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2 years ago

How much work does a 120 watt motor perform for 300 seconds?

Physics

1 answer:

MrRa [10]2 years ago

6 0

Answer:

36,000j

Explanation:

W= P•t

120•300= 36000

I checked my answer with P= W/t and got the right answer

hope this helps

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A 15.0 kg fish swimming at 1.10 m/s suddenly gobbles up a 4.50 kg fish that is initially stationary. Ignore any drag effects of

Nana76 [90]

Answer

given.

Mass of big fish = 15 Kg

speed of big fish = 1.10 m/s

mass of the small fish = 4.50 Kg

speed of the fish after eating small fish =?

a) using conservation of momentum

m₁v₁ + m₂v₂ = (m₁+m₂) V

15 x 1.10 + 4.50 x 0 = (15 + 4.5)V

16.5 = 19.5 V

V = 0.846 m/s

b) Kinetic energy before collision

$KE_1 = \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2$

$KE_1 = \dfrac{1}{2}\times 15 \times 1.1^2 + \dfrac{1}{2}m_2\times 0^2$

KE₁ = 9.075 J

Kinetic energy after collision

$KE_2= \dfrac{1}{2}(15+4.5)\times 0.846^2$

KE₂ = 6.98 J

Change in KE = 6.98 - 9.075 = -2.096 J

hence,

mechanical energy was dissipated during this meal = -2.096 J

5 0

3 years ago

The side length of a square is modeled by s=A, where A is the area of the square. Graph the function. What is the side length of

Naya [18.7K]

Answer:

√35 square units

Explanation:

Area of square = 35square units

Formula of area of square is a =s*s where a is area and s is side

So,

s*s = 35

s^2= 35

s = √35

tell me if this is the answer u r looking for

3 0

2 years ago

Apply: The earth's gravity is pulling on you. Are you pulling on the earth? Explain your

coldgirl [10]

Answer:

Yes, you are pulling on Earth. Reasoning. Third Newton's law of motion, action and reaction law, sates that for every action force, there is an equal (in magnitude) and opposite reaction force.

Explanation:

goo gle.

5 0

3 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$