Occurs when an object's velocity decreases

12. your friend with great excitement tells you about his newest idea to solve the energy crisis: he wants to use an electromoto

mixas84 [53]

I would tell him, in the kindest, most gentle way I could manage,
to fahgeddaboudit.

The total amount of energy doesn't change. Energy is never created,
and it never disappears. If you have some energy, then it had to come
from somewhere, and if you used some energy, then it had to go
somewhere.

You can never get more energy out of the electromotor than you put into it,
and in the real world, you can't even get THAT much out, because some
of it is always used on the way through.

Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
"Free Energy" on the internet, relax, and enjoy the show. They are all
fakes. They may not all be intentionally meant to fool you, but they are
all impossible.

4 0

3 years ago

Read 2 more answers

The momentum of a 5-kilogram object moving at 6 meters per second is-

goldfiish [28.3K]

I think the answer is 30 but I’m not sure

3 0

3 years ago

Explain: a camel walks easily on sand surface than a man inspite of the fact that a camel is much heavier than a man

Usimov [2.4K]

Camels have friction reduced on their feet making it easier to walk.

4 0

3 years ago

A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte

Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

$m_1u_1+m_2u_2=(m_1+m_2)v..................(1)$

were v is their common velocity after impact. If the second mass $m_2$ was at rest before the impact, then its initial velocity $u_2=0m/s$ . therefore $m_2u_2=0$ . Equation (1) then becomes;

$m_1u_1=(m_1+m_2)v..............(2)$

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

$m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?$

Substituting these values into (2), we get the following;

$0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)$

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

$R= vt............(4)$

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

$h=\frac{1}{2}gt^2...........(5)$

were g is acceleration due to gravity taken as $9.8m/s^2$ . Therefore;

$1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s$

We then substitute R and t into equation (4) to obtain v.

$3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s$

We now further substitute this value of v into (3) to obtain $u_1$ ;

$u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s$

4 0

4 years ago

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

IrinaVladis [17]

Answer:

The net gravitational force on the mass is $1.27\times 10^{-5}$

Explanation:

We have by Newton's law of gravity the force of attraction between masses $m_{1},m_{2}$

$F_{att}=G\frac{m_{1}m_{2}}{r^{2}}$

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

$F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N$

Force of attraction between 435 kg mass and 38 kg mass is

$F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N$

Thus the net force on mass 38.0 kg is $F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}$

8 0

3 years ago