The process by which metamorphic rock changes to igneous rock begins with?

A rocket is launched from a height of 3 m with an initial velocity of 15 m/s What is the maximum height of the rocket? When will

Fudgin [204]

If no extra acceleration is added to the rocket, then its velocity at time t is

v = 15 m/s - g t

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

Also, recall that

v² - u² = 2 a ∆x

where u is initial speed, v is final speed, a is acceleration, and ∆x is net displacement.

At the rocket's maximum height ∆x, the velocity is 0. So, the maximum height is

0² - (15 m/s)² = 2 (-g) ∆x

∆x = (15 m/s)² / (2 * (9.80 m/s²)) ≈ 11.48 m

But this assumes the rocket is launched from the ground. We're given that the rocket is launced from 3 m above the ground, so we need to add this to the height above. So the maximum height is closer to 14.48 m.

As mentioned before, this happens when vertical velocity is 0:

0 = 15 m/s - g t

t = (15 m/s) / (9.80 m/s²) ≈ 1.53 s

5 0

3 years ago

Two balls are dropped from rest and allowed to fall. If one ball is allowed to fall for 1 s and the other for 3 s compare the di

kirill115 [55]

The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

time of fall of the first ball, t = 1 s
time of fall of the second ball, t = 3 s

The distance traveled by each ball is calculated using the second equation of motion as shown below.

The distance traveled by the first ball is calculated as follows;

$h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m$

The distance traveled by the second ball is calculated as follows;

$h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m$

Thus, the second ball traveled a greater distance because it spent more time in motion.

Learn more here:brainly.com/question/5868480

3 0

2 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$