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belka [17]
3 years ago
6

Lil' Ricky is trying on his Halloween costume in front of a flat (plane) mirror. Lil Ricky stands 1.5 m from the mirror.

Physics
1 answer:
Flura [38]3 years ago
3 0

Answer:

The correct answer is A) It is upright and 1.5m behind the mirror

Explanation:

Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.

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When hockey players push the puck along the ice it slides causing heat which melts the ice causing the friction against the ice to be less.
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A boxer hits punching bag and gives it a change in momentum of 12 kg multiplied by m divided by s over 7.0ms what is the magnitu
mr_godi [17]

Answer: 1700

Explanation:

8 0
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70.5
ra1l [238]

Answer:5000000000hertz

Explanation:

Wavelength=6cm=6/100 m=0.06m

Frequency=velocity/wavelength

Frequency=(3×10^8)÷0.06

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5 0
3 years ago
In a clear cup there are three substances. Their densities are 3, 1, and 2. From top of the cup to the bottom, what would the or
Vesnalui [34]

Answer:

1, 2 and 3

Explanation:

The most dense substance will settle at the bottom of the cup

4 0
2 years ago
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

5 0
3 years ago
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