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3 years ago

If speed quadruples from 11 m/s to 44 m/s, what happens to the kinetic energy?

Physics

1 answer:

Ad libitum [116K]3 years ago

5 0

Answer:

16 fold increase.

Explanation:

As kinetic energy is a function of velocity squared, if the velocity quadruples, kinetic energy increases by a factor of 4² = 16

You need to <u>do</u> it in less than an hour because it is <u>due</u> soon.

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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

Electrons are added into the outermost what in groups 1 & 2

aivan3 [116]

S orbital.

Group 1 elements have a general configuration $ns^{1}$ , where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration $2s^{1}$ whereas Cesium has $6s^{1}$ . Both of them belong to Group 1 of Periodic Table.

Group 2 elements have a general configuration of $ns^{2}$ . For example, Magnesium has $3s^{2}$ as its outer shell configuration while Strontium has the same as $5s^{2}$ .

We see that in both the cases, the outermost S orbital is being filled.

3 0

3 years ago

Can someone explain to me #4.

nalin [4]

A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.

4 0

3 years ago

PLEASE HELP, PLEASE A CORRECT ANSWER!

LUCKY_DIMON [66]

Answer: I like your profile picture

Explanation:

6 0

3 years ago

Jaden has a mass of 45 kilograms on Earth. Jupiter has more gravity than the Earth. On Jupiter, Jaden's mass will be O more than

Dovator [93]

Answer:

m = 45 kg

Explanation:

Given that,

Mass of Jadan, m = 45 kg on Earth

Jupiter has more gravity than the Earth.

Mass of an object is the amount of matter contained inside an object. We need to tell about the mass of Jaden on Jupiter. The mass of the object remains same everywhere.It does not change in any of the location.

Hence, Jaden's mass will be 45 kg on Jupiter.

7 0

2 years ago