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il63 [147K]
1 year ago
5

Which of these has the most kinetic energy

Physics
2 answers:
Vesnalui [34]1 year ago
7 0
B. A 50g fish swimming in a fish tank.
garri49 [273]1 year ago
4 0

Answer:

A 50 kg boulder suspended from a cliff.

Explanation:

In this case, that totally depends on the velocity of the object, In case of car, that's just parked so the velocity is zero and a fish can't swim with higher velocity than a boulder which is suspended from a cliff.

Hope it helps!<3

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The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al
irina [24]

Answer:

a = 15.1 g

Explanation:

The relation between mass and acceleration is given by :

m\propto \dfrac{1}{a}

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂

So,

\dfrac{m_1}{m_2}=\dfrac{a_2}{a_1}\\\\a_2=\dfrac{a_1m_1}{m_2}\\\\a_2=\dfrac{0.8g\times 1510}{80}\\\\a_2=15.1g

So, the car's acceleration would be 15.1 g.

6 0
1 year ago
If you're ever standing on a mountaintop when a dark cloud passes overhead and your hair stands up, get off the mountain fast. H
OleMash [197]

Answer:

The hairs would have acquired charge by the passing of dry winds resulting in the loss of electron.

Explanation:

While standing on the top of a mountain if a person gets its hairs stand up after a cloud passes over, this might happen due to the static electric charges on the lower surface of the cloud  are opposite in nature to that of hairs which the hairs would have acquired by the passing of dry winds which would have resulted in the loss of electron from the hair tip.

Similar case happens when we rub a dry plastic ruler or a dry plastic comb on our hairs.

8 0
2 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                <em>Tₐ = 2π /ωₐ</em>

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

3 0
2 years ago
An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect
melisa1 [442]

Answer:

The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

W=\Delta K.E

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2

Here, final velocity is zero

W=0-\dfrac{1}{2}mv_{i}^2...(I)

We know that,

W=-Fd

W=-E\times e\times d

W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d...(II)

From equation (I) and (II)

-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d

Charge is negative for electron

\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}

Put the value into the formula

\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}

\lambda=-8.056\times10^{-9}\ C/m^2

Hence, The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

3 0
2 years ago
A radiator rests snugly on the floor of a room when the temperature is 10 oC. The radiator is connected to the furnace in the ba
s344n2d4d5 [400]
Answer: 10




Explanation subtract
8 0
2 years ago
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