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Lana71 [14]
2 years ago
11

Which type of curved mirror uses its inside as the reflecting surface? A. a convex mirror B. a plane mirror C. a virtual mirror

D. a concave mirror

Physics
2 answers:
ser-zykov [4K]2 years ago
6 0

D. a concave mirror

natali 33 [55]2 years ago
5 0

Answer:

Concave Mirror

Explanation:

1).Concave Mirror

here the reflecting surface is curved inside then it is always concave mirror

2). Convex mirror

here the curved surface outwards will be the reflecting surface and focus is taken as negative as it always forms virtual image of smaller size.

3). Plane Mirror

here the reflecting surface is always plane surface and straight surface and its focus is infinite and always forms virtual image of same size

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A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per
FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0
2 years ago
How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
salantis [7]

Answer:

Q = 4.63 \times 10^6 J

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

s = 240 J/kg C

now we will have

Q = ms\Delta T + mL

\Delta T = 961.8 - 20

\Delta T = 941.8 degree C

now from above equation

Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)

Q = 3.1 \times 10^6 + 1.53 \times 10^6

Q = 4.63 \times 10^6 J

3 0
2 years ago
A bowling ball accidentally falls out of the cargo bay of an airliner as it flies along in a horizontal direction. As seen from
valina [46]

The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

<h3>Path of the bowling ball</h3>

Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.

Thus, the path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

Learn more about horizontal direction here: brainly.com/question/2534565

#SPJ1

3 0
2 years ago
14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change
klio [65]

Lets se

And

\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}

\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}

So

\\ \rm\Rrightarrow k\propto m

If spring constant is doubled mass must be doubled

8 0
2 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
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