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3 years ago

If the kinetic energy of a given mass is to be doubled, its speed must be multiplied by

Physics

1 answer:

Tomtit [17]3 years ago

3 0

Answer:

Either B or D. The answer itself is 2.

Explanation:

The equation for the kinetic energy would be 1/2*mv^2.

When m is doubled, we can plug in 1 and 2 to compare our answers.

Plugging in 1 for mass would give us the answer 1/2*v^2.

Plugging in 2 for mass would give us v^2. This means that the velocity was multiplied by 2, meaning that the answer is it is multiplied by 2.

I am not sure which answer is correct since there seems to be two answer choices with 2 in it, but the answer is either B or D (I will call it ABCD because I do not want to cause confusion by saying 2 multiple times).

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Which statement is true for a sound wave entering an area of warmer air?

Sergio [31]

The answer is A........The sound is transferred by collisions of molecules. Therefore sound waves will travel faster on warm air because collisions of molecules of air in warm air are greater.

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3 years ago

Read 2 more answers

Light travels 300 000 000 m/s and one year has approximately 32 000 000 second a light year is the distance light travels in one

Lelu [443]

Explanation:

It is given that,

Speed of light, $v=300 000 000\ m/s=3\times 10^8\ m/s$

Seconds in 1 year, $t=32 000 000=32\times 10^6\ s$

We need to find the distance traveled by light in one year. Speed of an object is given by :

$v=\dfrac{d}{t}$

So,

$d=v\times t\\\\d=3\times 10^8\times 32\times 10^6\\\\d=9.6\times 10^{15}\ m$

Since,

$1\ \text{light year}=9.46\times 10^{15}\ m\\\\1\ m=\dfrac{1}{9.46\times 10^{15}}\ \text{ly}\\\\9.6\times 10^{15}\ m=\dfrac{9.6\times 10^{15}}{9.46\times 10^{15}}\\\\d=1.01\ \text{ly}$

So, the distance covered by light is 1.01 light years.

8 0

3 years ago

Suppose you want to operate an ideal refrigerator with a cold temperature of − 15.5 °C , and you would like it to have a coeffic

tensa zangetsu [6.8K]

Answer:

15.65 °C

Explanation:

cold temperature (Tc) = -15.5 degree C = 273.15 - 15.5 = 257.65 kelvin

minimum coefficient of performance (η) = 8.25

find the maximum hot reservoir temperature of such a generator (Th)

η = \frac{Tc}{Th-Tc}

Th = Tc x (\frac{1}{η} + 1)

Th = 257.65 x (\frac{1}{8.25} + 1)

Th = 288.8 K

Th = 288.8 - 273.15 = 15.65 °C

8 0

3 years ago

Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x

ludmilkaskok [199]

Answer:

$\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

Explanation:

In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

$\theta_1=tan^{-1}(\frac{0.6m}{0m})=90\° \\\\\theta_2=tan^{-1}(\frac{0.4m}{1.3m})=17.10\°$

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

$\vec{E} = E_x \hat{i}+E_y\hat{j}\\\\\vec{E}=(E_1cos\theta_1-E_2cos\theta_2)\hat{i}+(E_1sin\theta_1-E_2sin\theta_2)\hat{j}\\\\E_1=k\frac{q}{r_1^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{2.75*10^{-9}C}{(0.6m)^2}=68.67\frac{N}{C}\\\\E_2=k\frac{q}{r_2^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{3.40*10^{-9}C}{((1.3m)^2+(0.4m)^2)}=16.52\frac{N}{C}$

Hence, by replacing E1 and E2 we obtain:

$\vec{E}=[(68.67N/C)cos(90\°)-(16.52N/C)cos(17.10\°)]\hat{i}+[(68.67N/C)sin(90\°)-(16.52N/C)sin(17.10\°)]\hat{j}\\\\\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

hope this helps!!

3 0

3 years ago

a piece of space junk of mass 24kg is at a distance of 7.00 x 10(power 6)m from the centre of the Earth. What is the gravitation

My name is Ann [436]

Answer:

F = 195 N

Explanation:

F = GMm/d²

F = 6.67e-11(5.97e24)(24) / (7.00e6)²

F = 195 N

7 0

3 years ago