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3 years ago

11

If the angle of incidence of a light source to a shiny surface is 30 degrees, what will the angle

2 answers:

Vinvika [58]3 years ago

8 0

Answer : 30 degrees
*** HOPE THIS HELP YOU***

Virty [35]3 years ago

5 0

<h3>Answer: D) 30</h3>

Angle of incidence always equals angle of reflection. Think of a tennis ball being hit into a wall. The ball will bounce off at the same angle that it approached with. The angles mentioned are formed through the line called the "normal", which is the line perpendicular to the surface.

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To fill the medication prescription, what information must the pharmacy technician need to obtain? A. The name of the medication

shepuryov [24]

C. Patient info, name of med, dosage & route, special instructions, prescriber’s DEA#, and number of refills

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3 years ago

Water (density = 1 ´ 103 kg/m3) flows at 10 m/s through a pipe with radius 0.030 m. the pipe goes up to the second floor of the

Hatshy [7]

density of water = $1000 kg/m^3$

velocity of flow = $10 m/s$

radius of pipe = $0.030 m$

Height of second floor = $2 m$

Now we can use here Bernuoli's Equation to find the speed of water flow at second floor

$P_1 + 1/2\rho v_1^2 + \rho g h_1= P_2 + 1/2 \rho v_2^2 + \rho g h_2$

$P + 1/2 * 1000 * 10^2 + 1000* 9.8 * 0 = P + 1/2 * 1000 * v^2 + 1000*9.8*2$

$v = 7.8 m/s$

Now in order to find the radius of pipe we can use equation of continuity

$A_1 v_1 = A_2 v_2$

$\pi *0.030^2 * 10 = \pi * r^2 * 7.8$

$r = 0.034 m$

So radius of pipe at second floor is 0.034 meter

3 0

3 years ago

Perform the operation of 147.02 / 0.338, and report the result with the proper number of significant figures

Ede4ka [16]

Answer:

434.97041

Explanation:

4 0

2 years ago

A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s

Explanation:

It is given that,

Mass of the car, $m_1=1.1\times 10^3\ kg$

Mass of the truck, $m_2=2.3\times 10^3\ kg$

Initial velocity of the car, $u_1=22\ m/s$

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2$

$v_2=15.7\ m/s$

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0

3 years ago

Help plz I’ll mark brainliest

slavikrds [6]

Thinner at edges and its thick in the middle

3 0

3 years ago

Read 2 more answers