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2 years ago

River water held behind a dam is a form of ________. group of answer choices

Physics

1 answer:

WINSTONCH [101]2 years ago

8 0

Potential energy is present in the river water behind a dam.

<h3>How is potential energy defined?</h3>

Potential energy is a form of stored energy that is dependent on the relationship between different system components.
When a spring is compressed or stretched, its potential energy increases.
If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy.
The object's mass, its mass in relation to some reference points, and the gravitational field it is in all affect the potential energy of the object.
Joule is the SI unit of potential energy as a result (J)
As weight and height increase, gravitational potential energy rises.

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Define density and give its SI unit

3241004551 [841]

Answer:

Density is defined as mass per unit volume. its SI unit is: kg/m³

Explanation:

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2 years ago

Read 2 more answers

What is the momentum of a 200 kg truck travelling at 20 m/s?

geniusboy [140]

Answer:

p = 4000 kg-m/s

Explanation:

Given that,

The mass of a truck, m = 200 kg

Speed of the truck, v = 20 m/s

We need to find the momentum of the truck. The formula for momentum is given by :

p = mv

so,

$p=200\times 20\\\\p=4000\ kg-m/s$

So, the momentum of the truck is equal to 4000 kg-m/s.

8 0

3 years ago

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

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4 0

4 years ago

A capacitor with a very large capacitance is in series with a capacitor that has a very small capacitance. what can we say about

Misha Larkins [42]

<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.

The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)

The capacitance of a series combination is

1 / (1/A + 1/B + 1/C + 1/D + .....) .

If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is

(product of the 2 individuals) / (sum of the individuals) .

In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is

(1000 x 1) / (1000 + 1)

= (1000) / (1001)

= 0.999 000 999 . . .

which is smaller than the smaller individual.

It'll always be that way. </span>

5 0

3 years ago

If we want to describe work , we must have what ?

ivanzaharov [21]

An external force that is being applied in the direction of the displacement

4 0

3 years ago