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2 years ago

What is the energy equivalent of an object with a mass of 2.5 kg?

Physics

2 answers:

Bad White [126]2 years ago

8 0

This is the answer on Edgenuity 2.25*10^17

N76 [4]2 years ago

5 0

Answer : Energy equivalent of an object with a mass of 2.5 kg is $E=2.25\times 10^{17}\ J$ .

Explanation :

It is given that mass of an object is, m = 2.5 kg

It is required to find energy equivalence.

According to mass energy equivalence relation :

$E=mc^2$

Where,

m is the mas of an object.

c is the speed of light $c=3\times 10^8\ m/s$

$E=2.5\ kg\times (3\times 10^8)^2\ m/s$

$E=22.5\times 10^{16}\ J$

$E=2.25\times 10^{17}\ J$

Hence, this is the required solution.

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In the diagram below, a heavy box is being lifted. What is the function of the pulley in this situation?

zepelin [54]

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2 years ago

Read 2 more answers

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

3 0

2 years ago

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t=0.030 s$ is the time interval

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

$v=281 km/h =78.1 m/s$ is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N$

7 0

2 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

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2 years ago

9. (02.04 LC)

ycow [4]

Answer:

Molecules speed up

Explanation:

This is caused because of the temperature increasing. The temperature increase is telling us that the thermal energy of the reaction is increasing. When the energy is increased molecules increase their speed, because they have more energy in them

7 0

3 years ago