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2 years ago

6

I need Help ASAP I will give you brainlest and plus 5 stars ✨✨and plus thanks and 50 points EVeryONe need help this is due today

so plz help me I need to turn it in today so plz help I will give everything if you do

1 answer:

nikklg [1K]2 years ago

7 0

Answer:

12 is temperature, and 13 is artic and Antarctic oceans

Explanation:

I'm sure about 13, not so sure about 12

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What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT

sergiy2304 [10]

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

8 0

3 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago

How does inertia affect a person who is not wearing a seatbelt during a collision?

elena55 [62]

<span>The rule of inertia states that an object in motion will stay in motion unless another force has acted upon it. Because the person doesn't have their seatbelt on, they will keep moving. But if they were wearing a seatbelt, that would work as the force that is supposed to stop the person from flying forward.

Hope this helps :)
Please give brainliest</span>

6 0

3 years ago

A high diver dives into a swimming pool. His potential energy at the top is 10,000 J (relative to the surface of the pool). What

Semmy [17]

Answer:

Kinetic energy of diver at 90% of the distance to the water is 9000 J

Explanation:

Let d is the distance between the position of the diver and surface of the pool.

Initially, the diver is at rest and only have potential energy which is equal to 10000 J.

As the diver dives towards the pool, its potential energy is converting into kinetic energy due to law of conservation of energy, as total energy of the system remains same.

Energy before diving = Energy during diving

(Potential Energy + Kinetic Energy) = (Kinetic Energy + Potential Energy)

When the diver reaches 90% of the distance to the water, its kinetic energy

is 90% to its initial potential energy, as its initial kinetic is zero,i.e.,

K.E. = $\frac{90}{100}\times10000$

K.E. = 9000 J

6 0

2 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

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

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago