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2 years ago

N which order did the events forming our solar system occur?

Physics

2 answers:

Ugo [173]2 years ago

6 0

Answer:

Explanation:

brilliants [131]2 years ago

4 0

Answer:

The solar nebula became hot and dense because of that it pulling in more gas. This flattened into a rotating disk. It spun faster and faster, forming the Sun.

Explanation:

hope this helps

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The mass of a large car is 1000 kg. How much force would be required to accelerate the car at a rate of 3 m/sec2?

Lubov Fominskaja [6]

Answer: 3000 Newton

Explanation:

Mass of large car = 1000 kg.

Acceleration of the car = 3 m/sec2

Force = ?

Recall that force is the product of mass of an object and the acceleration by which it moves.

Thus, Force = Mass x Acceleration

Force = 1000 kg x 3 m/sec2

= 3000 Newton

Thus, 3000 Newton of force would be required to accelerate the car

8 0

3 years ago

A charge of 3.20 μC is held fixed at the origin. A second charge of 3.20 μC is released from rest at the position (1.25 m, 0.570

siniylev [52]

Answer:

a) $v = 7.137 m/s$ and b) $r = 1.832\ m$

Explanation:

We have to analyze this problem from the point of view of energy conservation. In this case there are two kind of energy, electric potential energy and kinetic energy. First, at $t = t_1$ there isn't relative movement between the two charges, so kinetic energy is zero and the total energy ( $E_T = E_p + E_k$ ) is just potential.

$E_T = E_p = \frac{k q_1 q_2}{r}$

where $k$ is the Coulomb constant, $q_1$ and $q_2$ are the two interacting charges, and $r$ is the distance between them.

Considering the fixed charge at (x1,y1) = (0,0) and the second one at (x2, y2) = (1.25, 0.57), the initial distance is

$r = \sqrt{(x_2-x_1)^2 + (y_2- y_1)^2} = \sqrt{(1.25)^2 + (0.57)^2} = 1.374\ m$ , then if

$k =8.987\times 10^9 N m^2/C^2$ and $q_1 = q_2 =3.2\times 10^{-6} C$ ,

$E_T = E_p = \frac{k q_1 q_2}{r} = \frac{8.987\times 10^9 * 3.2\times 10^{-6}*3.2\times 10^{-6}}{1.374} = 6.698 \times 10^{-2} Nm$ .

Now, at $t = t_2$ , $r \rightarrow \infty$ and $E_p \rightarrow 0$ . This means all the energy is kinetic

$E_T = E_k = \frac{1}{2}mv_f^2$ , so

$v_f = \sqrt{2E_T/m} =\sqrt{2 *6.698 \times 10^{-2} /0.00263} = 7.137 m/s$ (mass in Kg).

That would be the velocity when the second charge moves infinitely far from the origin.

For the second part we have that $v = v_f/2$ , so kinetic energy is

$E_k = \frac{1}{2}mv^2 = 1.674 \times 10^{-2} Nm$

and potential energy is

$E_p = E_T - E_k = 6.698 \times 10^{-2} - 1.674 \times 10^{-2} = 5.024 \times 10^{-2} Nm$

so the distance is

$r = \frac{k q_1 q_2}{E_p} = 1.832\ m$

6 0

3 years ago

Why does a change in the food web affect the balance of an ecosystem?

Deffense [45]

Answer: C) In an ecosystem, prey and predator populations affect each other.

Think of cause and effect: The wolves keep the deer population in check because they hunt them. If the wolf population increases, the deer population goes down. If you reduce the number of wolves than the deer population goes up.

Hope this helps :)

7 0

3 years ago

Most of the energy we use on earth comes from the sun—how does that energy(light and thermal)end up?

Lady bird [3.3K]

Most of the energy we use on earth comes from the sun—how does that energy(light and thermal)end up?

As ____energy in our food

8 0

2 years ago

Water leaves a fireman's hose (held near the ground) with an initial velocity v0=19.5v0=19.5 m/s at an angle θθ= 34 degrees abov

valina [46]

Answer:

tmax = 1.11 s, d = 35.98 m

Explanation:

Here is the complete question

Water leaves a fireman's hose (held near the ground) with an initial velocity

v 0 = 19.5 m/s at an angle θ = 34 degrees above the horizontal. Assume the water acts as a projectile that moves without air resistance. Use a Cartesian coordinate system with the origin at the hose nozzle position.

Using v 0 , θ , and g, write an expression for the time, t max, the water travels to reach its maximum vertical height. At what horizontal distance d from the building base, should the fireman place the hose for the water to reach its maximum height as it strikes the building? Express this distance, d, in terms of v 0 , θ , and g.

Solution

Since the water leaving the hose is considered to be a projectile motion with initial velocity,v₀ = 19.5 m/s and an angle θ = 34. The time tmax it takes the water to reach maximum height is given by

tmax = v₀sin θ/g = 19.5 × sin34/9.8 = 19.5 × 0.5592/9.8 = 10.904/9.8 = 1.113 s ≅ 1.11 s

The horizontal distance,d from the base of the building at which the hose must be placed to reach maximum height is the range of the projectile and is given by

d = v₀²sin2θ/g = 19.5²sin(2×34)/9.8 = 19.5²sin68/9.8 = 380.25 × 0.9272/9.8 = 352.562/9.8 = 35.98 m

3 0

3 years ago