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4 years ago

Why is it more difficult to slide a crate starting from rest?

2 answers:

7 0

Because the friction on the surface you are sliding on and the friction of the crate, go against each other. So once you start moving the friction will still be there but you now have the momentum to push past the starting point.

MatroZZZ [7]4 years ago

7 0

Because you are not full of as much energy as if you were awake for a certain period of time

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if a body of mass 2kg travels with 2 m/s and other body of mass 2kg travels with 1m/s then find v1 and v2 in given statement

KIM [24]

Answer:

Your answer is here,

Explanation:

v1 = 4 kg m/s

v2 = 2 kg m/s

8 0

3 years ago

Solving a series circuit, did I do this correctly?

nirvana33 [79]

The total resistance in the circuit is 16 Ohms.
The total current in the circuit is 0.5 Ampere.
The current at $R_1$ is 0.5 Ampere.
The current at $R_3$ is 0.5 Ampere.
The voltage drop at $R_1$ is 4 volts.
The voltage drop at $R_2$ is 2.5 volts.
The voltage drop at $R_3$ is 1.5 volts.
The total power consumed by the circuit is 4watts
The power consumed at $R_1$ is 2 watts
The power consumed at $R_2$ is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

$R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega$

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

The total resistance in the circuit = $R_{eq}$

$R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega$

The total current in the circuit = I

$V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A$ (Ohm's law)

For series combinations, the current in each resistor remains the same.

So, the current in $R_1, R_2 \&R_3$ :

$I_1= I_2= I_3=I=0.5 A\\$

The voltage drop across at $R_1$ = $V_1$

The current across $R_1$ = I = 0.5 A

$V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V$

The voltage drop across at $R_2$ = $V_2$

The current across $R_2$ = I = 0.5 A

$V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V$

The voltage drop across at $R_3$ = $V_3$

The current across $R_3$ = I = 0.5 A

$V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V$

The total power consumed by circuit:

$P= V\times I \\\\= 0.5 A\times 8 V = 4 watt$

Power consumed at $R_1$ :

$P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt$

Power consumed at $R_2$ :

$P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt$

Power consumed at $R_3$ :

$P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt$

Learn more about, current, voltage, resistance, and power of the circuit here:

brainly.com/question/11683246?referrer=searchResults

brainly.com/question/1430450?referrer=searchResults

5 0

2 years ago

Based on what you have read what will be another example of potential energy

almond37 [142]

The answer is A because potential energy is stored up energy which means that it’s not moving

4 0

3 years ago

Read 2 more answers

On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs.

BaLLatris [955]

I go inside and take a nap normally that will help maybe play music

3 0

3 years ago

As the distance of a sound wave from its source quadruples, how is the intensity changed?

timurjin [86]

Through the 1/r law, it was concluded that for the sound intensity or pressure is directly proportional to distance or radius. That is,
p = k/r
where k is the proportionality constant. If the distance of a sound wave is quadrupled then, the intensity of the sound is decreased to 1/4 of its original value.

8 0

3 years ago

Read 2 more answers