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Anna [14]
3 years ago
15

Light from a distant star is collected by a concave mirror that has a radius of curvature of 150 cm. How far from the mirror is

the image of the star?
Physics
1 answer:
andrey2020 [161]3 years ago
4 0
The focal point of a mirror is half of the radius of curvature. We can use the formula,                                                                                                                   1/f = 1/v + 1/u                                                                                                          where f is the focal length , v is the image distance and u is object distance        The distance of the star is assumed to be incredibly far away and 1 divided by a really big number is approximately zero, thus;                                                     1/f = 1/v  = 1/75 = 1/v                                                                                             therefore; the image is formed 75 cm infront of the mirror
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tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
What morbid structure traditionally has thirteen steps?
tatiyna

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Explanation:

6 0
3 years ago
A high powered rifle can shoot a bullet at a speed of 1500 mi/hr. On the moon, with almost no atmosphere and an acceleration due
Amiraneli [1.4K]

Answer:

Explanation:

On the Moon :----

1500 x 1.6 = 2400 m /s is initial velocity of bullet .

g = 1.6 m /s²

v = u - gt

0 = 2400 - 1.6 t

t = 1500 s

This is time of ascent

Time of decent will also be the same

Total time of flight = 2 x 1500 = 3000 s

On the Earth : ---

v = u - a₁ t

0 = u - a₁ x 18

u = 18a₁

v² = u² - 2 x a₁ x 2743.2

0 = (18a₁ )² - 2 x a₁ x 2743.2

a₁ = 16.93

For downward return

s = ut + 1/2 a₂ x t²

2743.2 = 0 + .5 x a₂ x 31²

a₂ = 5.7 m /s²

If d be the deceleration produced by air

g + d = 16.93 ( during upward journey )

g - d = 5.7

g = (16.93 + 5.7) / 2  

= 11.315 m / s

d = 5.6 m /s²

So air is creating a deceleration of 5.6 m /s².

6 0
3 years ago
The kinetic energy of a body of mass 15 kg is 30 joule. What is its momentum?
lys-0071 [83]

This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum.  So here they are:

Kinetic energy = (1/2) · (mass) · (speed²)

Momentum = (mass) · (speed)

So, now ... We know that

==> mass = 15 kg,  and

==> kinetic energy = 30 Joules

Take those pieces of info and pluggum into the formula for kinetic energy:

Kinetic energy = (1/2) · (mass) · (speed²)

30 Joules = (1/2) · (15 kg) · (speed²)

60 Joules = (15 kg) · (speed²)

4 m²/s² = speed²

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THAT's all you need !  Now you can find momentum:

Momentum = (mass) · (speed)

Momentum = (15 kg) · (2 m/s)

<em>Momentum = 30 kg·m/s</em>

<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum.  When I saw this, I wondered whether that's always true.  So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>

8 0
3 years ago
A body of mass 2kg moves round a circle of radius 5m with a constant speed of 10m/s. Calculate the force toward the centre of th
antiseptic1488 [7]

Answer:

F = 5

Explanation:

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5 0
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