What are the units for electric current

7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

wlad13 [49]

Answer:

a)

$N_{top}=9.8N\\N_{bottom}=33.4N$

b) $v_{min}=4.4m/s$

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is $a_{cp}=\frac{v^2}{R}$ . At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

$-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg$

Which means:

$N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N$

The limit for falling off would be $N_{top}=0$ , so the minimum speed would be:

$0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s$

3 0

3 years ago

Which object will move faster?

Katarina [22]

Answer:

depends on how big the car is and what force is moving them.

Explanation:

the 3lb box will go slower because it is the lightest. the 10lb box is the 2nd slowest. the car will be smaller than the 18 wheeler on the road. but it has less cargo so it will probably go faster. if you are dropping it, then the 18 wheeler will go faster. but if you are seeing if it will go faster on the ground, the car will because it has less cargo or wheels to weigh it down.

4 0

3 years ago

Read 2 more answers

In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law t

RSB [31]

Answer:

The answers and workings is in the Explanation section

Explanation:

In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps.

According to Ohm’s law V =I*R

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

 If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω. The current in the circuit is then __________ Amps.

Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series R₂ = 6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

The current is 1 Amps

Answer = 6Ω and 1 Amps

If a third identical lamp is connected in series, the total resistance is now _________Ω. The current through all three lamps in series is now _________ Amps.

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ = 9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

Answer = 9Ω and 0.67 Amps

The current through each individual lamp is __________ Amps.

Since all 3 lamps are connected in series, the same current will flow through each of the 3 lamps, and that current is I₃

The current through each individual lamp is 0.67 Amp

Answer = 0.67 Amp

What is the power when a voltage of 120 volts drives a current of 3 amps through a device?

The formula for power P = I*V =120*3 = 360 Watts

power P = 360 Watts

Answer = 360 Watts

What is the current when a 90-W light bulb is connected to 120 V?

From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

Current= 0.75 Amps

Answer = 0.75 Amps

How much current does a 75-W light bulb draw when connected to 120 V?

Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

Voltage V =P/I =6/3 =2 Volts

Answer = 2 Volts

If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work.

24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

Answer = $4.32

4 0

3 years ago

A Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia In one set of experimen

balu736 [363]

Answer:

Coefficient of static friction = 0.37

Explanation:

At the point the the quoll slides, quoll attains its maximum velocity.

So Ne = (mv^2)/r ....equa 1

And N =mg....equ 2

Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s

Sub equ 2 into equ 1

Mge= (mv^2)/r ...equa3

Simplfy equ3

e = v^2/(gr)...equ 4

Sub figures above

e = 5.76/(9.8*1.6)

e = 0.37

7 0

3 years ago

A raccoon falls out of a tree from a height of 1.2. Which equation can you use to calculate the time it takes for the raccoon to

Inga [223]

Answer:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2 \cdot 1.2 m}{9.81 m/s^2}}=0.50 s$

Explanation:

The equation that we can use to calculate the time it takes for the raccoon to fall to the ground is:

$t=\frac{2S}{g}$

where S=1.2 m is the height of the tree and g=9.81 m/s^2 is the acceleration due to gravity. This equation is derived from the equation of the distance in a uniformly accelerated motion, which is given by

$S=S_0 + v_0 t + \frac{1}{2}at^2$

where S0 is the initial position, v0 is the initial velocity and t the time. In this problem, we can put S0=0 (we can take the initial position as the initial position of the raccoon) and v0=0 (the raccoon starts from rest), so the equation becomes

$S=\frac{1}{2}at^2$

and since the motion is a free fall, the acceleration is equal to the acceleration of gravity, so a=g:

$S=\frac{1}{2}gt^2$

And by re-arranging it, we find

$t=\sqrt{\frac{2S}{g}}$

By substituting numbers, we find

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2 \cdot 1.2 m}{9.81 m/s^2}}=0.50 s$

3 0

3 years ago