Answer: *The inner planet that has two moons is Mars * the only planet in the Solar System with clockwise rotation is Venus * The surface of Mercury is covered with ancient magma ( which is similar to the spews of volcanoes on Earth as well) * The planet that is closest to Earth is Mercury * Venus has more volcanoes than any other planet * Earth's moon formed when a(n) can sometimes be called the Big Splash or can be called the Theia impact (Luna the moon formed from the ejecta of the collision between the Proto- Earth and Mars sized planet) * Mars is called the red planet because it's soil contains the element iron oxide ( which is the compound that gives blood and rust hue)

Explanation: Hope this helped :)

8 0

3 years ago

A certain circuit is composed of two series resistors. The total resistance is 10 Ohms. One of the resistors is 4 Ohms. The othe

Mariulka [41]

<h3>Given :- </h3>

A certain circuit is composed of two series resistors
The total resistance is 10 ohms
One of the resistor is 4 ohms

We have to find the value of other resistor?

<h3>Let's Begin :-</h3>

We know that,

In series combination,

When a number of resistances are connected in series, the equivalent I.e resultant resistance is equal to the sum of the individual resistances and is greater than any individual resistance

That is,

Rn in series = R1 + R2 + R3.....So on

Therefore,

According to the question,

We have,

R1 + R2 = 10 Ω

4 + R2 = 10Ω

R2 = 10 - 4

R2 = 6Ω

Hence, The value of R2 resistor in series is 6Ω

4 0

2 years ago

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

$a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2$

Let d be the breaking distance. It can be calculated using third equation of motion as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m$

So, the required distance covered by the car is 44.64 m.

4 0

3 years ago