt

V = dS(t)/dt = 36t - 6t² miles

Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.

V = 36(3) -6(3)²

V= 108 - 72

Velocity = 36mi/hr

b) for Velocity at time = 7hrs

V(7) = 36(7) - 6(7)²

V(7) = 252 - 294

V(7) = -42mi/hr

The velocity at t = 7hrs is -42mi/hr

c) Acceleration is the rate of change of velocity.

a(t) = dV(t)/dt

Given v(t) = 36t - 6t²

a(t) = 36 - 12t

Acceleration at t=1 is given as:

a(1) = 36 -12(1)

a(1) = 24mi/hr²

4 0
3 years ago
A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n.
Gekata [30.6K]

Answer:

ΔTita = 4205.6 rad

Explanation:

w_{i} means initial angular velocity, which is 0 rev/min

w_{f} means final angular velocity, which is 2.513*10^{4}rev/min

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

w_{f} = (2π/60) *2.513*10^{4} rad/s

w_{f}= 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

a = 821.40 rad/s^{2}

so the rotational motion about a fixed axis is:

w^{2} _{f} =w^{2} _{i} + 2αΔTita    where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita = \frac{w^{2} _{f}-w^{2} _{i}  }{2a}

ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40

ΔTita = 4205.6 rad

7 0
4 years ago
Read 2 more answers
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