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3 years ago

The layers of the atmosphere are divided into layers based on changes in __________.

Physics

1 answer:

Umnica [9.8K]3 years ago

6 0

A. altitude should be correct

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A missile is fired from a jet flying horizontally at Mach 1 (1100 ft/s). The missile has a horizontal acceleration of 1000 ft/s2

nadezda [96]

Answer: 11,100 ft/s^2

1) Constant acceleration=> uniformly accelerated motion.

2) Formula for uniformly accelerated motion:

Vf = Vo + at

3) Data:

Vo = 1,100 ft/s
a = 1,000 ft/s^2
t = 10.0 s

4) Solution:

Vf = 1,100 ft/s + 1,000 ft/s^2 * 10.0 s = 1,100 ft/s + 10,000 ft/s

Vf = 11,100 ft/s

6 0

3 years ago

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A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after

Delicious77 [7]

$\Delta L= \alpha L_0 (T_f-T_i)$

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D

5 0

3 years ago

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What would whales communicate about?

jeyben [28]

Where danger is and how to avoid it as well as to locate other whales

4 0

3 years ago

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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

7 0

3 years ago

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THE ANSWER!!! Please

Anuta_ua [19.1K]

Answer:

I think -7 N. Netforce is 3N-10N= -7N

Explanation:

5 0

3 years ago