How much voltage would it take to push 0.5 amps through a 2 kilowatt toaster?
1000 Volts
2000 Volts
3000 Volts
4000 Volts
5000 Volts
6000 Volts

Answers

Answer 1

Given :

Current, I = 0.5 A.

Power of toaster, P = 2 kW = 2000 W.

To Find :

How much voltage is required.

Solution :

We know, voltage is given by :

[tex]V = \dfrac{P}{I}\\\\V = \dfrac{2000\ W}{0.5 A}\\\\V = 4000 \ volts[/tex]

Therefore, 4000 Volts are required.

Answer 2

Answer:

4000

Explanation:

2000/0.5=4000


Related Questions

The final velocity of a car is 30m/s. The car is accelerating at a rate of 2.5m/s2
over an 2.8m/s2
8 second period of time. What is the initial velocity of the car?

Answers

Answer:

10m/s

Explanation:

Vf=v0+at

30m/s=v0+2.5m/s^2*8s

Solve for v0=10m/s

Sorry for the vague explanation I'm on my phone right now. Hope this still helps!

Initial velocity of the car where final velocity of a car is 30m/s and it is accelerating at a rate of 2.5 [tex]m/s^{2}[/tex] over an 2.8 [tex]m/s^{2}[/tex] 8 second period of time is 10 m/s.

To find the initial velocity of the car, we can use the kinematic equation:

Final velocity (v) = Initial velocity (u) + (Acceleration (a) * Time (t)).

Given the final velocity (v) as 30 m/s, acceleration (a) as 2.5 [tex]m/s^{2}[/tex], and time (t) as 8 seconds, we can plug these values into the equation:

30 m/s = u + (2.5 [tex]m/s^{2}[/tex] * 8 s).

Now, solve for the initial velocity (u):

30 m/s = u + 20 m/s.

Subtract 20 m/s from both sides:

u = 30 m/s - 20 m/s = 10 m/s.

Therefore, the initial velocity of the car is 10 m/s.

To know more about Initial velocity, here

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what are the 3 types of family goals​

Answers

be healthy, nice bond, achieving goals

Below is a diagram of the orbits of the inner planets examine the diagram and answer questions three and four 

Answers

Answer:

C is the answer

Explanation:

Earth is the third planet from the sun

you are cheating on your test, I know

Answer:

C..

Explanation:

I'm not sure about answer...

Question 6 of 59
A wave has a wavelength of 5 m and a speed of 65 m/s. What is the frequency of this wave in Hz?

Answers

Answer:

13 Hz Ans..

Explanation:

Data:

wavelength = 5 m

v = 65 m\s

f = ?

Formula:

v = f w

f= v\w

Solution:

f = 65 \ 5

f = 13 Hz Ans .........

Which of these is not a scalar quantity?

Answers

Answer:

options?!?

Explanation:

give us some options

which of these is NOT a metamorphic rock?

Basalt
Slate
Marble​

Answers

Explanation:

Basalt is not a metamorphic rocks

basalt is the answer

4. Determine the net force for the free-body
diagram
Fnorm = 40N
Ffrict =28N
Fapp =44N
Fgrav = 40N

Answers

Answer:

Fapp =44N is the correct answer

Explanation:

Please mark brainliest

4. Premature wrinkling due to overexposure to the Sun, what am I?​

Answers

Premature Aging

Mark as brainlist

write down the expressions for the heat produced per second in a resistor in terms of potential difference and resistance​

Answers

Answer:

[tex]H=\frac{V^2}{R}[/tex]

Explanation:

Let H be  heat produced per sec

Time, t=1 sec

Resistance of resistor=R ohm

We have to write down the expression for the  heat produced per second in a resistor in terms of potential difference and resistance​.

We know that

Heat, [tex]H=I^2 Rt=\frac{V^2}{R}t[/tex]

Substitute the value

Then, we get

Heat,[tex]H=\frac{V^2}{R}\times 1[/tex]

[tex]H=\frac{V^2}{R}[/tex]

This is required expression for the heat produced per second in a resistor  in terms of potential difference and resistance​.

1.) A projectile is launched at 32.1 m/s and 52.6 º above the horizontal. Determine the:
a.) Maximum height of the projectile.
b.) The time it takes to reach the maximum height
c.) Time of flight
d.) Range

Answers

Answer:

a.) Maximum height of the projectile. =32.5m

b.) The time it takes to reach the maximum height=2.6s

c.) Time of flight=2.6s

d.) Range=99.5m

Explanation:

in a nuclear plant, 1 * 10 ^27 J of energy is available from mass conversion How much mass was lost ?

