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Harlamova29_29 [7]
3 years ago
10

If a 20 gram bird and a 0.001 gram mosquito are moving at the same speed, which has the higher kinetic energy?why?

Physics
1 answer:
MissTica3 years ago
8 0

Answer:

The bird has 20,000 times higher kinetic energy than the mosquito.

Explanation:

The formula for kinetic energy shows that the amount of energy is proportional to the mass:

E_k=\frac{1}{2}mv^2

The velocity of both objects being the same, the mass is the determiner. The ratio of the kinetic energy of the bird and the mosquito then becomes the ratio of their masses, i.e., 0.02kg/0.00001kg. The bird's energy is 20,000 times higher than that of the mosquito. This proportion becomes more believable if you imagine the event of a distracted errant bird colliding with your head, versus a confused mosquito colliding with your head.

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Answer:

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hope it's helpful

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7 0
2 years ago
Presence of _______- in retina gives us night vision that is help us to see in dim light.Required to answer. Single choice. Imme
qwelly [4]

Answer: Rods

Explanation:

The rod cells in the retina are the reason we are able to see at night and in dim light. They exist on the edges of the retina which is why they are also very useful for the peripheral vision of a human.

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7 0
2 years ago
A length change 0.08 m will occur for an object that is L= 56 m long. If the coefficient of thermal expansion is5.3 x 10 /C and
ArbitrLikvidat [17]

Answer:

Increase in temperature =  269.54 °C

Explanation:

We have equation for thermal expansion

          ΔL = LαΔT

Change in length, ΔL = 0.08 m

Length, L = 56 m

Coefficient of thermal expansion, α = 5.3 x 10⁻⁶ °C⁻1

Change in temperature, ΔT = T - 253

Substituting

          0.08 = 56 x 5.3 x 10⁻⁶ x (T - 253)

         (T - 253) = 269.54

           T = 522.54 °C

Increase in temperature =  269.54 °C      

4 0
2 years ago
A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F
Marrrta [24]

Answer:

Part a)

x = 0.4 m

Part b)

v_i = 11.7 m/s

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

\frac{1}{2}mv^2 = \frac{1}{2}kx^2

\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2

x = 0.4 m

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

W_f = -7( \pi R)

W_f = -7(\pi \times 1.00)

W_f = -22 J

work done against gravity is given as

W_g = - mg(2R)

W_g = -(0.500)(9.8)(2\times 1)

W_g = -9.8 J

Now by work energy equation we have

\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2

\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2

v_f = 4.1 m/s

Part c)

now minimum speed required at the top is such that the normal force must be zero

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = 3.13 m/s

so here we got speed more than the required speed so it will reach the top

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3 years ago
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Nat2105 [25]

Answer: Potassium iodide

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2 years ago
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