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Aliun [14]
3 years ago
14

Can you help quick in science please

Physics
1 answer:
nikdorinn [45]3 years ago
4 0
Uhhhh...you should have paid attention in class, just saying...
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A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati
slega [8]

Answer:

29223.6J

Explanation:

Given parameters:

Mass of Piano = 852kg

Height of lifting  = 3.5m

Unknown:

Gravitational potential energy  = ?

Solution:

The gravitational potential energy of a body can be expressed as the energy due to the position of a body;

  G.P.E  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

 Now insert the given parameters and solve;

   G.P.E  = 852 x 9.8 x 3.5 = 29223.6J

6 0
2 years ago
Estimate the power you produce in running up a flight of stairs. Give your answer in horsepower (1 hp = 746 W). Suppose you clim
GarryVolchara [31]
The first thing you should do is calculate the work done when climbing the stairs. This work by definition will be given by:
 W = F * d
 W = (m * g) * (d)
 W = ((71) * (9.8)) * (3) = 2087.4J
 Then, you can calculate the power that in this case is given by
 P = W / t
 P = (2087.4) / (10) = 208.74W
 To have the result in HP we use the fact that 1HP = 746W
 P = (208.74) / (746)
 P = 0.28 HP
 answer
 the power you produce in running up a flight of stairs is 0.28 HP
3 0
3 years ago
Read 2 more answers
I need help with number 6 and 7 please and I’ll mark brainliest
arsen [322]

Answer:

n.6 is T

Explanation:

because mass always stays the same where ever you are but weight changes depending on the gravity

3 0
3 years ago
Read 2 more answers
A hydrogen atom has its electron in the n = 6 level. the radius of the electron's orbit in the bohr model is 1.905 nm.
gayaneshka [121]
Are there any options or is it not multiple choice.
5 0
3 years ago
When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to
Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

F=k\frac{q_1 q_2}{d^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

d=\sqrt{k\frac{q_1 q_2}{F}}

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}

so the correct option is E.

8 0
3 years ago
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