Answers

Answer:

[tex]1.2\times 10^{10}\ kg[/tex]

Explanation:

Given that,

The energy of the nuclear plant, [tex]E=10^{27}\ J[/tex]

We need to find how much mass wast lost.

The relation between energy and mass is given by :

[tex]E=mc^2[/tex]

Where

c is speed of light

[tex]m=\dfrac{E}{c^2}\\\\m=\dfrac{10^{27}}{(3\times 10^8)^2}\\\\m=1.2\times 10^{10}\ kg[/tex]

So, [tex]1.2\times 10^{10}\ kg[/tex] of mass was lost.

Find the weight of an object of mass 5 kg
on
i. Surface of the earth
ii. b) Surface of the moon

Answers

Answer:

weight on earth is mg

which is 5*9.8

49 Newton

weight on moon is 1/6 th of weight on earth

1/6*49

8.166 Newton..

Answer:

1

For instance, on Earth, a 5.0-kg object weighs 49 N; on the Moon, where g is 1.67 m/s2, the object weighs 8.4 N.

2

The weight of the object on the surface of earth would be 5kg. The weight of the object on the surface of moon would be 5/6= 0.83 as the weight of any object on earth is six times than that on moon.

Which has more potential energy: a baseball held 1 m above the ground or a baseball held 2 m above the ground?

Answers

a baseball 1m above the ground i'm pretty sure

The baseball held 2 meters above the ground has more potential energy than the baseball held 1 meter above the ground.  

The potential energy can be calculated with the following equation:

[tex] E_{p} = mgh [/tex]

Where:

m: is the mass of the baseball

g: is the acceleration due to gravity

h: is the height

We can see from equation (1) that the potential energy depends on height, mass, and gravity. Since the question refers to the same baseball, the mass is constant besides the acceleration due to gravity, so:

Potential energy above 1 m

[tex] E_{p_{1}} = mg*1 = mg [/tex]  (1)

Potential energy above 2 m

[tex] E_{p_{2}} = mg*2 = 2mg [/tex]   (2)

By entering equation (1) into 2 we have:

[tex] E_{p_{2}} = 2mg = 2E_{p_{1}} [/tex]

Hence, the potential energy of the baseball held 2 m above the ground is twice that of the baseball held 1 m above the ground.

Therefore, the baseball held 2 m above the ground has more potential energy.

You can learn more about potential energy here: https://brainly.com/question/15813853?referrer=searchResults

I hope it helps you!

examples of non viscous liquid​

Answers

The fluids that have no resistance or zero resistance to internal friction are known as non- viscous fluids. • The fluid which has more resistance to flow is called viscous fluid. Viscosity is a property of a fluid, which is a degree of the amount of resistance between the fluid layers.

The period of the sound waves produced by the motor is 8.3 milliseconds.
Calculate the frequency of the sound waves.

Answers

Answer:

120.48hz

Explanation:

f=1/t..........

Either the wavelength, velocity, or wave period—the amount of time it takes for a wave cycle to complete—must be known. If you know the period, divide 1 by it and, if necessary, convert it to seconds. The outcome will be the frequency in Hertz.

What frequency of the sound waves?

As with all waves, the relationship between the speed of sound, its frequency, and its wavelength is vw=f, where vw is the speed of sound, f is the frequency, and is the wavelength.It is necessary to know either the wavelength, the velocity, or the wave period, which is the duration of a wave cycle.

Therefore, More oscillations are produced at higher frequencies. Hertz are the frequency measurement units (Hz). Noises between 20 Hz and 20,000 Hz can be heard by those with normal hearing. Ultrasound is a term for frequencies greater than 20,000 Hz.

Learn more about sound waves here:

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HELP please!!!!!!!!!!!!!!!​

Answers

Nolur acil lütfen yalvarırım sana

Anyone able to help with these questions, will give brainliest to first correct anwser

Answers

Answer:

b

Explanation:

I did this question earlier

A human cannonball act uses springy bungee cords instead of gunpowder to launch a person through a long cylindrical tube into the air. If a cannon is capable of launching a 55 kg test dummy 11 m into the air when launched straight up, how much energy is stored in the bungee cords? *

A. 539J
B. 605 J
C. 2965J
D. 5929 J

Answers

Answer:

Energy stored, E = 5929 J

Explanation:

Given that,

The mass of a test dummy, m = 55 kg

It is launched 11 m into the air.

We need to find the energy stored in the bungee cords. It can be calculated as follows :

[tex]E=mgh\\\\E=55\times 9.8\times 11\\\\E=5929\ J[/tex]

So, 5929 J of energy is stored in the bungee cords.

How could you measure the strength of static electricity compared to the strength of an
electromagnet?

