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3 years ago

(AKS 3al) Which graph best represents a moving object in a state of equilibrium

Physics

1 answer:

I am Lyosha [343]3 years ago

7 0

Where is the image? I am unable to find it.

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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.600 W/m2.

LenKa [72]

Answer:

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Explanation:

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5 0

3 years ago

A box is moving along the x-axis and its position varies in time according to the expression:

Colt1911 [192]

Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. $x = 6 * 3.2^2 = 61.44 m$

b) at t = 3.2 + Δt. $x = 6*(3.2 + \Delta t)^2$

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

$v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}$

$v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}$

$v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}$

$v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}$

$v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}$

$v = 38.4 + \Delta t$

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0

3 years ago

An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?

anyanavicka [17]

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

8 0

3 years ago

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What will happen if you light up a candle that is in a beaker?

salantis [7]

It will melt so it is c

8 0

3 years ago

A "spherical capacitor" is constructed of two thin, concentric spherical shells of conducting material. Let a be the radius of t

Shalnov [3]

Answer:

$C=\frac{ab}{k(b-a)}$

Explanation:

We can assume this problem as two concentric spherical metals with opposite charges.

We have also to take into account the formulas for the electric field and the capacitance. Hence we have

$C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\$

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating

$dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]$

Hence, the capacitance is

$C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}$

but R1=a and R2=b

$C=\frac{ab}{k(b-a)}$

HOPE THIS HELPS!!

3 0

3 years ago

Read 2 more answers