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2 years ago

As the distance between two charged objects increases, the strength of the electrical force between the objects

1 answer:

7 0

As the distance between two charged objects increases, the strength of the electrical force between the objects decreases.

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What is the range of a ball thrown horizontally at 12 m/s if its time of flight is 3.0 s?

vladimir1956 [14]

Answer:

Explanation:

because I did this assignment, :) your welcome

Next time do it by yourself, but here's the answer kid

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2 years ago

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The dark bands on the wall are from ______________ interference. and i will give brainlyest

Fittoniya [83]

Light can be seen as an electromagnetic wave.
What happens when two waves, with the same frequency, superpose is called interference.

If at a certain point two waves arrive both with a crest, we have constructive interference and the amplitudes sum up, reaching the maximum value, resulting in bright spots.

If at a certain point one of the waves arrives with a crest and the other wave arrives with a trough, we have destructive interference, and the two amplitudes cancel out, resulting in dark spots.

Therefore, the dark bands on the wall are from destructive interference.

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3 years ago

A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calcul

algol [13]

Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= Q

Mk ×DTk

=250 × 9 × 5

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = 25×10 ×100

1.082

=2500

1.082

= 23105.360 g/kj.

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2 years ago

Why do scientists believe that light is made of streams of particles?

irina [24]

I just need points to ask my own question.

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2 years ago

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You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int

blsea [12.9K]

1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor + (V_capacitor/RC) = (V_source/RC)

Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (30.5 ohms) * (89.9-mf) = 2.742 s

2) Charge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

V_capac (t) = (V_capac(0) - V_capac(∞))e ^(-t /T) + V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V

This means,

V_capac (t) = (-9V)e ^(-t /T) + 9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞) = 89.9mF*(9V) = 0.8091 C

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2 years ago