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3 years ago

What change should a rahul make

Physics

2 answers:

aksik [14]3 years ago

7 0

Answer:

Whats a rahul?

Explanation:

I'll edit this once I know :)

vovikov84 [41]3 years ago

7 0

Answer: its B

Explanation: just took the quiz

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If an element with an atomic number of 51 and an atomic mass of 121, how many neutrons does it have?

Llana [10]

121 -51=70 neutrons

7 0

4 years ago

Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20

timama [110]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

3 0

4 years ago

Read 2 more answers

Jason drives uphill from home to school, averaging 20 miles per hour. On the way back, he drives back downhill from school to ho

r-ruslan [8.4K]

Answer:

Average velocity = 30 miles per hour

Explanation:

Given that

Velocity for uphill = 20 miles per hour

Velocity for downhill = 60 miles per hour

We know that

Distance = Velocity x time

Total distance = Average velocity x total time

Lets take distance from home to school is s

Time taken from home to school= s/20 hr

Time taken from school to home= s/60 hr

Total time =s/20+s/60

Total distance = Average velocity x total time

s + s = Average velocity x (s/20+s/60)

So average velocity = 30 miles per hour

6 0

3 years ago

Which planet might best be described as a large dirty ice ball?

Diano4ka-milaya [45]

This planet would be known as pluto

3 0

3 years ago

Read 2 more answers

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning: