Write and solve a MA or efficiency problem

A boy throws a ball of mass 0.22 kg straight upward with an initial speed of 29 m/s. When the ball returns to the boy, its speed

maksim [4K]

Answer:

The work is -67.76 J

Explanation:

The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.

This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.

In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.

So, the loss of kinetic energy is $\frac{1}{2} *m*(vf^{2} -vi^{2} )$

You know:

mass=m=0.22 kg
Initial velocity of the ball: $vi= 29 \frac{m}{s}$

Final velocity of the ball: $vf= 15 \frac{m}{s}$

Replacing:

$\frac{1}{2} *0.22 kg*(15^{2} -29^{2} )$ = -67.76 J

Friction work is always negative because friction is always against displacement.

5 0

3 years ago

7. How much work is done in moving a charge of 10 micro coulombs 1 meter along an equipotential of 10 volts?

Neko [114]

It takes work to push charge through a change of potential.
There's no change of potential along an equipotential path,
so that path doesn't require any work.

8 0

3 years ago

Read 2 more answers

Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

b)

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago

Are chemical reactions necessary in life

Salsk061 [2.6K]

Yes they are necessary

7 0

3 years ago

Networks of interconnected wireless devices that are embedded into the physical environment to provide measurements of many poin

bazaltina [42]

Wireless sensor networks

Explanation:

Networks of interconnected wireless devices that are embedded into the physical environment to provide measurements of many points over a large spaces are called Wireless sensor networks.

They are very useful in obtaining real-time data and information about every day life.

The internet of things greatly relies on the use of wireless sensor networks in devices and gadgets to better and improve life.
They are constantly in use by various organizations and bodies.
Wireless sensor networks can be designed to collect specific scientific data or even more.

learn more:

Connecting IoT devices brainly.com/question/11028028

#learnwithBrainly

6 0

3 years ago