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3 years ago

5

Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a const

ant rate of 2 m/s. How fast is the area of the spill increasing when the radius is 21 m

1 answer:

vesna_86 [32]3 years ago

4 0

Answer:

84π m²/s or 263.76 m²/s

Explanation:

see the attached file

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Post WWII Germany was divided into four zones with the Americans occupying one zone and all of the following countries occupying

maxonik [38]

Answer:

b. Italy

Explanation:

Italy had sided with germany and lost in world war 2

After the Potsdam conference, Germany was divided into four occupied zones: Great Britain in the northwest, France in the southwest, the United States in the south and the Soviet Union in the east.

europeuncedu

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2 years ago

A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c

Romashka [77]

Answer:

Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

Substituting the values given

I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2

I = 1.67 × 10^-18 A

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3 years ago

By what percent must one increase the tension in a guitar string to change the speed of waves on the string from 301 m/s to 343

garri49 [273]

Answer:

29.8 %

Explanation:

7 0

3 years ago

It was a dark and stormy night, when suddenly you saw a flash of lightning. Three-and-a-half seconds later you heard the thunder

Gekata [30.6K]

We're going to multiply the time it took for you to hear thunder (3.5 seconds) by the speed of sound in air (340 m/s)

3.5 x 340 = 1190

The lightning bolt was 1,190 meters away.

3 0

3 years ago

An object is 15 cm in front of a diverging lens with a

Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}$

$q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm$

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

$h_i = - h_o \frac{q}{p}$

where $h_i, h_o$ are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that $h_i >0$ , which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

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3 years ago