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3 years ago

a battery is connected to a resistor. as charge flows, the chemical energy of the battery is dissipated as _________.

1 answer:

5 0

As charge flows, the chemical energy of the battery is dissipated as the time of charge flow increases.

Electric power is defined as the rate at which electric energy is dissipated.

The power of the current flowing through the circuit is calculated as follows;

P = IV

P = I x IR

P = I²R

where;

P is the power dissipated
I is the current flowing in the circuit
R is the resistance of the circuit

The electric energy is given as;

E = Pt

E = I²Rt

Thus, we can conclude that as charge flows, the chemical energy of the battery is dissipated as the time of charge flow increases.

Learn more here:brainly.com/question/21392880

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baptist walked 420 m in 17.5 min. what is his average speed in m/min? What is his average speed in m/s? (Hint: There are 60 s in

yulyashka [42]

Answer:

Average speed is 24 m/min

in another unit is is 0.4 m/s

Explanation:

As we know that the average speed is the ratio of total distance covered and total time

so here we know that total distance moved by the baptist is

$d = 420 m$

total time taken by it is given as

$t = 17.5 min$

so average speed is given as

$v = \frac{d}{t}$

$v = \frac{420}{17.5} m/min$

$v = 24 m/min$

now we have to find the speed in m/s unit also

so we have

$v = \frac{24}{60} m/s$

$v = 0.4 m/s$

8 0

3 years ago

A 15.00 kg particle starts from the origin at time zero. Its velocity as a function of time is given by = 8t2î + 5tĵ where is in

Elden [556K]

Answer:

Explanation:

I will assume the equation reads:

v = 8t²î + 5tĵ

The velocity v is the time derivative of the position x.

$x = \int\limits^t_0 {v} \, dt = \int\limits^t_0 {8t^{2}\hat i + 5t\hat j} \, dt = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j |^t_0 = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j - \frac{8}{3} \hat i - \frac{5}{2} \hat j\\ x = \frac{8}{3} (t^{3} - 1 )\hat i + \frac{5}{2} (t^{2} - 1 )\hat j$

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4 years ago

What is the relationship between electric field lines and equipotential lines that you observed in doing the lab

lidiya [134]

Answer:

Explained below

Explanation:

Generally speaking, we know in physics that Electric field lines are lines which usually start at positive charges and deflect away from them to terminate at the negative charges. Meanwhile Equipotential lines are lines that are used to connect points located on the same electric potential.

Finally, in conclusion, electric field lines are usually lines that go through in a perpendicular manner across every equipotential lines.

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3 years ago

The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d

r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let $F_1\ and\ F_2$ are two forces. The resultant of two forces acting on the same point is given by :

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}$

Where $\theta$ is the angle between two forces

When $\theta=0$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}$

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}$

When $\theta=90^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}$

$F_R=\sqrt{F_1^2+F_2^2}$

When $\theta=180^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}$

$F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}$

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

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3 years ago

2. A 200.0-kg bear grasping a vertical tree slides down at constant velocity. What is the friction force between the tree and th

Neporo4naja [7]

Answer: 1960 N

Explanation:

The bear is sliding down at constant velocity: this means that its acceleration is zero, so the net force is also zero, according to Newton's second law:

$F_{net}=ma=0$

There are two forces acting on the bear: its weight W, pulling downward, and the frictional force Ff, pulling upward. Therefore, the net force is given by the difference between the two forces:

$F_{net}=W-F_f=0$

From the previous equation, we find that the frictional force is equal to the weight of the bear:

$F_f=W=mg=(200.0 kg)(9.8 m/s^2)=1960 N$

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3 years ago