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vredina [299]
1 year ago
6

If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many images wo

uld you see in the viewer?
Physics
1 answer:
sammy [17]1 year ago
6 0

Explanation:

The question was incomplete. Here is the complete question.

If you punched thousands of holes in the aluminium foil of the scope, how many images would you see in the viewer.

a) Hundreds of small images

b) a few bright images

c) one large blurry image

d) no images at all.

So the correct option is c) one large blurry image.

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A boat moves through the sea.
sergiy2304 [10]

Answer:

dont you have to times it

Explanation:

4 0
2 years ago
the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of
nata0808 [166]
To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
c = 2000/ 10
c = 200

Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1
3 0
2 years ago
An electric heater draws a steady current = 20.0 A on a 120-V line. (a) Calculate how much power does it require.
babymother [125]

Answer:

The heater power required is 2400 W. The power in the heater can be calculated as the product of the voltage line and the steady current:

P=V.I

P=120 V * 20 A = 2400 VA = 2400 W

Explanation:

8 0
2 years ago
(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the
Anastaziya [24]
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7 0
2 years ago
Read 2 more answers
You have found a treasure map that directs you to start at a hollow tree, walk 300 meters directly north, turn and walk 500 mete
Dmitry_Shevchenko [17]

Answer:633 m

Explanation:

First we have moved 300 m in North

let say it as point a and its vector is 300\hat{j}

after that we have moved 500 m northeast

let say it as point b

therefore position of b with respect to a is

r_{ba}=500cos(45)\hat{i}+500sin(45)\hat{j}

Therefore position of b w.r.t to origin is

r_b=r_a+r_{ba}

r_b=300\hat{j}+500cos(45)\hat{i}+500sin(45)\hat{j}

r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}

after this we moved 400 m 60^{\circ} south of east i.e. 60^{\circ} below from positive x axis

let say it as c

r_{cb}=400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=r_{b}+r_{cb}

r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}

magnitude is \sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}

=633.052

for directiontan\theta =\frac{250\sqrt{2}+300-200\sqrt{3}}{250\sqrt{2}+200}

tan\theta =\frac{307.139}{553.553}

\theta =29.021^{\circ} with x -axis

7 0
1 year ago
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