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zavuch27 [327]
1 year ago
10

Adam observed properties of four different waves and recorded observations about the frequency and volume of each one in his cha

rt. Which statement could Adam declare based on the data in the table
Physics
1 answer:
butalik [34]1 year ago
3 0
I believe this answer is B but my believing sucks so
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A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
2 years ago
In a study, the data you collect is Habits on a Always/Sometimes/Never scale.What is the level of measurement?
zzz [600]

Answer:

Ordinal

Explanation:

There are four levels of measurement which include the nominal, ordinal, interval, and ratio. The data collected above is ordinal data as it qualifies the data and still indicates the ordering of the data. It gives the observer an idea of the range of data collected or its rating although mathematical calculations may not be done with it.

The other forms of data include the nominal which simply qualifies the data, the interval which qualifies the data but which the differences between the data can be obtained, and of course the data has no starting point. The ratio scale which is similar to the interval scale but which the ratios between the data obtained can be compared.

8 0
2 years ago
The drawing shows a side view of a swimming pool. The pressure at the surface of the water is atmospheric pressure. The pressure
Nikitich [7]

Answer:

A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

Explanation:

Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is

    P =P_{atm} + ρ g y

Where P_{atm} is the atmospheric pressure, ρ  the water density, and 'y' the depth measured from the surface.

Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth

Now we can examine the claims

A) To true. State agreement or with the equation above

B) False. Pressure changes with atmospheric pressure

C) False. It's the opposite

D) False. They are directly proportional

7 0
2 years ago
Read 2 more answers
A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diamete
d1i1m1o1n [39]

Answer:

I = 69.3  μA

Explanation:

Current through the straight wire, I = 3.45 A

Number of turns, N = 5 turns

Diameter of the coil, D = 1.25 cm

Resistance of the coil, R = 3.25 \mu ohms

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

B_{1} = \frac{\mu_{0}I }{2\pi d}

B_{1} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d } \\

B_{2} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

Induced current, I = \frac{E}{R}

E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ =  -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3  μA

3 0
2 years ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C is the magnitude of the charge

\Delta V = 6.0 kV = 6000 V is the potential difference between point P and infinity

Substituting into the equation, we find

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
2 years ago
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