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3 years ago

Suggest a reason why the earth can be compared to a bar magnet

Physics

1 answer:

Vadim26 [7]3 years ago

6 0

Because it’s a man fetus

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What does this circuit do? T is the clock All flip flops are D-Flip-Flop What is the function of the circuit if Dser is ‘1’ alwa

Harlamova29_29 [7]

Answer:

That is true

Explanation:

6 0

3 years ago

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency?(b) What is its wavelength in glass t

inn [45]

(a) $4.74 \cdot 10^{14}Hz$

The frequency of a wave is given by:

$f=\frac{v}{\lambda}$

where

v is the wave's speed

$\lambda$ is the wavelength

For the red laser light in this problem, we have

$v=c=3\cdot 10^8 m/s$ (speed of light)

$\lambda=632.8 nm=632.8\cdot 10^{-9} m$

Substituting,

$f=\frac{3\cdot 10^8 m/s}{632.8 \cdot 10^{-9} m}=4.74 \cdot 10^{14}Hz$

(b) 427.6 nm

The wavelength of the wave in the glass is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0 = 632.8\cdot 10^{-9} m$ is the original wavelength of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$\lambda=\frac{632.8\cdot 10^{-9}m}{1.48}=427.6\cdot 10^{-9}m=427.6 nm$

(c) $2.02\cdot 10^8 m/s$

The speed of the wave in the glass is given by

$v=\frac{c}{n}$

where

$c = 3\cdot 10^8 m/s$ is the original speed of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$v=\frac{3\cdot 10^8 m/s}{1.48}=2.02\cdot 10^8 m/s$

5 0

3 years ago

Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o

jolli1 [7]

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

$r = \frac{GM}{v^2}$

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

7 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$