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3 years ago

11

Someone pls help! My physics quiz is today:(

2 answers:

lisabon 2012 [21]3 years ago

8 0

Answer:

it should be -2m/s

Explanation:

Because it is going down from minutes 6-8. And down in most cases means negative. And since it is going down 2 it should be -2m/s.

Hope this helped!!

Vladimir [108]3 years ago

3 0

Answer:

0m/s^2

Explanation:

This is a position vs time graph, so the slop of the line represents the velocity (if you do not understand why that is then feel free to ask). Since the slope is constant (a straight line), we know that the velocity is not changing, in other words, the object is not accelerating.

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Which property of light is a constant in a vacuum?

faust18 [17]

Answer:

La velocidad de la luz en el vacío es una constante universal con el valor de 299 792 458 m/s (186 282,397 mi/s),aunque suele aproximarse a 3·108 m/s. Se simboliza con la letra c, proveniente del latín celéritās (en español, celeridad o rapidez).

¿Cuál es la consecuencia que a velocidad de la luz sea constante?

Respuesta. En modificaciones del vacío más sutiles, como espacios curvos, efecto Casimir, poblaciones térmicas o presencia de campos externos, la velocidad de la luz depende de la densidad de energía de ese vacío.

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3 years ago

What sort of waves exhibit the doppler effect?

Alexus [3.1K]

The waves that occur when you drop and stone into the water when skipping rocks

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3 years ago

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

8 0

4 years ago

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = $\sqrt{2gh}$

V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

= 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

= - 0.329 J

Hence, kinetic energy is not conserved.

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3 years ago

Which picture correctly shows the path of the reflected light rays given an object outside the focal point?Select one:a. Ab. Bc.

nignag [31]

We will have that the graph that describes the scenario is given by graph B.

6 0

1 year ago