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3 years ago

A coin is dropped. What is its displacement after 0.30 s.

1 answer:

3 0

Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.

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A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 6 m/s at the floor. Next, the p

Anna71 [15]

Answer:

Explanation:

Let h be the height .

initial velocity in first case u = 0

final velocity v = 6 m /s

acceleration due to gravity g = 9.8 m /s²

v² = u² + 2 g h

6² = 0 + 2 x 9.8 x h

h = 1.837 m .

For second case u = 3 m /s

v² = u² + 2 gh

= 3² + 2 x 1.837 x 9.8

= 9 + 36

= 45 m

v = 6.7 m /s

8 0

3 years ago

Which best summarizes a concept related to the work-energy theorem?

Maurinko [17]

Answer:

When work is positive, the environment does work on an object.

Explanation:

According to the work-energy theorem, the net work done by the forces on a body or an object is equal to the change produced in the kinetic energy of the body or an object.

The concept that summarizes a concept related to the work-energy theorem is that ''When work is positive, the environment does work on an object.''

4 0

3 years ago

amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i

Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,

weight of equipment = 6 kN

coefficient of kinetic friction = 0.05

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²

Work done by the camper = ?

actual acceleration acting a'

m a = m a' - μ mg

a' = a + μ g

a' = 0.2 + 0.05 x 9.8

a' = 0.69 m/s²

Work done = Force x distance

F = m a'

$F = \dfrac{6000}{9.8} \times 0.69$

F = 422.44897 N

Work done = F x d

Work done = 422.44897 x 1000

Work done = 422449 J

Work done = 422.45 kJ

4 0

3 years ago

WILL GIVE BRILLIANTIST

sleet_krkn [62]

Answer: A. foliated

Explanation: took the test!

7 0

3 years ago

Read 2 more answers

A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci

Vlad [161]

Answer:

Approximately $\rm 2.5\; m \cdot s^{-1}$ .

Explanation:

Let the increase in the rocket's velocity be $\Delta v$ . Let $v_0$ represent the initial velocity of the rocket. Note that for this question, the exact value of $v_0$ doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

Mass of the rocket with the 5 kg of fuel: $1000$ .
Initial velocity of the rocket and the fuel: $v_0$ .
Hence the initial momentum of the rocket: $1000\,v_0$ .

Mass of the rocket without that 5 kg of fuel: $1000 - 5 = 995$ .
Final velocity of the rocket: $v_0 + \Delta v$ .
Hence the final momentum of the rocket: $995\,(v_0 + \Delta v)$ .

Mass of the 5 kg of fuel: $5$ .
Final velocity of the fuel: $v_0 - 500$ (assuming that the the 500 m/s in the question takes the rocket as its reference.)
Hence the final momentum of the fuel: $5\,(v_0 - 500)$ .