If it takes 100 N to move a box 5 meters, what is the work done on the box?

What happen to the weight of a body when it is falling freely under the action of gravity?

mr Goodwill [35]

Answer:

Explanation:

where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object.

6 0

3 years ago

How much work must I do to assemble a charge distribution consisting of three point charges of -1.00 nC, 2.00nC, and 3.00 nC, lo

leonid [27]

Answer:

$W_{total}=4.5*10^{-8}J$

Explanation:

Remember that electric potential can be written as:

$V=\frac{kQ}{r}$ ,

Where V is the electric potential, k is Coulomb's constant, Q is a point charge, and r is the distance from the point charge. Also, we can write the electric potential as:

$V=\frac{W}{q}$ ,

where W is the work made to move a charge from infinitely far apart to a certain distance, and q the point charge were are moving.

From all this we can get an expression for the work:

$W=\frac{kQq}{r}$

We are going to let

$q_{1}=-1.00nC\\q_{2}=2.00nC\\q_{3}=3.00nC$

To take the first charge $q_{1}$ from infinitely far apart to one of the vertices of the triangle, since there is no electric field and charges, we make no work.

Next, we will move $q_{2}$ . Now, $q_{1}$ is a vertice of the triangle, and we want $q_{2}$ to be 20.cm apart from $q_{1 }$ so the work we need to do is

$W_{12}=\frac{kq_{1}q_{2}}{(0.20)}\\\\W_{12}=\frac{(9*10^{9})(-1*10^{-9})(2*10^{-9})}{0.20}\\\\W_{12}=-9*10^{-8}J$

Now, we move the last point charge. Here, we need to take in account the electric potential due to $q_{1}$ and $q_{2}$ . So

$W=W_{13}+W_{23}\\\\\\W=\frac{kq_{1}q_{3}}{0.20}+\frac{kq_{2}q_{3}}{0.20}\\\\W=q_{3}k(\frac{q_{1}}{0.20}+\frac{q_{2}}{0.20})\\\\W=(3*10^{-9})(9*10^{9})(\frac{-1*10^{-9}}{0.20}+\frac{2*10^{-9}}{0.20})\\\\$

$W=1.35*10^{-7}J$

Now, the only thing left to do is to find the total work, this can be easily done by adding $W_{12}$ and $W$ :

$W_{total}=W_{12}+W\\W_{total}=1.35*10^{-7}-9*10^{-8}\\W_{total}=4.5*10^{-8}J$

8 0

3 years ago

1. At which point is the vertical velocity equal to zero?

Alex17521 [72]

#1

in projectile motion when object reached to the highest point then its velocity in vertical direction becomes zero

So its velocity in vertical direction is zero at POINT B

#2

During first half part of motion ball goes up high in air so its vertical speed will decrease due to gravity

while when it will drop back in next half where its vertical speed will increase

So its speed will increase between point B and C

7 0

3 years ago

The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at

Volgvan

The working equation to be used here is written below:

Q = kA(T₁ - T₂)/Δx
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area of the wall
T₁ - T₂ is the temperature difference between the sides of the wall
Δx is the thickness of the wall

The solution is as follows:

Q = (0.69 W/m²·°C)(5 m × 6 m)(50°C - 20°C)/(30 cm * 1 m/100 cm)
Q = 2,070 W/m

4 0

3 years ago

What are 2 systems that work with the muscle system? and explain how they work.

Leto [7]

The muscular system is responsible for controlling the skeletal muscles and the muscles of the internal organs, including the heart, stomach and intestines. It connects with the nervous system and receives input from efferent neurons, allowing it to respond to external or internal stimuli.

6 0

3 years ago