Explain how the biosphere,hydrosphere,geosphere, and atmosphere are interconnected

How did the horizontal velocity vector component change during the flight of the cannonball in the simulation

arlik [135]

Answer:

The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.

Explanation:

First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight

7 0

4 years ago

Elephants are loaded into containers that are pulled up a ramp an inclined plane at 15 degrees to the ground .The container is h

tankabanditka [31]

Given:

Inclined at 15 degrees

Tension of 5000N

Mass of container 800kg

Required:

Mass of elephant

Solution:

F = mgcos(theta)

5000 = m (9.8m/s2)cos(15)

M = 528.2kg

8 0

4 years ago

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

Alexandra [31]

Answer:

The maximum no. of electrons- $2.25\times 10^{22}$

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

$I = \frac{Q}{t}$

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

$Q = 1.0\times 3600 = 3600\ C$

Maximum number of electrons, n is given by:

$n = \frac{Q}{e}$

where

e = charge on an electron = $1.6\times 10^{- 19}\ C$

Thus

$n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}$

3 0

4 years ago

NASA launches a satellite into orbit at a height above the surface of the Earth equal to the Earth's mean radius. The mass of th

Rzqust [24]

Answer:

a) 3h 59’ b) 5,590 m/s c) 1,910 N

Explanation:

The only force acting on the satellite (neglecting the influence of any other body) is just the attractive force from Earth, which is given by the Universal Law of Gravitation:

Fg = G* ms*me / (res² (1)

where: G = 6.67* 10⁻¹¹ N.m² / kg² ms = 780 kg, me= 5.97* 10²⁴ kg, and

res = 2* 6.38* 10⁶ m (distance between the center of the Earth and the satellite).

Fg = 1,910 N (answer c)

At the same time, this force, is the centripetal force that keeps the satellite in orbit, and that can be written as follows:

Fg = ms * v² / res (2)

By definition of velocity, we can say that the constant speed at which the satellite orbits, can be expressed as the quotient between the distance travelled around the Earth once, and the time needed to do that, which is called the period of the orbit:

v = 2*π*res / T (3)

We can solve for v first, taking the right sides of (1) and (2), as follows:

G* ms*me / (res)²= ms * v² / res

v = √(G*me/r) = 5,590 m/s (answer b)

Once obtained the value of v, we can replace in (3), and solve for T:

T = 2* π*res / v = 3 h 59 min (answer a)

7 0

3 years ago

Give me three examples of fossil fuel-based energy sources pls

svet-max [94.6K]

Gasoline, kerosene, and coal.

4 0

4 years ago