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S_A_V [24]
3 years ago
15

Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth

?
Physics
1 answer:
zimovet [89]3 years ago
6 0

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

The work done on an object is the force applied to lift a body in a specific direction.

   Work done = force x distance

  Weight is a force in the presence of gravity;

  Work done = weight x height of lifting

Height of lifting = \frac{work done }{weight}

 Height of lifting = \frac{250}{50} = 5m

The vertical height through which the object was lifted is 5m

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

distance ( d ) = 11.6 meter                                       ( m = mass of the car )

Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0
3 years ago
What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

6 0
3 years ago
A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res
VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

KE = PE

\frac{1}{2} mv^2 = mgh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(2(55.8*10^{-2}))}

v = 4.67m/s

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

4 0
3 years ago
A boy on rollerskates is travelling along at 8 m/s. He has a mass of 60 kg and is carrying his
Brrunno [24]

Answer:

6m/s

Explanation:

the original momentum = mass x velocity = 8x (60+10) = 560

momentum after = mass x velocity of the school bag + mass x velocity of the boy = 10x20 + 60x A

200+60A = 560

A=6

5 0
3 years ago
rs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she use to produce the
anyanavicka [17]

Answer:

Mg and HCl

Explanation:

Here, we want to get the combination that could be used in the production of the needed hydrogen

An important chemical property of inorganic acids is that when they react with metals, they give off hydrogen gas in conjunction with the formation of a salt

HCl is a mineral acid while Mg is a metallic substance

So the reaction between this metal and the mineral acids will give the needed hydrogen gas to be produced

4 0
3 years ago
Read 2 more answers
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