All categories

3 years ago

1) mechanical

1 answer:

7 0

I am pretty sure about these answers.Thermal goes in the 4th blank.
Mechanical goes in the 2nd blank.
Electrical goes in the 3rd blank.
I think chemical goes in the 1st blank and light goes in the 5th blank
Hope this helps

You might be interested in

In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a

Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m² ------------------------- (2)

from (1) and (2)

B²q²R² / m² = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V

7 0

3 years ago

How much energy to be removed to drop the temperature of 5.7kg of wood from 20degrees to 7degrees. #100points

Neporo4naja [7]

see i was trying to figure out the answer but i didn't understand it so i took the time to research and work it out but i still didn't understand i found one that was close to it and i got the same one as the other person which is D but idk if it is that type of question if it is than it is d if not then idk

5 0

3 years ago

Read 2 more answers

Why centre of mass equal to centre of gravity

Maksim231197 [3]

Because mass and distance determine gravity, so the more mass you have, the more gravity.

4 0

3 years ago

Read 2 more answers

Benjamin Franklin has convinced his hapless assistant Mike Piepan to participate in an experiment on electiricty. Ben has set up

ad-work [718]

Answer:

Mike will receive an electric shock

Explanation:

Human body is a conductor of electricity. When lightning strikes rod and give it negative charge, the rod will dissipate its charge as soon as it comes in contact with earth via conduting material. Mike will receive a severe electric shock as negatice charge pass through his body to other rod and to the ground.His body will feel numb. He may also get unconscious.

5 0

3 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago