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Deffense [45]
3 years ago
8

What is the net force on a bag pulled down by gravity with a force of 18 newtons and pulled upward by a rope with a force of 18

newtons?
Physics
1 answer:
VARVARA [1.3K]3 years ago
4 0

Answer:

0 N.

Explanation:

Force: This can be defined as the product of mass and the acceleration of the body. The S.I unit of force is Newton (N).

The expression of net force when both force act in the different direction is given as

F' = W-F ........................ Equation 1

Where F' = Net force on the bag, W = gravitational force on the bag, F = Force acting upward on the bag

Given: W = 18 N, F = 18 N.

Substitute into equation 1

F' = 18-18

F' = 0 N.

Hence the net force = 0 N.

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\frac{kQ}{r^2} r^

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3 years ago
The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr
Dvinal [7]

Answer:

T_A_v_g=9.918192559$^{\circ}C

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

T(t)=10-5*sin(\frac{\pi}{12t})

The average temperature over a given interval can be calculated as:

T_a_v_g=\frac{T_o+T_f}{2}

Where:

T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C

T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C

Hence, the average temperature between noon and midnight is:

T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C

3 0
3 years ago
Two children of about the same weight are playing at the playground. They both climb up to the top of a small tower. One slides
olya-2409 [2.1K]

Answer:

D

Explanation:

D) The overall work done by gravity is zero  

This statement is correct .

If m be the mass of each of the children and h be the height of tower

work done by gravity on the boys in going up = - mgh

it is so because force applied by gravity = mg downwards and displacement

is upwards

work done will be negative = - mgh

Work done by gravity on boys when they come down = + mgh because both force and displacement are downwards .

Hence total work done = - mgh + mgh = 0.

The children will have same kinetic energy as the inclined surface is friction-less so no energy will be dissipated hence addition of energy to boys in both the cases will be same.

4 0
4 years ago
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