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3 years ago

12

Which characteristic of the gas giants decreases with increasing distance from the sun?

2 answers:

hodyreva [135]3 years ago

7 0

Equator diameter and Number of moons. So both a and c would be the answer :)

anyanavicka [17]3 years ago

4 0

Gas giant are plants with mass 10 times bigger than the mass of the Earth. They are also called outer planets. There are four gas giant planets in our solar system: Jupiter, Saturn, Uranus and Neptune. The farther away from the Sun they are (the distance from the Sun is increased) the longer they orbit the Sun because of the great distance.

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2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

<em>weight of the load, w = 557 N</em>
<em>force applied , F = 410 N</em>
<em>angle of force, = 15°</em>
<em>coefficient of kinetic friction = 0.46</em>
<em>distance moved, d = 6.5 m</em>

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

<em>m is the mass of the recycling bin</em>
<em /> $F_k$ <em> is the frictional force </em>

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0

2 years ago

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) = B0 cos(ωt) ˆz.

Alexxandr [17]

Answer:

Explanation:

Given that,

B(t) = B0 cos(ωt) • k

Radius r = a

Inner radius r' = a/2 and resistance R.

Current in the loop as a function of time I(t) =?

Magnetic flux is given as

Φ = BA

And the Area is given as

A = πr², where r = a/2

A = πa²/4

Then,

Φ = ¼ Bπa²

Φ(t) = ¼πa²Bo•Cos(ωt)

Then, the EMF is given as

ε(t) = -dΦ/dt

ε(t) = -¼πa²Bo • -ωSin(ωt)

ε(t) = ¼ωπa²Bo•Sin(ωt)

From ohms law,

ε = iR

Then, i = ε/R

I(t) = ¼ωπa²Bo•Sin(ωt) /R

This is the current induced in the loop.

Check attachment for better understanding

7 0

2 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

Helga [31]

Answer: $3.23 m/s^2$

Explanation:

Given

$\mu_s =0.330$

Frictional Force is balanced by force due to car acceleration

Frictional force $F_s$

$F_s=ma_{min}$

$\mu_sN=ma_{min}$

$\mu_s\cdot mg=ma_{min}$

$a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2$

6 0

2 years ago

8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?

Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

$q_1=q_2=+1C$ are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

$F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N$

8 0

3 years ago

What must exist for the engineering process to begin?

Vlad1618 [11]

Answer:

d

Explanation:

rzp-yyib-oiv

plèase jóin ón góógle mèèt

5 0

2 years ago

Read 2 more answers