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IrinaK [193]
3 years ago
4

Which equation relates charge, time, and current? answer A/ l= triangle q/t

Physics
2 answers:
TiliK225 [7]3 years ago
8 0

Answer:

I=\frac{\Delta q}{\Delta t}

Explanation:

The equation that relates charge, time and current is the following:

I=\frac{\Delta q}{\Delta t}

where

I is the current intensity, measured in Amperes (A)

\Delta q is the amount of charge that passes through a given point in a certain interval of time \Delta t. \Delta q is the measured in Coulombs (C) while \Delta t in seconds (s), so the Ampere is defined as

1 A = \frac{1 C}{1 s}

Looking at the equation, we see that the current intensity represents the "rate of flow of charge through a given point".

KIM [24]3 years ago
3 0
Yes. It is I = delta q / t

If you think about it, current’s definition is the flow of charge.

This equation then means that current is the rate at which charge flows (coulombs per second).
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A 65.0 kg skier is moving at 6.85 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 4.00
marusya05 [52]

Answer:

v = 4.58 m/s

Explanation:

In order to calculate the speed of the skier when she gets the bottom of the hill, you have to calculate the speed of the skier when she crosses the rough patch.

To calculate the velocity at the final of the rough patch you take into account that the work done by the friction surface is equal to the change in the kinetic energy of the skier:

W_f=\Delta K\\\\-N\mu_kd=\frac{1}{2}mv^2-\frac{1}{2}mv_o^2=\frac{1}{2}m(v^2-v_o^2)        (1)

Where the minus sign means that the work is against the motion of the skier.

Wf: friction force

m: mass of the skier = 65.0kg

N: normal force = mg

g: gravitational acceleration = 9.8m/s^2

d: distance of the rough patch = 4.00m

v: speed at the end of the rough patch = ?

vo: initial speed of the skier = 6.85m/s

μk: coefficient of kinetic friction = 0.330

You replace the expression for the normal force in the equation (1), and solve for v:

-mg\mu_kd=\frac{1}{2}m(v^2-v_o^2)\\\\-g\mu_kd=\frac{1}{2}(v^2-v_o^2)\\\\v=\sqrt{-2g\mu_kd+v_o^2}\\\\v=\sqrt{-2(9.8m/s^2)(0.330)(4.00m)+(6.85m/s)^2}=8.53\frac{m}{s}=4.58\frac{m}{s}

Then, the speed fot he skier at the bottom of the hill is 4.58m/s

3 0
4 years ago
For a freely falling object dropped from rest, what is the acceleration at the end of the fifth second of fall?
stepan [7]
Near the Earth's surface, a freely falling body has constant acceleration at every instant of its fall ... 9.8 meters per second^2.
3 0
3 years ago
A 190 g rock at 20°C is completely immersed in 600 g of water at 80°C. Which one of the following statements is true right after
Umnica [9.8K]

Rock is completely immersed in hot water. By the second law of thermodynamics, thermal energy or heat is transferred from substance with higher temperature to substance with lower temperature until they come to thermal equilibrium i.e. both at same temperature.

It is given here that rock is at 20°C which is at lower temperature than water at 80°C. ∴Heat or thermal energy flows from water to rock. So, right choice is-

A. The water gives the rock thermal energy and gets no thermal energy in return.

8 0
3 years ago
What is the electric force acting between two charges of 0.0042 C and −0.0050 C that are 0.0030 m apart?
Black_prince [1.1K]

Answer:

A.−2.1 × 10^10 N

Explanation:

Using the formula;

E = k Q1Q2/d²

Where;

E is the electrical force  

k is the constant  

Q1, Q2 are the two charges  and

d is the distance between the two charges

Therefore;

E = (9 x 10^9) × (0.0042) × (-0.0050) / (0.0030)²  

 = -2.1 x 10^10 N

Therefore; electrical force acting between the two charges is -2.1 x 10^10 N.

6 0
4 years ago
Select all that apply.
lyudmila [28]
B and d r both answers.
6 0
3 years ago
Read 2 more answers
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