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1 year ago

A 600 kg car is at rest, and then accelerates to 5 m/s.

Physics

1 answer:

Crank1 year ago

3 0

Answer:

7500J

Explanation:

Given parameters:

Mass of car = 600kg

Velocity = 5m/s

Unknown:

Original kinetic energy = ?

Final kinetic energy = ?

Work used = ?

Solution:

The kinetic energy of a body is the energy due to the motion of a body.

It can be solve mathematically using expression below;

K.E = $\frac{1}{2}$ m v²

where m is mass

v is velocity

original kinetic energy;

The car started at rest and v = 0, therefore K.E = 0

Final kinetic energy;

K.E = $\frac{1}{2}$ x 600 x 5² = 7500J

Work done:

Work done = Final K.E - Initial K.E = 7500 - 0 = 7500J

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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

2 years ago

The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an obj

aliina [53]

Answer:

Follows are the solution to this question:

Explanation:

By checking the value in which we have calculated by performing its differentiation of $\frac{a}{3}t^3+bt$ , the correct form of its integer value is calculating with regard to t, that also provides as expected $at^2+b = F(t)$ .

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1 year ago

The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length L, what fractio

Alexxandr [17]

Answer:

4/10 of L.

Explanation:

A stopped pipe is a pipe with one closed end and one opened end. it is also called a closed pipe.

The fundamental mode of a stopped pipe is also called its fundamental frequency, and is f₁=v/4L.

Where f₁=fundamental frequency, v= velocity of sound, L= Length of pipe.

The fifth harmonic of the stopped pipe f₅ =5v/4L .................(1)

For an open pipe,

Fundamental mode is also called fundamental frequency f₁₀=v/2l₀ .......... (2)

Where f₁₀ = fundamental frequency of a closed pipe, v= velocity of sound and l₀=length of the resulting open pipe.

from the question, the fundamental mode of the resulting open pipe = The fifth harmonic of the original stopped pipe.

∴ f₅=f₁₀

⇒5v/4L = v/2l₀

Equating v from both side of the equation,

⇒ 5/4L = 1/2l₀

Cross multiplying the equation,

5×2l₀ = 4L× 1

10l₀ = 4L

Dividing both side of the equation by the coefficient of l₀ i.e 10

10l₀/10 = 4L/10

∴ l₀ = 4/10(L)

∴ 4/10 of L must be cut off

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2 years ago

Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

Explanation:

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0

2 years ago

Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

Arturiano [62]

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s$

So, its velocity at the bottom of the ramp is 2.21 m/s.

4 0

1 year ago