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Ira Lisetskai [31]
4 years ago
10

What is a plane mirror ? ​

Physics
1 answer:
g100num [7]4 years ago
6 0

Answer:

A plain mirrior is a mirrior with flat reflective surface.

hope it is helpful for you.

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Craig Terrill has many _________. Press enter to interact with the item, and press tab button or down arrow until reaching the S
iren2701 [21]

Answer:

is there anything that is a hint? like a picture or a story?

Explanation:

7 0
3 years ago
Two coins, made of different material, are placed next to each other so that they are touching. Both coins are 15mm in diameter.
nekit [7.7K]

The mass of the second coin for the given gravitation force is 20.9 g.

The given parameters;

  • <em>distance between the two coins, r = 15 mm = 0.015 m</em>
  • <em>mass of one coin, m₁ = 6.5 g = 0.0065 kg</em>
  • <em>gravitational force between the coins =  4.03 × 10⁻¹¹ N</em>

The mass of the second coin is determined by applying Newton's law of universal gravitation as shown below;

F = \frac{Gm_1 m_2 }{R^2} \\\\m_2 = \frac{FR^2}{Gm_1} \\\\m_2 = \frac{(4.03 \times 10^{-11}) \times (0.015)^2}{(6.67 \times 10^{-11}) \times (0.0065)} \\\\m_2 = 0.0209 \ kg\\\\m_2 = 20.9 \ g

Thus, the mass of the second coin for the given gravitation force is 20.9 g.

Learn more here:brainly.com/question/13590473

5 0
3 years ago
If the load is 13 cm from the fulcrum, how much effort is needed to lifth the load ? Help me.
snow_tiger [21]
That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:

-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.

-- Then the effort needed to lift the load is

(the weight of the load) x (13 / the distance between the fulcrum and the effort)
7 0
3 years ago
Modern oil tankers weigh over a half-million tons and have lengths of up to one-fourth mile. Such massive ships require a distan
sergiy2304 [10]

(a) -1.46\cdot 10^{-4} m/s^2

The average acceleration of the ship is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Here we have:

u=34 km/h =9.44 m/s is the initial velocity

v = 0 is the final velocity

t=18 min =64800 s is the time elapsed

Substituting, we find

a=\frac{0-9.44 m/s}{64800 s}=-1.46\cdot 10^{-4} m/s^2

(b) 4.72 m/s

Assuming the acceleration is uniform, the average velocity of the ship is given by:

v_{avg} = \frac{v+u}{2}

where

v is the final velocity

u is the initial velocity

Here we have:

v = 0

u = 9.44 m/s

So the average velocity of the ship is

v_{avg} = \frac{0+9.44 m/s}{2}=4.72 m/s

6 0
4 years ago
How long does it take to travel?<br> 120 miles at 40mph
lianna [129]

Answer:

it takes 3 hours to travel

Explanation:

120 ÷ 40 = 3

7 0
2 years ago
Read 2 more answers
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