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viktelen [127]
2 years ago
12

How high must you lift a 25 Newton book for it to have the same increase in potential energy as a 20 Newton book that was lifted

to 0.5 meters?
Physics
1 answer:
Annette [7]2 years ago
4 0

Given :

An object with weight 20 N was lifted to 0.5 meters.

To Find :

How high must you lift a 25 Newton book for it to have the same increase in potential energy as the given book.

Solution :

Since both have same potential energy :

P.E_2 = P.E_1\\\\W_2h_2 = W_1h_1

Putting all given values in above equation :

25h_2 = 20\times 0.5\\\\h_2 = \dfrac{20\times 0.5}{25}\\\\h_2 = 0.4\ m

Therefore, book with same potential energy is at a height of 0.4 m.

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How much work is done when you lift a 5kg bag of groceries 1.5 m?
olga55 [171]

Answer:

7.5J

Explanation:

5x1.5=7.5

8 0
3 years ago
What are the seven types of electromagnetic waves?
Sunny_sXe [5.5K]
Radio waves. Giant satellite-dish antennas pick up long-wavelength, high-frequency radio waves. ...
Microwaves. Because cosmic microwaves can't get through the whole of Earth's atmosphere, we have to study them from space. ...
Infrared. ...
Visible light. ...
Ultraviolet light. ...
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5 0
3 years ago
An ice chest at a beach party contains 12 cans of soda at 4.05 °C. Each can of soda has a mass of 0.35 kg and a specific heat ca
BARSIC [14]
Below is the solution:

Heat soda=heat melon 
<span>m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater </span>
<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
<span>15960(t-5)=27300(27-t) </span>
<span>15960t-136500=737100-27300t </span>
<span>43260t=873600 </span>
<span>t=873600/43260 </span>
<span>t=20.19 deg celcius</span>
4 0
3 years ago
Pleaseeeee HELPPPP THIS IS TIMED ALSO,
TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

8 0
2 years ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0
3 years ago
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