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boyakko [2]
3 years ago
5

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

Wtoaster , which is used for 6.00 minutes, and then calculate the amount of energy that an 11.0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 9.00 hours .
Physics
1 answer:
Nikolay [14]3 years ago
3 0

Explanation:

It is given that,

Power consumed by toaster oven, P=1.4\ kW

Time taken, t = 6 minutes = 0.1 hour

Energy used by toaster, W_{toaster}=P\times t

W_{toaster}=1.4\ kW\times 0.1\ h=0.14\ kWh

Similarly,

Power consumed by toaster oven, P=11\ W

Time taken, t = 9 minutes = 0.15 hour

Energy used by toaster, W_{toaster}=P\times t

W_{toaster}=11\ W\times 0.15\ h=1.65\ Wh=0.0016\ kWh

Hence, this is the required solution.

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A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one
Fittoniya [83]
I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)
8 0
3 years ago
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A body is oscillating up and down at the end of a spring. Let’s consider when the body is at the top of its up-and-down motion.
Klio2033 [76]

The velocity of the body is zero; option A

<h3>What is the motion of an oscillating body?</h3>

The motion of an oscillating body is known as simple harmonic motion.

Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.

For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.

In conclusion, oscillating bodies undergo simple harmonic motion.

Learn more about simple harmonic motion at: brainly.com/question/24646514

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3 0
1 year ago
If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft
Marina CMI [18]

Explanation:

The equation of motion of an object is given by :

h(t)=-16t^2+112t+128

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

-t^2+7t+8=0

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

8 0
3 years ago
What is the period of a wave that has a frequency of 300 hz?
Butoxors [25]

Answer:

T = 0.003 s

(Period is written as T)

Explanation:

Period = time it takes for one wave to pass (measured in seconds)

frequency = number of cycles that occur in 1 second

(measured in Hz / hertz / 1 second)

Period : T

frequency : f

So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period

T =  \frac{1}{f}    f = \frac{1}{T}

to find the period (T) of this wave, we need to plug in the frequency (f) of 300

T = \frac{1}{300}

T = 0.00333333333

So, the period of a wave that has a frequency of 300 Hz is 0.003 s

[the period/T of this wave is 0.003 s]

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What is the minimum time that a warm up should last
Vladimir [108]
5 minutes is the minimum time
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3 years ago
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