Answers

Answer:

Lines of Force around an Electromagnet

Explanation:

The magnetic field strength of an electromagnet is therefore determined by the ampere turns of the coil with the more turns of wire in the coil the greater will be the strength of the magnetic field

Why does it take a long time for the earth to rotate?

Answers

24hrs or i full year as in 365 days

Answer:

earth takes a long time to rotate due to the tidal effects the moon has on earth's rotation

Explanation:

I hope this helps!

Which of the following determines the identity of an atom?

A. atomic mass

B. mass number

C. atomic number

D. charge

Answers

I believe the answer is C

Which one of the following best describes a fuse?

Answers

A material that allows current to travel through a circuit

An electric iron is connected to the mains power supply of 220 V. When the electric iron is

adjusted at 'minimum heating' it consumes a power of 360 W but at 'maximum heating' it takes a

power of 840 W. Calculate the current and resistance in each case.​

Answers

Answer:

Given: V = 220V, Pmin = 360W, Pmax = 840W

For minimum heating case:

We know that

Pmin = VI

360 = 220 X I

I = 1.63 amp

R = V/I

R = 220/1.63

R = 134.96ohms

For maximum heating case:

We know that

Pmax = VI

840 = 220 X I

I = 3.81 amp

R = V/I

R = 220/3.81

R = 57.74 ohms





4. A substance is cut into several piece and compared to the density of the
original sample. The density of each piece will be:-
areater
bl the same​

Answers

Answer:

The density pf each piece will be the same, since density is defined as a mass per unit volume, so even when you cut an object it will not change

A skateboarder coasts a distance of 20 meters in 4 seconds. What is the person's speed?

Answers

Answer:

5 m/s

Explanation:

The formula of speed is : distance/time taken.

The skateboarder covered a distance of 20 meters in 4 seconds. So,

Distance = 20 meters

Time Taken = 4 seconds

Speed = 20/4 = 5 m/s

Which graph shows a vehicle traveling at a constant speed?
(look at the attached photo)

Answers

b because it’s a straight line

1/2 means that the voltage (or charge) of the system will increase to half more of what is left in a time equal to t1/2 seconds. Therefore if a system is already at half charge (t1/2 seconds after starting) then after t1/2 more seconds the system will be charged to 50% plus half of 50%. That is 25% more, or 75% of the entire charge. Let's say that four t1/2's have gone by. That means that the charge (or voltage) is at (50% + 1/2*50% + 1/2*1/2*50% + 1/2*1/2*1/2*50%) = 93.75% of maximum charge. Yikes! Now look in your manual for a more simple mathematical derivation of this concept. Given t1/2 to be 0.4406 seconds, how long should it take to reach 75% of maximum charge? answer in seconds.

Answers

Answer:

The answer is "0.047".

Explanation:

Given value:

[tex]\to t_{\frac{1}{2}}= 0.4406\\\\\to V=0.9375 V_{max}[/tex]

Calculating the capacitance:

[tex]\to V=V_{max} (1-e^{\frac{-t}{Rc}})[/tex]

In this, the t = time, which is taken to calculates its maximum voltage.

[tex]\to 0.9375 V_{max} = V_{max}(1-e^{- \frac{t}{103\times(165.279\times10^{-6})}})\\\\\to e^{- \frac{t}{103\times(165.279\times10^{-6})}}= 0.0625\\\\\to - \frac{t}{103\times(165.279\times10^{-6})} = \ln(0.0625)\\\\\to -t= 0.01702\times(-2.77258) \\\\ \to -t = -0.04719 \\\\ \to t= 0.04719 \approx 0.047 \ s[/tex]

HELP ASAP PLS


A locomotive with a rocket engine is being tested on a smooth horizontal track. The mass of the

locomotive is unknown. Starting from rest, the engines are fired for 20 s. During this time

they expel 500 kg of oxidized kerosene. The kerosene particles are expelled at an average

velocity of 1200 m/s. At the end of the 20-s period the speed of the locomotive is 50 m/s.

What is its mass?

