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Anna35 [415]
2 years ago
11

A ray diagram shows that an object is placed in front of a plane mirror. What are the characteristics of the image produced by t

he object? inverted, larger than object, real upright, smaller than object, virtual inverted, same size as object, real upright, same size as object, virtual
Physics
2 answers:
Scrat [10]2 years ago
4 0

Answer:

D. upright, same size as object, virtual

Explanation:

it is a plane mirror, it has no changes

romanna [79]2 years ago
3 0
Same size as object should be the answer, it is a “plane” mirror
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The answer is 59.4 degrees.
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A material that transmits nearly all the light in a ray because it offers little resistance to the light is ________.
Nina [5.8K]

Answer:

D. transparent.

Explanation:

A material that transmits nearly all the light in a ray because it offers little resistance to the light is <u>transparent.</u>

A transparent material allows light to pass through them with little or no resistance enabling them see-through. A material that transmits nearly all the light rays that pass through it because it offers little resistance to the light is TRANSPARENT. Examples of transparent materials are water,  glass (flint and crown), air, and diamond.

4 0
3 years ago
A ball is thrown w a speed of 30m/s at an angle of 10. When is the vertical component of the velocity equal to zero
irga5000 [103]

Now the vertical velocity of the ball thrown at an angle 10° is given as

Voy(initial vertical velocity)= 30m/s x sin 10

Voy(initial vertical velocity)= 5.2m/s

Now the ball is decelerating with an acceleration due to gravity equivalent to 9.8m/s^2.

Let Vy be the final velocity and that is equal to zero in this case.

Now

Vy= Voy- tx9.8

Where t is the time at which the vertical velocity becomes 0.

Substituting the values we get

0= 5.2-tx9.8

9.8t=5.2

t=0.53 secs



5 0
2 years ago
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A heated iron is pressed against a shirt, warming it. This is an
Nady [450]

Answer:

Convection? I'm pretty sure that's it

6 0
2 years ago
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
2 years ago
Read 2 more answers
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