All categories

3 years ago

If a simple machine reduces the strength of a force, what must be increased?

1 answer:

6 0

The distance that the force moves" is the one among the following choices given in the question that must be increased, if a simple machine reduces the strength of a force. The correct option among all the options that are given in the question is the first option or option. I hope the answer helped you.

You might be interested in

The inner planets formed:

Ivan

Answer:

a. by collisions and mergers of planetesimals.

Explanation:

Inner planets are planets within 1.5 AU distance from the sun. These are called terrestrial planets because they are somewhat similar to Earth, mainly made of rocks.

The main ingredient of these planets are solar nebula and interstellar dust condensation of which leads to formation of small rock particles. These particles come close to each other under in the influence of gravity and other forces. As the mass of the particles increase they form planetesimals, these planetesimals eventually merge to form planets.

8 0

3 years ago

What is the scientific revolution??

Basile [38]

The scientific revolution is a concept which explains how the developments of science (biology, chemistry, physics and etc.) changed the way we (society) think about nature. I hope this helps! :)

4 0

4 years ago

Look at the quote: "Newton's laws of motion can be summarized in two simple ideas...." According to Newton's laws of motion, the

sertanlavr [38]

Bell math hajawonbejwbwbanananwnw w a. Sor

6 0

3 years ago

(3) What is the weight of a 50-kg astronaut (a) on Earth (b) On the Moon ,(g=1.7m/s2), (c) on Mars (g=3.7m/s2) (d)in outer space

artcher [175]

Answer:

a) On Earth

490N

b) On the Moon

85N

c) On Mars

185N

d)in outer space traveling with constant velocity.

Explanation:

The weight is defined as:

$W = mg$ (1)

Where m is the mass and g is the gravity

a) On Earth $g = 9.8m/s^{2}$

Then, equation 1 can be used:

$W = (50Kg)(9.8m/s^{2})$

$W = 490Kg.m/s^{2}$

but 1N = Kg.m/s^{2}

$W = 490N$

Hence, the weight of the astronaut on Earth is $490N$

b) On the Moon $g = 1.7m/s^{2}$

$W = (50Kg)(1.7m/s^{2})$

$W = 85N$

Hence, the weight of the astronaut on the Moon is $85N$

c) On Mars $g = 3.7m/s^{2}$

$W = (50Kg)(3.7m/s^{2})$

$W = 185N$

Hence, the weight of the astronaut on Mars is $185N$

(d) in outer space traveling with constant velocity.

Tanking into consideration that the astronaut is traveling in outer space at a constant velocity, it can be concluded that the acceleration will be zero.

Remember that the acceleration is defined as:

$a = \frac{v_{f} - v_{i}}{t}$

Since the acceleration is the variation of the velocity in a unit of time.

Therefore, from equation 1 is gotten.

$W = (50kg)(0)$

Remember that g is the acceleration that a body experience as a consequence of the gravitational field.

$W = 0$

5 0

3 years ago

Suppose that you hear a clap of thunder 16.2 s (seconds) after seeing the associated lightning stroke. The speed of sound waves

spayn [35]

Since the speed of sound is very high, we consider the time period of thunder and lightening negligible.

Now, d = vt

thus, d = 343 x 16.2

thus, d = 5556.6 metres

Thus, we are 5556.6 metres away from lightening ( approximately)

5 0

3 years ago