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3 years ago

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A joule is an amount of energy, and a watt is a rate of using energy, defined as 1 W = 1 J / s. How many joules of energy are re

quired to run a 100 W light bulb for one day?
Coal has about 30E6 J of energy per kg, but a power plant can only use 30% of this energy to make electricity (i.e., it is 30% efficient). How many kilograms of coal have to be burned to light a 100 W light bulb for one day?

1 answer:

Naily [24]3 years ago

4 0

How many joules of energy are required to run a 100 W light bulb for one day?

<span><span><span>A</span><span>100 </span>joules</span><span><span>B</span>100<span>W </span><span>× </span>24<span>hr </span>joules</span><span><span>C</span>100<span>W </span><span>× </span>24<span>hr </span><span>× </span>60<span>min∕hr </span>joules</span><span><span>D</span>100<span>W </span><span>× </span>24<span>hr </span><span>× </span>60<span>min∕hr </span><span>× </span>60<span>s∕min </span>joules</span></span>

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Find the current in the 12 ohm resistor.

disa [49]

Answer:

1.5 A

Explanation:

V =I.R

18 = I × 12

I = 18/12

= 3/2 = 1.5 A

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3 years ago

I NEED HELP ON THIS QUESTION!

Leto [7]

The second option is the correct one. m/s^2

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3 years ago

a rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of 22.9 m/s2. however, after 50.0

xxTIMURxx [149]

The acceleration of a rocket engine is given here, and after 50 seconds of flight, the engine fails, and we must determine the altitude of the rocket at the time the engine fails. Because the rocket starts from rest, the time taken is 50 seconds, the initial velocity is zero, and the acceleration is 22.9 meters per second square. So we use the kinamatics equation s equal to v. I t plus half 8 square. There is no acceleration at the start. 22.9 and t is 50 seconds, so displacement 2.86 times 10 to the power 4 is met. This is the rocket's displacement in 50 seconds, so this is the rocket's altitude when the engine fails.

<h3>What exactly is accelerate?</h3>

In mechanics, acceleration is defined as the rate of change of an object's velocity with respect to time. Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.

In his second law of motion, Sir Isaac Newton (1642-1727) defined acceleration as the ratio of a force acting on an object to its mass: a = f/m.

Accelerate is a verb that means to speed up. When you press the gas pedal, the car accelerates. If you know someone who works at the consulate, you can speed up the process.

Acceleration is the rate at which velocity changes over time, both in terms of speed and direction. A point or object moving in a straight line is accelerated if it accelerates or decelerates.

Even if the speed is constant, motion on a circle is accelerated because the direction is constantly changing. Both effects contribute to acceleration in all other types of motion.

Hence, There is no acceleration at the start. 22. This is the rocket's displacement in 50 seconds, so this is the rocket's altitude when the engine fails.

To know more about Accelerate refer to:

brainly.com/question/460763

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1 year ago

8. An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object? _______________

Sholpan [36]

Answer:

Mass of object is 0.5kg

Explanation:

Given the following data;

Force = 6N

Acceleration = 12m/s²

Mass =?

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

$F = ma$

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making mass (m) the subject, we have;

$Mass (m) = \frac{F}{a}$

Substituting into the equation;

$Mass (m) = \frac{6}{12}$

Mass, m = 0.5kg.

Therefore, the mass of the object is 0.5kg

5 0

2 years ago