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stepladder [879]
3 years ago
8

How much work is done when a hoist lifts a 290-kg rock to a height of 7 m?

Physics
1 answer:
Zepler [3.9K]3 years ago
5 0
Given: Mass m = 290 Kg;  Height  h = 7 m

Required: Work = ?

Formula: Work = Force x distance  but, Force F = mg

W= fd

W = mgh

W= (290 Kg)(9.8 m/s²)(7 m)

W = 19,894 Kg.m²/s²

W = 19,894  J




  
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Do you think it is possible to control the magnetic properties of a magnet? Can a magnet be turned on and off?
Sunny_sXe [5.5K]

Answer:

Yes it is possible to control to some extent.

Explanation:

In general there are two types of magnets : permanent and temporary (electromagnets).

Electromagnets can be controlled since it basically depends on electricity. By switching on and off the electric supply the magnets also can be switched on and off respectively. We can also control the intensity of magnetic power.

On the other hand permanent magnet cannot be switched on and off but the magnetic properties can be altered event to an extent when it loses all its magnetic properties. It can be caused by high temperature, physical impact and also exposure to other magnetic fields. For every element there is a point of temperature called curie temperature above which the permanent magnet loses its magnetic properties. This can be brought back again by induced magnetism. The only issue is that induced magnetism work in most cases but not in all.

6 0
3 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
The chart shows data for an object moving at a constant acceleration. Which values best complete the chart? Time (s) Velocity (m
dangina [55]

Answer:

A.

x: 0

y: 0

z: 0

Explanation:

3 0
3 years ago
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Which major planet has the largest . . . A. semimajor axis? B. average orbital speed around the Sun? C. orbital period around th
Yuliya22 [10]
<h2>Mercury, Neptune, and Jupiter </h2>

Explanation:

  • Mercury has the largest semimajor axis that is 5.791 x 107 in km.
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  • Neptune has the longest orbital speed around the sun of any planet in the Solar System which is equivalent to 164.8 years (or 60,182 Earth days)
  • Jupiter has the largest eccentricity.

Hence, the answer is Mercury, Neptune, and Jupiter respectively.

6 0
3 years ago
P2O5 is a covalent compound used to purify sugar. What is the name of this compound?
AfilCa [17]

Answer:

B) Diphosphorus pentoxide

Explanation:

8 0
2 years ago
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