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Liono4ka [1.6K]

3 years ago

13

Light traveling through air strikes the surface of the four different materials shown. Which material reflects light but does no

t refract it? 1. Diamond 2. water being poured 3. triangular glass prism 4. 2 silver spoons

2 answers:

AysviL [449]3 years ago

7 0

Answer:

The answer is 4. 2 silver spoons

Explanation:

Hello! Let's solve this!

Many materials refract light in addition to reflecting it. This means that the light passes through the object at a certain angle.

In this case, of the given options, the correct option is 4. 2 silver spoons, since it is the only material that is not transparent of the dice.

We conclude that the answer is 4. 2 silver spoons

Elza [17]3 years ago

4 0

It would be{ 4) 2 silver spoons}<span>
that the material reflects light but does not refract it</span>

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A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h

katen-ka-za [31]

Answer:

$v_r=5.89\ m.s^{-1}$

Explanation:

Given:

mass of rocket, $m_r=50\ g$

time of observation, $t=2\ s$

mass lost by the rocket by expulsion of air, $m_a=10\%\ of m_r=5\ g$

velocity of air, $v_a=53\ m.s^{-1}$

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

$m_a.v_a=(m_r-m_a)\times v_r$

$5\times 53=(50-5)\times v_r$

$v_r=5.89\ m.s^{-1}$

7 0

3 years ago

A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done

Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

$W = F*d$

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

$F = W_{x} = mgsin(\theta)$

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²

θ: is the degree angle to the horizontal = 30°

$F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N$

Now, we can find the work:

$W = F*d = 24.53 N*20 m = 490.6 J$

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0

3 years ago

3 m/s north is an example of a(n) ____.

alexandr1967 [171]

Rate of speed (3 m/s north is three miles per second north, so it's a rate of speed)

4 0

3 years ago

Read 2 more answers

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago

A boulder with a mass of 50 kg, sits on a cliff that is 50 m high. The boulder is pushed off of the cliff. How much kinetic ener

rosijanka [135]

50 +50 =100 Since it’s sitting on a 50m cliff that’s high with a mass of 50 kg it would be adding because once it goes down it’s adding speed

8 0

3 years ago