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3 years ago

8

A battery is used to power a flashlight. When the flashlight is in use, what type of energy is lost during energy transformation

? *
chemical energy
electrical energy
radiant/light energy
thermal energy

2 answers:

noname [10]3 years ago

8 0

It’s either chemical or thermal energy

But most chemical

Hope this helps! :)

diamong [38]3 years ago

4 0

Answer:

The answer is chemical energy

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Read the passage.

Paladinen [302]

<h2>Answer: B)Scientists’ understanding of cells continually improved as the results of studies built upon each other over time and formed the cell theory.</h2>

Explanation:

Nowadays we know <u>cells are essential microscopic units that make up the living beings, capable of reproducing independently. </u>

However, this is the result of a long process of discoveries and studies made since the 19th century, in which the continuous improvement of new technologies was helpful.

In fact, it is wel known the English scientist Robert Hooke was the first to discover the existence of cells by looking through a compound microscope at a cork sheet, realizing that it was made up of small polygonal holes (like those of a honeycomb) that reminded him of the chambers in which the monks stayed (called cells). Then, during the next centuries more studies were made until we had the current knowledge about the structure of a cell.

7 0

3 years ago

Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

denis-greek [22]

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

$Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts$

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

$Time Period = 3 div * 5msec /div\\Time Perod = 15 msec$

Since,

$Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz$

8 0

3 years ago

A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t

gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

$p_i = (M+m)u$

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

$p_f = MV+mv$

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

$p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s$

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

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5 0

4 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

An Astronaut lands on an Earthlike planet and drops a small lead ball with a mass from the top of her spaceship. The point of re

bearhunter [10]

First we have to find out the gravity on that planet. We use Newton second equation of motion. It is given as,

s = ut +(gt^2)/2

Distance s = 25m

Time t = 5 s

Velocity u = 0

By putting these values,

25 = 1/2.g.(5)²

g = 2

So the gravity on that planet is 2. Lets find out the weight of the astronaut.

Mass of the astronaut on earth m = 80 kg

Weight of astronaut on earth W = mg = (80)(9.8) = 784 N

Weight of astronaut on earth like planet = (80)(2) = 160 N

x = 160N

5 0

3 years ago