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2 years ago

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An object launched from Earth must attain a speed of 7,900 m/s to achieve a low orbit. What happens if the object’s maximum spee

d is less than 7,900 m/s?

1 answer:

Bingel [31]2 years ago

5 0

It will not achieve a state of low orbit and will not go into space.

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SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi

tino4ka555 [31]

Answer:

i = 0.477 10⁴ B

the current flows in the counterclockwise

Explanation:

For this exercise let's use the Ampere law

∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

L₀ = (y2-y1)

L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

L₁ = 3-0

L₁ = 3cm

B L = μ₀ I

B 2L = μ₀ I

i = 2 L B /μ₀

i= 2 0.03 / 4π 10⁻⁷ B

i = 4.77 10⁴ B

The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise

8 0

2 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

2 years ago

If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system

$W$ is the work done by the system on the surrounding

In this problem, the work done by the system is

$W=-225 kcal=-941.4 kJ$

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

$Q=-5 \cdot 10^2 kJ=-500 kJ$

with a negative sign as well because it is released by the system.

Therefore, by using the initial equation, we find

$\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ$

8 0

2 years ago

Describe how a metamorphic rock might become a sedimentary rock over time.

andriy [413]

<span>Metamorphic rock undergoes weathering, erosion; the particles are deposited and undergo lithification.</span>

7 0

2 years ago

A little girl riding a train rolls a ball toward the back of the train at 1.25 m/s NE. The train is traveling at a velocity of 1

Lelechka [254]

Answer:

Answer: 2.70m/s NE

Explanation:

Just did it.

4 0

3 years ago