All categories

3 years ago

Electricity is converted into other forms of _________ such as heat, sound, or light.

Physics

1 answer:

mestny [16]3 years ago

4 0

Answer:

chemical i think

I hope this helped

You might be interested in

What is the relation between centre of gravity and stability

maw [93]

Explanation:

tilting it will raise the height of its center of gravity.

8 0

3 years ago

A basketball player can jump 1.6 m off the hardwood floor. With what upward velocity did he leave the floor?

irinina [24]

First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2

= 31.36

Sq root = 5.6

3 0

3 years ago

Read 2 more answers

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ

ra1l [238]

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

$\Delta U = q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

For a proton,

$q=+e=1.6\cdot 10^{-19}C$

And since $\Delta V=100 V$

The energy gained by the proton is

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

For an alpha particle,

$q=+2e=3.2\cdot 10^{-19}C$

Therefore, the energy gained is

$\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J$

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

$q=+e=1.6\cdot 10^{-19}C$

So the energy gained is the same as the proton:

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

6 0

3 years ago

A 1-kilogram ball has a kinetic energy of 50 joules the speed of the ball is

sveta [45]

Answer:

10 m/s

Explanation:

3 0

3 years ago

A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

4 0

3 years ago