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jeka94
3 years ago
15

You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use

Physics
1 answer:
Yuri [45]3 years ago
3 0
P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]
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A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac
Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

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energy (work) of friction 352.8 - 250 = 102.8 J

8 0
3 years ago
Which provides the best analogy for an electron in an atomic orbital?
Svetllana [295]
<h2>Answer: a bee trying to escape from a closed jar  </h2>

In an atom the electrons will occupy orbitals so that their energy is as small as possible. That is why the orbitals are ordered based on their energy level in an increasing order, which is associated with a particular range of energy based on its distance from the atom nucleus.

In this sense, an electron "jumps" from one level to another in the atom in the same way a bee tries to escape from a closed jar.

3 0
3 years ago
According to Coulomb’s Law, the force between two charged objects is related to _____.
Natasha_Volkova [10]

Answer:

A.) the inverse of the square of the distance separating them

Explanation:

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Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

5 0
3 years ago
A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
3 years ago
"how do the fundamental laws of physics make manifest that space has 3 dimensions?"
bazaltina [42]
If you are asking for a proof on having at least 3 dimensions in space, you can find the physical proof anywhere in your daily life activities. Just the fact that solids have volumes is a proof already that we live in a three-dimensional space. We can move forwards, backwards, sidewards and in all other directions possible.

When you go right into detail, the fundamental laws governing these proofs are very technical. They have differential equations to show as proof. It is too detailed to discuss here. The important things is that, these fundamental laws are what explains the science in our basic activities and natural phenomena:

*Gravitation and planetary motion
* Translation, rotation, magnetic field, forces
* Integrals of equations: 
8 0
3 years ago
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