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3 years ago

I need help i don’t want to go to summer school

Physics

1 answer:

Sonbull [250]3 years ago

7 0

Answer:

makes you hungry is not a science based benefit of meditation

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An eagle carries a fish up 50 m into the sky using 90 N of force. How much work did the eagle do on the fish? (Work: W = Fd)

Leno4ka [110]

<span>Work: W = Fd. 50(distance) multiplied by 90(force) would equal 4500 J or, answer D</span>

4 0

3 years ago

Read 2 more answers

The bulk modulus of water is 2.2 x 109 Pa, and its density is 1.0 x 103 kg/m3. What is the speed of sound in water?

yarga [219]

hi guys! Lets be friends!

7 0

2 years ago

You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 130-turn coil that has

dimaraw [331]

Answer:

B = 0.129 T

Explanation:

Given,

frequency, f = 60 Hz

maximum emf = 5200 V

Number of turns, N = 130

Area per turn = 0.82 m²

We know,

ω = 2 π f

ω = 2 π x 60 = 376.99 rad/s

now, Magnetic field calculation

$B =\dfrac{\epsilon_{max}}{NA\omega}$

$B =\dfrac{5200}{130\times 0.82\times 376.99}$

B = 0.129 T

Hence, the magnetic field is equal to B = 0.129 T

3 0

3 years ago

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil

MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!

6 0

2 years ago

An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i

crimeas [40]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

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

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

$\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}$

$q=\frac{1}{0.047 cm^{-1}}=+21.3 cm$

The sign of q is positive, so the image is real.

Now let's also write 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

By substituting p = 50 cm and q = 21.3 cm, we find

$h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o$

So we notice that:

$|h_i| < |h_o|$ : this means that the image is smaller than the object

$h_i < 0$ : this means that the image is inverted

so, the correct option is:

It is real, inverted, and smaller than the object.

7 0

2 years ago