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2 years ago

PLEASE HELP! ASAP

Physics

2 answers:

hoa [83]2 years ago

8 0

Explanation:

Specifically, small amounts of mass are turned into energy from the breaking up (fission) or combination (fusion) of the nuclei of atoms. Even spontaneous radioactive decay converts a bit of mass into incredible amounts of energy.

Damm [24]2 years ago

5 0

Answer:

A. Uranium breaks apart into krypton, barium, and three neutrons.

Explanation:

did the test !!

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The resistivity of gold is 2.44 x 10-8 ohms.m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries

harkovskaia [24]

Answer:

E=0.036 V/m

Explanation:

Given that

Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m

d= 0.9 mm

L= 14 cm

I = 940 m A = 0.94 A

We know that electric field E

E= V/L

V= I R

R=ρL/A

So we can say that

E= ρI/A

Now by putting the values

$E=\dfrac{ 2.44\times 10^{-8}\times 0.94}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}$

E=0.036 V/m

6 0

2 years ago

A sinusoidal traveling wave is generated on a string by an oscillating source that completes 116 cycles per minute. What is the

slava [35]

Answer:

$\lambda =6.32\ cm$

Explanation:

given,

number of cycle complete (f) = 116 cycles per minute

wavelength observed at 11 m in 1.5 m.

$v = \dfrac{distance}{time}$

$v = \dfrac{11}{1.5}$

v = 7.33 m/s

$\lambda = \dfrac{v}{f}$

$\lambda = \dfrac{7.33}{116}$

$\lambda =0.0632\ m$

$\lambda =6.32\ cm$

The wavelength of the wave is equal to $\lambda =6.32\ cm$

6 0

2 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

2 years ago

Leia is presenting her project about gravity to her class. She says that both mass and distance affect how strong the gravitatio

DaniilM [7]

Answer:

No, because as distance increases, gravitional force decreases.

5 0

3 years ago

How much pressure is created when you apply a

Alekssandra [29.7K]

Pressure
= Force/Area

Area = π(d^2)/4
= π(0.4^2)/4
=0.126 m2
Pressure
= 50/0.126
= 396.825 Pa

5 0

1 year ago