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OleMash [197]
2 years ago
10

A car of weight 300N moved through a distance of 10m when pushed by 3 students

Physics
1 answer:
Sati [7]2 years ago
3 0

The work done by the three students is 3,000 J.

The energy transferred in the process is 3,000 J.

<h3>What is work done?</h3>
  • Work done is the product of force and distance moved by the object.

W = Fd

The work done by the three students is calculated as follows;

W = 300 x 10

W = 3,000 J

<h3>What is energy transfer?</h3>
  • This is means by which energy is converted from one form to another.

The energy transferred in the process is determined by work energy theorem.

E = W

E = 3,000 J

Learn more about work-energy theorem here: brainly.com/question/22236101

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pochemuha

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A 3 kg object is attached to a 1000 N/m spring. The spring is compressed 0.10 m and then the spring launches the object horizont
Viktor [21]

Answer:

1.826m/s

Explanation:

E=1/2*k*(∆L)^2

E=1/2*mV^2

6 0
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A 2.3 kg cart is rolling across a frictionless, horizontal track towards a 1.5 kg cart that is initially held at rest. The carts
Inga [223]

Answer:

total momentum = 8.42 kgm/s

velocity of the first cart is 3.660 m/s

Explanation:

Given data

mass m1 = 2.3 kg

mass m2 = 1.5 kg

final velocity V2 = 4.9 m/s

final velocity V3 = - 1.9 m/s

to find out

total momentum  and velocity of the first cart

solution

we know mass and final velocty

and initial velocity of second cart V1 = 0

so now we can calculate total momentum that is m1 v2 + m2 v2

total momentum =  2.3 ×4.9 + 1.5 ×(-1.9)

total momentum = 8.42 kgm/s

and

conservation of momentum  is

m1 V + m2 v1  = m1 v2  + m2 v3

put all value and find V

2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)

V = 8.42 / 2.3

V = 3.660 m/s

so velocity of the first cart is 3.660 m/s

8 0
3 years ago
According to the Big Bang Theory, the density and temperature of the Universe is ___________ now than in the past.
34kurt

Answer:

According to the Big Bang Theory, the density and temperature of the Universe is <u>lower</u> now than in the past.

<em>Hope</em><em> this</em><em> answer</em><em> correct</em><em> </em><em>:</em><em>)</em>

5 0
3 years ago
Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila
Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

or

\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}

substituting the values in the above question we get

\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}

or

\frac{A_1}{A_2} }=0.3659

6 0
3 years ago
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