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Elenna [48]
3 years ago
12

Is number six correct? If so why???

Physics
1 answer:
Yuki888 [10]3 years ago
3 0
In question #5, the one where you have to label the three pictures,
all of your labels are correct !

Look at the 3rd picture, where the light is coming down slanted,
hitting some kind of surface, and bouncing away slanted.
You labelled that picture "reflection", and that's correct.

Now, look at the two angles marked in the picture .... the angle
between the arriving light and the surface, and the angle between
the departing light and the surface.

There is a statement in Science called the "Law of Reflection".
It simply says that those two angles are EQUAL.
I don't think we have to get into WHY right now.  I think that
at this point, your class is just trying to teach you what the law says. 
If you go on and study more physics in High School, then you'll learn
WHY the angles are equal.

If the light ray arrives and hits the mirror at a 40° angle,
then it'll leave the mirror at a 40° angle.  It's that simple.

The two angles are the ones marked equal in the picture
where you wrote 'reflection'.

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When you double your speed, it takes about _____ times as much distance to stop?
Zigmanuir [339]
4 times as much distance to stop
8 0
3 years ago
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quester [9]
Answer is (c), the latent heat of fusion. That is by definition the heat that 1 kg of a substance must absorb to melt in the vicinity of its melting point.
4 0
2 years ago
An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible
harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

3 0
3 years ago
How much watt is called 1 H. P​
agasfer [191]

Answer:

746 watts

Explanation:

7 0
3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
4 years ago
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