(a). Derive the mathematical formula of Newton's second law of motion. Mention the SI unit of Force.

what would the mass be of an object that was moving at a velocity of 35 m/s and has a kinetic energy of 500j be?

bearhunter [10]

Answer:

0.816 kg

Explanation:

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

so $m=\frac{2E_k}{v^2}=\frac{2\times500}{35^2}=0.816 kg$

5 0

2 years ago

Sound

Ann [662]

Longitudinal, because the sound can only travel at one direction

3 0

3 years ago

Read 2 more answers

Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro

aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

$S = \frac{Q}{T}$

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

$\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}$

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy $\Delta S_{hot}$ is smaller than $\Delta S_{cold}$

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0

3 years ago

The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.

kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

$P = 300\ N \ \times \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\
P = 300 \ N \times 22.22 \ m/s\\\\
P = 6,666 \ W$

The power of the engine at the given efficiency is calculated as follows;

$E = \frac{P_{out}}{P _{in}} \times 100\%\\\\
60\% = \frac{P_{out}}{6,666} \times 100\%\\\\
0.6 = \frac{P_{out}}{6,666} \\\\
P_{out} = 3,999.6 \ W$

Learn more about efficiency here: brainly.com/question/15418098

8 0

2 years ago

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

4 0

3 years ago