Which best illustrates projectile motion

How is cold front formation different from stationary front formation?

padilas [110]

Answer:

A

Explanation:

4 0

3 years ago

Identify and sketch all the external forces acting on the chair. because the chair can be represented as a point particle of mas

SOVA2 [1]

There are two external force acts on the chair.

1. The force due to earth gravity, acts in the downward direction.

2. Reaction force of the gravity, which acts in the Upward direction (Normal Force).

On every object, there is a force acts due to gravity of earth, which pulls the object towards the centre of earth, known as gravity force, always acts in the downward direction. Mathematically it's given as

F=mg

here, m is the mass of the object, and g is the acceleration of gravity.

To balance this gravity force, a counter force acts in the opposite direction, whose magnitude is equal to the force of gravity

8 0

3 years ago

Read 2 more answers

A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper nee

baherus [9]

Answer:

120.125 m

Explanation:

Density = Mass/volume

D = m/v .............................. Equation 1.

Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.

Making v the subject of the equation,

v = m/D............................... Equation 2

Given: m = 76.5 kg,

Constant: D = 8960 kg/m .

Substituting into equation 2

v = 76.5/8960

v = 0.0085379 m³

Since the copper sphere is to be drawn into wire,

Volume of the copper sphere = volume of the wire

v = volume of the wire

Volume of wire = πd²L/4

Where d = diameter of the wire, L = length of the wire.

Note: A wire takes the shape of a cylinder.

v = πd²L/4 ........................ equation 3.

making L the subject of the equation,

L = 4v/πd²..................... Equation 4

Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095 and π = 3.14

Substitute into equation 4

L = 4×0.0085379/(3.15×0.0095²)

L = 0.0341516/0.0002843

L = 120.125 m.

L = 120.125 m

Thus the length of the wire produced = 120.125 m

4 0

3 years ago

As SCUBA divers go deeper underwater, the pressure from the weight of all the water above them increases tremendously which comp

faltersainse [42]

Answer: The volume of gas expands because of the decrease in pressure as he tries to exit the water body, therefore he must take necessary precaution.

Explanation:

Using Boyle's law which states that the the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature

ie P1VI=P2V2

A diver absorbs compressed nitrogen gas when he dives into the water body, As he ascends out of the water body having less pressure, the volume of nitrogen gas which he absorbs will tend to expand following Boyle's Law. Therefore a scuba driver should not rises quickly but slowly to the surface or else the expanding nitrogen gas can cause tiny bubbles in his blood and tissue to form together with joints pains and eventually cause decompression sickness needing medical attention.

5 0

3 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$