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2 years ago

What is the work required to lift a 10 kg box off the ground to a height of 5.0 m?

Physics

1 answer:

FinnZ [79.3K]2 years ago

7 0

Answer:

490J

Explanation:

Given parameters:

Mass of box = 10kg

Height = 5m

Unknown:

Work done = ?

Solution:

Work done can be derived from the product of force and distance;

Work done = force x distance

Force here is applied by gravity and it is the weight

The distance is the height

Work done = weight x height

Work done = mass x acceleration due to gravity x height

Work done = 10 x 9.8 x 5 = 490J

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A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each

nataly862011 [7]

The centripetal acceleration is $13.0 m/s^2$

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference ( $2\pi r$ ) and the period of revolution (T), so it can be rewritten as

$v=\frac{2\pi r}{T}$

Therefore we can rewrite the acceleration as

$a=\frac{4\pi^2 r}{T^2}$

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

$T=\frac{1}{4}s = 0.25 s$

Substituting into the equation, we find the acceleration:

$a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2$

Learn more about centripetal acceleration:

brainly.com/question/2562955

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8 0

2 years ago

Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 kg Jan

Brut [27]

Complete part of Question: What is Jane's (and the vine's) angular speed just before she grabs Tarzan

Answer:

Jane's (and the vine's) angular speed just before she grabs Tarzan, w = 1.267 rad/s

Explanation:

According to the law of energy conservation:

Total change in kinetic energy = Total change in potential energy

Mass of Jane = 60 kg

Mass of the vine = 32 kg

Mass of Tarzan = 72 kg

Height of Tarzan = 5.50 m

Length of the vine = 8.50 m

Jane's change in gravitational potential energy,

$U_J = 60 * 9.8 * 5.5\\U_J = 3234 J$

Vine's gravitational potential energy,

$U_v = Mgh/2\\U_v = 32*9.8*5.5/2\\U_v = 862.4J$

Vine's Kinetic energy :

$KE_V = 0.5 I w^{2} \\I_V = \frac{ML^2}{3} = \frac{32 * 8.5^2}{3} = 770.67 kg m^2\\ KE_V = 0.5 *770.67 * w^{2}\\KE_V = 385.33 w^{2}$

Jane's Kinetic energy:

$KE_J = 0.5m(wL)^2\\KE_J = 0.5*60(w * 8.5)^2\\KE_J = 2167.5 w^2$

$U_J + U_V = KE_J + KE_V$

3234 + 862.4 = 2167.5w² + 385.33w²

4096.4 = 2552.83w²

w² = 4096.4/2552.83

w² = 1.605

w = √1.605

w = 1.267 rad/s

5 0

2 years ago

A piece of candy has 5 calories (or 5000 calories). if it could be burned, leaving nothing but carbon dioxide and water, how muc

posledela

A) 500 calories
B)5000 joules
C) 5 kilocalories

Answer is C

8 0

2 years ago

Read 2 more answers

How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is

mafiozo [28]

1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

<h3>What is Speed ?</h3>

Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s

In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.

What are the parameters to consider ?

The parameters are;

The distance S = 3.85 × $10^{5}$ km

The Speed of Light C = 3 × $10^{8}$ m/s

The time taken t = ?

Speed = distance S ÷ Time t

Convert kilometer to meter by multiplying it by 1000

C = S/t

3 × $10^{8}$ = 3.85 × $10^{8}$ / t

Make t the subject of formula

t = 3.85 × $10^{8}$ / 3 × $10^{8}$

t = 1.2833

t = 1.3 s

Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

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1 year ago

If you are driving 90 km/hkm/h along a straight road and you look to the side for 2.8 ss , how far do you travel during this ina

pashok25 [27]

Answer:

Distance = 70 meters

Explanation:

<u>Given the following data;</u>

Speed = 90 km/h

Time = 2.8 seconds

<u>Conversion:</u>

90 km/h to meters per seconds = 90 * 1000/3600 = 90000/3600 = 25 m/s

To find the distance covered during this inattentive period;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

$Speed = \frac{distance}{time}$

Making distance the subject of formula, we have;

$Distance = speed * time$

Substituting into the above formula, we have;

$Distance = 25 * 2.8$

<em>Distance = 70 meters</em>

5 0

2 years ago