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liberstina [14]

3 years ago

15

An archer pulls a bowstring back a distance of 20 cm with an average force of 75 N. The arrow has a mass of 20.0 g. When he rele

ases the string, what is the velocity of the arrow when it leaves the bow? (A) 1.2 m/s (B) 22 m/s (C) 32 m/s (D) 39 m/s (E) 42 m/s

1 answer:

Masja [62]3 years ago

8 0

To solve this problem we will apply the concepts related to the work theorem for which it is defined as the product of Force and distance. In turn, we will use the energy conservation theorem for which the applied work must be equivalent to the total kinetic energy on the body.

The work is defined as

$W = Fd$

Here,

F = Force

d = Displacement

Replacing with our values we have that

$W = 75*0.2$

$W = 15J$

Now by conservation of energy,

$W = KE$

$W = \frac{1}{2}mv^2$

$15 = \frac{1}{2} (20*10^{-3}kg)(v^2)$

Solving for v,

$v = 10\sqrt{15}$

$v = 38.72 \approx 39$

Therefore the correct answer is D.

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Describe the principle of conversation of energy:

exis [7]

Answer: Energy can neither be created nor destroyed, rather it is converted from one form to another

Explanation:

The principle of conversation of energy explains how energy is conserved in nature by being converted from one form to another such that no energy is created nor destroyed.

Practical examples include:

- electrical pressing iron that converts electrical energy to heat energy

- solar panels that converts solar energy to electrical energy

- Car batteries that converts chemical energy to light energy etc

7 0

3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

5 0

3 years ago

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder

beks73 [17]

Answer:

$d=1.67mi$

Explanation:

Assuming the light takes essentially no time to reach you, the distance at which the lightning occurred can be calculated by multiplying the speed of sound by the time it takes to hear the thunder:

$d=v_st\\d=344\frac{m}{s}(7.8s)\\d=2683.2m*\frac{1mi}{1609.34m}=1.67mi$

4 0

2 years ago

An airplane cabin is pressurized to 570 mmhg. what is the pressure inside the cabin in atmospheres?

abruzzese [7]

1 atm corresponds to 760 mmHg, so we can set up a simple proportion to find how many atmospheres correspond to 570 mmHg:
$1 atm: 760 mmHg = x: 570 mmHg$
and from this, we find
$x= \frac{1 atm \cdot 570 mmHg}{760 mmHg} =0.75 atm$

8 0

3 years ago

Read 2 more answers

A capacitor is charged to a potential difference of 3 volt it delivers 30% store energy to lamp what is the final potential diff

nexus9112 [7]

Answer:

3.98V

Explanation:

Given

Pontential difference V as 3v

Energy delivered is 30%,

Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)

So E=V^2

E=3^2=9

At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3

But E=V^2

✓E=V

✓6.3=3.98V

4 0

2 years ago