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3 years ago

ANSWER ASAP AND WILL GIVE BRAINLIEST AND POINTS

Physics

2 answers:

denis-greek [22]3 years ago

8 0

Answer:

Explanation:

Mars is the planet between Earth and Jupiter, or the 4th planet from the Sun.

34kurt3 years ago

5 0

Answer:

D is the correct answer

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What is the connection between electrons and protons

TEA [102]

Answer:

Electrons and protons are connected with one another in term of charge and size.

Explanation:

The size of electron and proton is always same in any atom but they possess opposite charge.
Electron in any atom carries negative charge where as proton carries the positive charge.
In any neutral atom the charge between electron and proton is balanced along with the size.
The nucleus of any atom bounds only proton and neutron but the electron is present revolving around the nucleus.

7 0

3 years ago

What is resonance in sound

Sveta_85 [38]

Answer: Resonance in sound is when one object is vibrating at the same frequency to the second object of forces to the second frequency.

Explanation:

"Acoustic resonance is a phenomenon in which an acoustic system amplifies sound waves whose frequency matches one of its own natural frequencies of vibration (its resonance frequencies)." wikipedia I hope this helps you!

6 0

3 years ago

A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on

klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, $q=1.60218\times 10^{-19}\ C$

Mass of a proton, $m=1.67262 \times 10^{-27}\ kg$

The cyclotron frequency is given by :

$f=\dfrac{qB}{2\pi m}$

$f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}$

f = 2286785.40 Hz

$\omega=14368296.44\ rad/s$

$\omega=1.43\times 10^7 rad/s$

Hence, this is the required solution.

8 0

3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago

A cat leaps to catch a bird. if the cat's jump was at 60.0° off the ground and its initial velocity was 2.74 m/s, what is the hi

user100 [1]

Answer:

0.29 m

Explanation:

7 0

3 years ago