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posledela
3 years ago
12

HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!​

Physics
2 answers:
statuscvo [17]3 years ago
8 0

Answer:

2500j/s

Explanation:

sleet_krkn [62]3 years ago
3 0

Answer:

a. 25000 J

b. 2500 J/s

Explanation:

Given,

Distance ( s ) = 50 m

Force ( f ) = 500 N

a.

To find : -

Work done ( W ) = ?

Formula : -

W = fs

W

= 500 x 50

= 25000 J

Therefore,

the work done by the force the horse exerts is

25000 J.

b.

To find : -

Power ( P ) = ?

Formula : -

W = Pt

P = W / t

P

= 25000 / 10

= 2500 J/s

Therefore,

the power produced if the movement took 10 s

is 2500 J/s.

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Coasting due west on your bicycle at 8 m/s, you encounter a sandy patch of road 7.2m
jeka57 [31]

Answer:

V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s     average speed

t = 7.2 / 7.25 = .993 sec      time to cross patch

a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2     or 1.5 m/s^2

8 0
3 years ago
Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso
TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

3 0
3 years ago
19. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to
Aleks [24]

Recall this gas law:

\frac{P₁V₁}{T₁} = \frac{P₂V}{T₂}

P₁ and P₂ are the initial and final pressures.

V₁ and V₂ are the initial and final volumes.

T₁ and T₂ are the initial and final temperatures.

Given values:

P₁ = 475kPa

V₁ = 4m³, V₂ = 6.5m³

T₁ = 290K, T₂ = 277K

Substitute the terms in the equation with the given values and solve for Pf:

\frac{475*4}{290} = \frac{P₂*6.5}{277}

<h3>P₂ = 279.2kPa</h3>
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3 years ago
If we take away a lot of energy from a liquid what phase will it change to ?
DanielleElmas [232]

The liquid would change to gas

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All except the first one. I just took the test.
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