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1 year ago

6

A vertical black line crossed by a horizontal yellow line. A blue arrow strikes the crossing point from above at an angle from h

orizontal labeled angle of incidence. Another blue arrow points away from the crossing point at an angle labeled angle of reflection. If the angle of incidence is 30°, what is the value of the angle of reflection? °

2 answers:

Alex777 [14]1 year ago

8 0

Answer:

30

Explanation:

Paha777 [63]1 year ago

7 0

Answer:

30°

Explanation:

It will be the angle that it strikes an object at [30°].

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The dwarf planet praamzius is estimated to have a diameter of about 300km and orbits the sun at a distance of 6.4E12m . What is

RoseWind [281]

Answer:

Time period of the planet is

$T_2 = 278.7 years$

Explanation:

As we know by Kepler's law of time period

square of time period of planet is proportional to the cube of its orbital radius

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

here we know for Earth

$T_1 = 1 year$

$R_1 = 1.5 \times 10^{11} m$

now we have

$\frac{1^2}{T_2^2} = \frac{(1.5 \times 10^{11})^3}{(6.4 \times 10^{12})^3}$

$T_2^2 = 7.77 \times 10^4$

$T_2 = 278.7 years$

8 0

1 year ago

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

$0 = m(3.5 sin30) - mv_y$

$v_y = 1.75 m/s$

Now similarly in X direction we will have

$m(4.5) = m(3.5 cos30 ) + mv_x$

$v_x = 1.47 m/s$

now final kinetic energy of both puck after collision is given as

$KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)$

$KE_f = 8.73 m$

initial kinetic energy of both pucks is given as

$KE_i = \frac{1}{2}m(4.5^2) + 0$

$KE_i = 10.125 m$

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0

2 years ago

A certain slide projector has a 100 mm focal length lens. How far away is the screen, if a slide is placed 103 mm from the lens

Vlad [161]

Answer:

3.43 m

Explanation:

f = 100 mm

u = - 103 mm

Let v be the distance between the screen and the lens of the projector.

Use lens equation

1 / f = 1 / v - 1 / u

1 / 100 = 1 / v + 1 / 103

1 / v = 1 / 100 - 1 / 103

1 / v = (103 - 100) / (100 x 103)

1 / v = 3 / 10300

v = 3433.33 mm = 3.43 m

4 0

1 year ago

Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. The electric field at a point P, which

krok68 [10]

Answer:

$E_{max}=41666.66\ N/C$

Explanation:

Given that,

The radius of sphere, r = 0.3 m

Distance from the center of the sphere to the point P, x = 0.5 m

Electric field at point P, $E_P=15000\ N/C$ (radially outward)

The maximum electric field is at the surface of the sphere. We know that the electric field is inversely proportional to the distance. So,

$\dfrac{E_{max}}{E_p}=\dfrac{0.5^2}{0.3^2}$

$\dfrac{E_{max}}{15000}=\dfrac{0.5^2}{0.3^2}$

${E_{max}}=\dfrac{0.5^2}{0.3^2}\times 15000$

$E_{max}=41666.66\ N/C$

So, the magnitude of the electric field due to this sphere is 41666.66 N/C. Hence, this is the required solution.

6 0

1 year ago

An ambulance moving at 42 m/s sounds its siren whose frequency is 450 hz. a car is moving in the same direction as the ambulance

Korvikt [17]

(a) Since the ambulance and the car are moving one relative to each other, we have to use the general formula of the Doppler effect, which gives us the shift of the frequency of the siren as heard by an observer in the car:
$f'=( \frac{v+v_o}{v+v_s} )f$
where
f' is the apparent frequency as heard by the observer in the car
v is the velocity of the wave
$v_o$ is the velocity of the observer (positive if it is moving towards the source, negative if it is moving away)
$v_s$ is the velocity of the source (positive if the source is moving away from the observer, negative if is is moving towards it)
f is the real frequency of the sound

In the first part of the problem:
$v=343 m/s$ (speed of the sound wave)
$v_o =-25 m/s$ (the car is moving away from the ambulance)
$v_s = -42 m/s$ (the ambulance is moving towards the car)
$f=450 Hz$ (original frequency of the sound)

If we plug the numbers into the formula, we find
$f'=( \frac{343 m/s-25 m/s}{343 m/s-42 m/s} )(450 Hz)=475 Hz$

b) This time, the ambulance passes the car, so the ambulance is now moving away from the car; this means that $v_s$ must be positive:
$v_s=+42 m/s$
Moreover, the car is now moving towards the ambulance, so we should reverse also the sign of $v_o$ :
$v_o=+25 m/s$
All the other data do not change, so if we use the same formula as before, we find
$f'=( \frac{343 m/s+25 m/s}{343 m/s+42 m/s} )(450 Hz)=430 Hz$

8 0

1 year ago