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2 years ago

A 40W lamp wastes 34 J of energy every second by heating its surroundings.

Physics

1 answer:

Artemon [7]2 years ago

4 0

Answer:

$15\%$ .

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is $40\; \rm W$ (same as $40\; \rm J \cdot s^{-1}$ ,) meaning that $40\; \rm J$ of energy is supplied to this lamp every second.

The question states that $34\; \rm J$ out of that $40\; \rm J$ of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the $(40\; \rm J - 34\; \rm J) = 6\; \rm J$ of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive $40\; \rm J$ of energy input and would outputs $6\; \rm J$ of useful work. The efficiency of this lamp would be:

$\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}$ .

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Bob will not develop a good habit if he _____. works out for only one week develops a consistent schedule of exercise works out

Charra [1.4K]

Answer:

works out for only one week

Explanation:

In order to develop a good habit, there are certain things that you can do like:

-Develop a consistent schedule of exercise which allows you to work everyday over a period of time which helps you to create a routine that you would get used to it creating a habit.

-Work out with a friend which allows you to have someone to motivate you and that would go everyday with you which would help to be consistent.

-Use success strategies that are ways in which you can achieve your goal of creating a good habit to exercise.

According to this, the option that would not help Bob develop a good habit is if he works out for only one week as doing something everyday is what would help you to create the habit.

3 0

3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of

IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0

3 years ago

Consider the video tutorial you just watched. Suppose that we duplicate this experimental setup in an elevator. What will the sp

zloy xaker [14]

Explanation:

if the elevator is moving upward with the constant speed the spring scale will read 18 N which is the mass of each of the two blocks attached by separate springs to the scale at opposite ends.