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Advocard [28]
3 years ago
5

The pictures shows what a student observed when he looked through a hand lens at a penny.

Physics
2 answers:
igomit [66]3 years ago
7 0
The answer is A. Refracted
sergeinik [125]3 years ago
6 0
A. The light is refracted
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Elodia [21]

Answer: just do the same thing, but the problems are different

Explanation: try you best

Download pdf
3 0
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I NEED HELP WITH THE LAST QUESTION I’L MARK U AS BRANLIEST
velikii [3]

Answer:

gap junctions,tight junctions,and desmosomes

6 0
3 years ago
What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t
Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

= 500 x 5625

= 2812500 joules

Thus, the car travels with a kinetic energy of 2812500 joules

5 0
3 years ago
A wave traveling in water has a frequency of 500.0 Hz and a wavelength of 3.00 m. What is the speed of the wave?
Sergeu [11.5K]

Answer:

1500 m/s

Explanation:

Recall that for a wave,

Speed = frequency x wavelength

here we are given frequency = 500 Hz and wavelength = 3m

simply substitute into above equation

Speed = 500 Hz x 3m

= 1500 m/s

6 0
3 years ago
. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the
lana66690 [7]

Answer:

Outlet Velocity = 192 m/s

Outlet Pressure = 510 kPa

Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure,  P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂  (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π  (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s  (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂  (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁  + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000  + (1/2)(1000) (12-192)

= 600,000  + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

8 0
3 years ago
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