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ANEK [815]
3 years ago
7

How much work does an athlete do if she raises a 5N kettle bell 2 m off the ground?

Physics
1 answer:
Degger [83]3 years ago
6 0

Answer:

<h2>The answer is 10 N</h2>

Explanation:

To find the work done by an object given the distance covered by the object and it's force we use the formula

<h3>work done = force × distance</h3>

From the question

force = 5 N

distance = 2 m

We have

work done = 5 × 2

We have the final answer as

<h3>10 N</h3>

Hope this helps you

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Answer:

Space probes are made to conduct science experiments. They do not have people on them. Space probes have helped scientists get information about our solar system. Most probes are not designed to return to Earth. Some have landed on other planets! Others have flown past the planets and taken pictures of them for scientists to see. There are even some space probes that go into orbit around other planets and study them for a long time. The information they gather is used to help us understand the weather and other changes which happen on planets other than the Earth. This information is important in helping to plan other space missions such as ones to Mars and to Saturn.

Explanation:

7 0
3 years ago
A student who weighs 750 newton’s runs up the steps (which have a height of 8 meters) in 13.5 seconds. How much work did the stu
nikklg [1K]

Answer:

B

Explanation:

6 0
3 years ago
Gravitational forces is ------ force​
mel-nik [20]

Answer:

Gravitational force is <u>noncontact</u> force

Explanation:

Contact force occurs due to the contact between two different objects. Non-contact force occurs due to either attraction or repulsion between two objects such that there is no contact between these objects. There is no field linked with the contact force. ... Gravitational force is an example of a non-contact force.

7 0
3 years ago
Please help!
andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0
3 years ago
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0
3 years ago
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