All categories

2 years ago

A box of mass 21 kg sits on an inclined surface with an angle of 10°. What is

Physics

2 answers:

N76 [4]2 years ago

7 0

36 N is the correct answer

alexandr1967 [171]2 years ago

4 0

The component of the weight of the box along the surface is 34.9N.

<u>Explanation:</u>

Given that

Mass of the box m=21kg

Angle of inclination θ=10°

Acceleration due to gravity g=9.8 $m/s^{2}$

Weight of an object is the force with which the earth attracts the object. Since weight is a force its unit is Newton(N). When an object is placed on a surface that has no inclination, its weight is given by the expression

W=mg

where m denotes mass of the object and g denotes the acceleration due to gravity.

When the surface has an inclination, the weight of the object can be resolved into two components that are parallel and perpendicular to the surface.

The component of weight parallel to the surface is given by

W=mgsinθ

where θ is the angle of inclination

The component of weight perpendicular to the surface is given by

W=mgcosθ

In the given problem, we have to find out the parallel component of weight

W=mgsinθ

$W=21\times 9.8 \timessin10\°\\=21\times9.8\times0.17\\=34.9N$

The component of the weight of the box along the surface is 34.9N

You might be interested in

PLEASE HELP ME ALREADY!!! MAX POINTS RANDOM ANSWERS WILL BE REPORTED!!!

horsena [70]

Answer:

C. Energy is used for life processes (e.g. Movement, breathing, etc.)

Explanation:

As we pass from one trophic level to the next, only 10% of energy is transferred from the first trophic level to the next. This is because a lot of energy is lost to the surroundings and rest is utilised by the organism.

3 0

1 year ago

Read 2 more answers

The solar system is made up of eight planets, numerous comets, asteroids and moons, and the Sun. The force that holds all of the

s2008m [1.1K]

Your answer is Gravity

Hope this could help :)

5 0

2 years ago

A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its

larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

$F_n_e_t=ma$

Plug in what you know, and solve:

$18=m(12)\\m=1.5kg$

Find matching soluation:

C.) 1.5 kg

5 0

2 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

2 years ago

What happens to a theory as new evidence is found?

algol13

New evidence may support the theory -> then nothing

New evidence conflicts with the theory => rework theory / create a new one

6 0

2 years ago