Which characteristic of the planets in our solar system increases as the distance from the sun increase?

zhannawk [14.2K]

The answer is D. The temperature obviously doesnt rise slower or faster, and if you are heating an object, it would make no sense to say that less heat is being transferred.

8 0

2 years ago

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Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 19.0 cmcm and carries

Lemur [1.5K]

Answer:

Explanation:

magnetic field due to circular wire

= μ₀ i / 2r

i is current and r is radius of coil .

Magnetic fields due to inner coil

μ₀ x 20 / (2 x 9.5 x 10⁻²)

Magnetic field due to outer coil

= μ₀ x I / (2 x 19 x 10⁻²) , I is the current to be calculated

Total field

μ₀ x 20 /( 2 x 9.5 x 10⁻²) +μ₀ x I / (2 x 19 x 10⁻²) = 0

20 + I /2 = 0

I = - 40 A

Current required is 40 A , and it will be in opposite direction.

7 0

2 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

2 years ago

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block

MariettaO [177]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

7 0

2 years ago

A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti

Pavel [41]

If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

d = √(120)²+(100)²
d = 156.2 km

So, the ship will have to travel 156.2 km to the northwest direction.

8 0

2 years ago