Answers

Answer:

The mass of the locomotive is 288,000 kg

Explanation:

The question relates to the law of conservation of energy

The given parameters are;

The form of engine on the locomotive = Rocket engine

The mass of the locomotive = m

The initial velocity of the locomotive, v = 0 m/s

The duration at which the engine is fired = 20 s

The mass of the kerosene expelled, m₁ = 500 kg

The average speed with which the kerosene is expelled, v₁ = 1,200 m/s

The final speed of the locomotive engine after 20-s, v₂ = 50 m/s

Kinetic energy, K.E. = 1/2·m·v²

The total kinetic energy of the kerosene particles, K.E.[tex]_k[/tex] = 1/2 × 500 × 1,200² = 360,000,000

∴ K.E.[tex]_k[/tex] = 360,000,000 J

By the conservation of energy, the kinetic energy of the expelled kerosene particles is equal to the kinetic energy gained by the moving locomotive

Therefore, we have;

K.E.[tex]_L[/tex] = K.E.[tex]_k[/tex]

K.E.[tex]_L[/tex] = 1/2 × m × v₂² = 1/2 × m × 50²

∴ K.E.[tex]_L[/tex] = 1/2 × m × 50² = K.E.[tex]_k[/tex] = 360,000,000 J

1/2 × m × 50² = 360,000,000 J

m = 360,000,000 J/(1/2 × (50 m/s)²) = 288,000 kg

m = 288,000 kg

The mass of the locomotive, m = 288,000 kg

. A boy wishes to throw a ball through a house via two small openings, one in the front and the other in

the back window, the second window being directly behind the first. If the boy stands at a distance of 5m

in front of the house and the house is 6m deep and if the opening in the front window is 5m above him

and that in the back window 2m higher, calculate the velocity and the angle of projection of the ball that

will enable him to accomplish his desire.

Answers

Answer:

1) The angle of projection of the ball is approximately 52.496° above the horizontal

2) The velocity with which the boy throws the ball to enable him accomplish his desire is approximately 14.769 m/s

Explanation:

In the projectile motion of the ball, the parameters are;

The distance the boy stands from the front of the house = 5 m

The depth of the house, front to back = 6 m

The height of the opening of the front window above him = 5 m

The height of the back window = 2 m higher than the front window

Therefore, the height of the back window = 5 m + 2 m = 7 m

1) The general formula for the vertical height of a parabolic projectile motion is given as follows;

y = a·x² + b·x + c

At y = 0, x = 0, therefore, c = 0

At y = 5, x = 5, we have;

5 = a·5² + b·5

∴ 5 = 25·a + 5·b...(1)

At y = 7, x = 11, we have;

7 = a·11² + b·11

7 = 121·a + 11·b...(2)

Making 'a' the subject of equation (1) and (2) and equating both values of 'a' gives;

For equation (1);

a = (5 - 5·b)/25 = 1/5 - b/5

a = 1/5 - b/5

For equation (2);

a = (7 - 11·b)/121 = 7/121 - b/11

a = 7/121 - b/11

∴ 1/5 - b/5 = 7/121 - b/11

1/5 - 7/121 = b/5 - b/11

86/605 = 6·b/55

b = (86/605) × (55/6) = 43/33

b = 43/33

a = 1/5 - b/5 = 1/5 - (43/33)/5 = -2/33

∴ y = (-2/33)·x² + (43/33)·x

The slope of the curve = dy/dx = d((-2/33)·x² + (43/33)·x)/dx = -4/33·x + 43/33

The slope of the curve at any point on the projectile = -4/33·x + 43/33

The slope at the origin is given by plugging x = 0  as follows

The slope at the origin = -4/33 × 0 + 43/33 = 43/33

The slope at the origin = tan(θ) = 43/33

Where;

θ = The angle of projection of the ball

∴ θ = arctan(43/33) ≈ 52.496°

The angle of projection of the ball, θ ≈ 52.496° above the horizontal

2) At the back widow, the equation for the vertical height, 'h', is given as follows;

[tex]h = (tan \angle \theta) \cdot x - \left(\dfrac{g}{2 \cdot v_1^2 \cdot cos^2 \angle \theta } \right )\cdot x^2[/tex]

Where;

h = The vertical height of the back window = 7 m

tan∠θ = 43/33

x = The horizontal distance of the back window from the boy = 5 m + 6 m = 11 m

g = The acceleration due to gravity = 9.8 m/s²

v₁ = The velocity with which the boy throws the ball

cos²∠θ = cos²(arctan(43/33)) = 1089/2938

Plugging in the values gives;

[tex]7 = \left (\dfrac{43}{33} \right) \times 11 - \left(\dfrac{9.8}{2 \cdot v_1^2 \cdot \dfrac{1089}{2938} } \right )\cdot 11^2[/tex]

[tex]\therefore v_1^2 = \dfrac{71981}{330}[/tex]

[tex]\therefore v_1 = \sqrt{\dfrac{71981}{330}} \approx \pm14.769[/tex]

The velocity with which the boy throws the ball to enable him accomplish his desire, v₁ ≈ 14.769 m/s.

What do seismic waves and sound waves have in common?

Answers

They're both a type of mechanical wave, which require a medium for propagation.
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