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TEA [102]
3 years ago
10

A scientist is trying to determine whether a newly observed phenomenon can be described as a wave. Which of the following questi

ons would the scientist want to ask about the phenomenon?
A.Can you see it?
B.Does it transfer energy?
C.Is it solid?
D.How quickly does it travel?
Physics
1 answer:
Step2247 [10]3 years ago
7 0

Answer:

B. does it transfer energy

A wave is a disturbance in a medium. Particles vibrate and TRANSFER their ENERGY to their neighbors.

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A lamp is on a dimmer switch. As you turn the switch from the lowest setting to the highest setting, which statement would best
avanturin [10]

Answer:

D

Explanation:

When something that needs power that you are turning off you will not need power when its off.

8 0
3 years ago
Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside
jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0
1 year ago
A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that h
RoseWind [281]

Answer:

  a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

          F -W = m a

Force is electrical force

         F = k q₁ q₂ / r²

         k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

         q₁ = q₂ = q

         a = (k q² / r² - m g) / m

         a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

        m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

        q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

        r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

        a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

        a = -7.29 m / s²

The negative sign indicates that the direction of this acceleration is downward

6 0
3 years ago
A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.
kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current I = 1 A

Voltage V = 12 V

For finding the resistance,

  V = IR

  R = \frac{V}{I}

  R = \frac{12}{1}

  R = 12Ω

(B)

For finding power delivered,

  P = I^{2} R

  P = (1) ^{2} \times 12

  P = 12 Watt

(C)

For finding the potential difference,

   V = IR

   V = 5 \times 10^{-3} \times 2

   V = 10 \times 10^{-3}

   V = 0.01 V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0
3 years ago
The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li
melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate Q of a hypodermic needle:

Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}  (1)

Where the dimensions of each one is:

Volume flow rate Q=\frac{L^{3}}{T}

Radius of the needle R=L

Length of the needle L=L

Pressures at opposite ends of the needle P_{2} and P_{1}=\frac{M}{LT^{2}}

Viscosity of the liquid \eta=\frac{M}{LT}

We need to find the value of n whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}  (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}  (3)

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}  (4)

As we can see n must be 4 if we want the exponent to be 3:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}  (5)

Finally:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}  (6)

8 0
3 years ago
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