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sineoko [7]
3 years ago
14

How do mirrorswork?​

Physics
2 answers:
stealth61 [152]3 years ago
8 0

Answer:

When photons - rays of light - coming from an object (your smiling face, for example) strike the smooth surface of a mirror, they bounce back at the same angle. Your eyes see these reflected photons as a mirror image.... If a smooth surface absorbs the photons, they can't bounce back and there will be no reflection.

Explanation:

*Hope this helps*

fiasKO [112]3 years ago
6 0

Answer:

the light relflects off them and creates a image

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If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th
Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

a=cs   (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

v^2=v_o^2+2a \Delta s   (2)

vo: initial  velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

v^2=2(cs)\Delta s

You do the constant c in the last equation, then you replace the values of v, s and Δs:

c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44

6 0
3 years ago
What is carried by a wave?
Nata [24]
Waves carry energy from one place to another. Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals. ... Some types of waves need to be transmitted through matter, either a solid, liquid or a gas. For example, water waves have to travel in water.
3 0
3 years ago
Read 2 more answers
A one-dimensional plane wall of thickness 2l= 100 mm experiences uniform thermal energy generation of q˙= 800 w/m3 and is convec
slega [8]

Answer:

The thermal conductivity of the wall = 40W/m.C

h = 10 W/m^2.C

Explanation:

The heat conduction equation is given by:

d^2T/ dx^2 + egen/ K = 0

The thermal conductivity of the wall can be calculated using:

K = egen/ 2a = 800/2×10

K = 800/20 = 40W/m.C

Applying energy balance at the wall surface

"qL = "qconv

-K = (dT/dx)L = h (TL - Tinfinity)

The convention heat transfer coefficient will be:

h = -k × (-2aL)/ (TL - Tinfinty)

h = ( 2× 40 × 10 × 0.05) / (30-26)

h = 40/4 = 10W/m^2.C

From the given temperature distribution

t(x) = 10 (L^2-X^2) + 30 = 30°

T(L) = ( L^2- L^2) + 30 = 30°

dT/ dx = -2aL

d^2T/ dx^2 = - 2a

4 0
3 years ago
baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0
3 years ago
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