All categories

3 years ago

40 POINTS I’ve been stuck on this for hours please help :(

Physics

2 answers:

kondaur [170]3 years ago

8 0

.................................. uh ya

Norma-Jean [14]3 years ago

7 0

Answer:

Explanation:

Particle Relative Mass Charge Location

Proton 1 + Nucleus

Electron 0 - Outside rings (not nucleus)

Neutron. 1 No Charge Nucleus

You might be interested in

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is

zysi [14]

Answer:

$q_2=2.47\times 10^{-4}\ C$

Explanation:

The charge on one object, $q_1=9.9\times 10^{-5}\ C$

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C$

So, the charge on the other sphere is $2.47\times 10^{-4}\ C$ .

7 0

3 years ago

Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified res

VLD [36.1K]

Answer:

$R_{eq}$ = 19 Ω, I = 0.105 A, V1 = 1.05 V and V2 = 0.95 V

Explanation:

The correct way to solve this type of problem is to find the current or voltage values for the equivalent resistance and from here find the other values.

For a series circuit the equivalent resistance is the sum of the resistance

$R_{eq}$ = R1 + R2

$R_{eq}$ = 10 +9

$R_{eq}$ = 19 Ω

Let's use the equation for the voltage

V = I $R_{eq}$

I = V / $R_{eq}$

I = 2/19

I = 0.105 A

In a series circuit the current is constant, so let's use the voltage equation for each resistor

V1 = I R1

V1 = 0.105 10