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bija089 [108]
3 years ago
15

How fast would a 2-kilogram object need to move to have the same kinetic

Physics
1 answer:
siniylev [52]3 years ago
8 0

Answer:

11.3 m/s

Explanation:

KE₁ = KE₂

½m₁v₁² = ½m₂v₂²

½ (2 kg) v² = ½ (4 kg) (8 m/s)²

v ≈ 11.3 m/s

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Elements that typically give up electrons
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Same answer as the first one above
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Which acceleration-time graph corresponds to the motion of the car if it moves toward the right, while slowing down at a steady
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We define acceleration as the rate of change of the velocity

Thus, if you have positive velocity and positive acceleration, your <u>speed increases.</u>

If you have positive velocity and negative acceleration, your speed decreases.

Now you get the idea, we will see that the correct option is graph 1.

We know that the car moves towards the right (let's define this as "the car has positive velocity") and we also know that te car is slowing down constantly (thus the acceleration needs to be negative and constant).

By looking at the graphs, the only one with these properties is graph 1.

If you want to learn more, you can read:

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2 years ago
Electro negativity increases when atoms __. (Apex)
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C. I took the test...........
3 0
3 years ago
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The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he
schepotkina [342]

Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

7 0
3 years ago
A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on
klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, q=1.60218\times 10^{-19}\ C

Mass of a proton, m=1.67262 \times 10^{-27}\ kg

The cyclotron frequency is given by :

f=\dfrac{qB}{2\pi m}

f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}

f = 2286785.40 Hz

or

\omega=14368296.44\ rad/s

\omega=1.43\times 10^7 rad/s

Hence, this is the required solution.

8 0
3 years ago
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