Ask question

Ask question

All categories

2 years ago

5

A car weighs 1323 N. what is its mass?

2 answers:

Shalnov [3]2 years ago

6 0

If the car is on the moon, its mass is about 817 kg.

If it's on the Earth, its mass is about 135 kg.

vesna_86 [32]2 years ago

6 0

Answer:

Mass of the car, m = 135 kg

Explanation:

The weight of the car is 1323 N. Let m is its mass. The relation between the mass and weight comes form Newton's second law of motion. We know that the weight of an object is the force that is exerted due to acceleration due to gravity.

The mathematical form of weight of an object is given by :

W = mg

g is the acceleration due to gravity

$m=\dfrac{W}{g}$

$m=\dfrac{1323\ N}{9.8\ m/s^2}$

m = 135 kg

So, the mass of the car is 135 kg. Hence, this is the required solution.

You might be interested in

A car engine burns gas at 495 K, and exhausts to the air at 293 K. If it ran at the highest possible efficiency, how much input

slamgirl [31]

The heat taken to do the 10,000 J of work is 4081 J

<u>Explanation:</u>

The ratio of the temperature difference between the reservoirs and the temperature of the hot reservoir is termed as maximum thermal efficiency.

Maximum thermal efficiency is

1 - (Tc / Th).

where Tc represents the absolute temperature of the cold reservoir

Th represents the absolute temperature of the hot reservoir.

1 - (293/495) = 0.408

If the input is 10,000 J, then 4081 J is used to do work. The remaining 5919 J is rejected as heat into the air.

4 0

2 years ago

How to calculate the average value of the indirect measurements of KE

Tom [10]

Answer:

1/2 m*(2(h)/t)*g

Explanation:

8 0

2 years ago

Fredrick did an investigation on mass and acceleration. He applied a force to a toy car and measured the distance it moved. He t

garik1379 [7]

Explanation:

B. More mass results in less acceleration.

6 0

2 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

<u>From the Fourier's law of conduction:</u>

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&<u>For Styrofoam-plywood interface from inside:</u>

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

2 years ago

How much power does it take to do 500 J of work in 10 seconds?

Dmitry_Shevchenko [17]

Power = work/time

= 500/10

= 50J/s or 50 watt

7 0

2 years ago