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Maru [420]
3 years ago
8

An elevator ascends at a constant speed of 4 m/s, how much time is required for the elevator in order to travel 120 m upwards?

Physics
2 answers:
bekas [8.4K]3 years ago
6 0

Hello, Reenation12

The correct and aprooved answer is 30 m/s

This answer is really easy to get all you need to do is shown below

120 divided by 4  equals 30

If my answer helped please rate me as brainliest leave a thanks and rate my answer 5 stars thank you and have theh best day ever!

Dominik [7]3 years ago
5 0

Divide 120 by 4, to get 30 seconds

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If you apply an equal force to a larger mass, it will take longer to accelerate. <br> Tru or false
Makovka662 [10]

Answer:

true

Explication:

The acceleration of an object depends on the mass of the object, and the amount of force applied

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2 years ago
How do microwaves use the behavior and characteristics of the electromagnetic wave to function?
serg [7]

Answer:

Radiation moves out of the microwave into waves causing heat.

Explanation:

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Hope it helped!

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3 years ago
A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc
Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

<h3>Fբ = 75 N</h3>

  • Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

<h3>Fₙ = 25 N</h3>

  • Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

<h3>a = 0.5 m/s²</h3>

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0
2 years ago
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece
Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

8 0
2 years ago
Globalization is the process of
Delvig [45]

Answer: Globalization is the process of interaction and integration among people, companies, and governments worldwide.

Explanation:

5 0
3 years ago
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