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2 years ago

Which of the angles shown in this picture is the angle of incidence? A B

Physics

2 answers:

Dmitry_Shevchenko [17]2 years ago

6 0

The pictures are not attached, therefore, I cannot give a specific choice.
However, I will try to help you out.

The angle of incidence is defined as the angle formed between the ray of light and the normal to the surface that the ray is falling on.

The angle of incidence can be shown in the attached image.

Therefore, for your question, choose the image on which the above description applies.

Hope this helps :)

dsp732 years ago

6 0

B. is the correct answer, hope this helps!

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What is a example of a real life scientific inquiry problem

Papessa [141]

What is an example of how you can use scientific inquiry to solve a real life problem.

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2 years ago

A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4

timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

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2 years ago

A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse

Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒ Ii*ωi = If*ωf <em>(I)</em>

where

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒ m = 35.98 Kg ≈ 36 Kg

5 0

3 years ago

The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A

steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

$\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}$

$\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}$

Net force

$\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

$\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}$

$\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}$

$\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}$

$\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}$

Work done is defined as

$W = \overrightarrow{F}.\overrightarrow{S}$

$W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k} \right )$

W = -1120 + 450 + 70

W = - 600 J

3 0

2 years ago

when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of

Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂ (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂

or, C₂=2X/m. T₂ (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂ /2X /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0

3 years ago