How does the mass of an object affect its motion through the air?

Answers

Answer 1

Answer:

The motion of an object through the air does not affect by its mass. The rate of fall of objects does not depend upon the mass.

What are free fall and air resistance?

Free fall is a motion of a body in which gravity is the only force acting upon it. An object moving upwards might not be considered to be falling. But if the object is under the effect of the force of gravity, it is said to be in free fall.

Free fall is a type of motion in which the force acting upon an object is only gravity. Objects are not encountering a significant force of airresistance as they are only falling under the sole influence of gravity. All objects under such conditions will fall with the same rate of acceleration, regardless of their masses.

As an object falls through the air, have gone through some degree of air resistance. Air resistance is the collisions of the object's leading surface with molecules present in the air. The two most common factors that have a direct effect on the amount of air resistance are the cross-sectional area of the object and the speed of the object.

Learn more about free-fall motion, here:

brainly.com/question/13297394

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Answer 2

Answer: Mass does not affect the speed of falling objects assuming there is only gravity acting on it. For example Both Bullets will strike the ground at the same time. The horizontal force applied does not affect the downward motion of the bullets — only gravity and friction (air resistance), which is the same for both bullets
(Sorry if u don’t get what I mean)

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A plank 2.00 cm thick and 13.0 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 68.4 kg is forced to stand on the very end. If the end of the board drops by 5.20 cm because of the man's weight, find the shear modulus of the wood.

Answers

Answer:

9.93 MPa

Explanation:

Given:

- mass of the man = 68.4 kg

- Deflection dx = 5.2 cm

- thickness of plank t = 2.0 cm

- width of plank w = 13.0 cm

- Length subtended L = 2.0 m

Find:

Shear Modulus of Elasticity S :

S = shear stress / shear strain

Shear stress = F / A

Shear stress = 68.4*9.81 / 0.02*0.13

Shear stress = 258078.4615 Pa

Shear strain = dx / L

Shear Strain = 0.052 / 2

Shear Strain = 0.026

Hence,

S = 258078.4615 / 0.026

S = 9.93 MPa

A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?

Answers

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

$S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n t=\sqrt{1.245*10^(-3) }\n t=0.035s$

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

$v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s$

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

$D=|vt|\nD=|0.346m/s*4.465s|\nD=1.54489m$

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

(a) How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A? (b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?

Answers

Explanation:

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

$=(v)/(c)\n v=(550-450)/(550)*3*10^(8)\n v=5.4545*10^(7)m/s$

For Part (b)

Since the apparent wavelength increases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

$=(v)/(c)\n v=(700-550)/(550)*3*10^(8)\n v=8.1818*10^(7)m/s$

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion. What is the car's velocity after 2.52s?
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?

Answers

Answer:

19.1 m/s

58.1 m

8.60 s

Explanation:

Take north to be positive and south to be negative.

Use Newton's second law to find the acceleration.

∑F = ma

-7850 N = (2500 kg) a

a = -3.14 m/s²

Given:

v₀ = 27.0 m/s

a = -3.14 m/s²

Find: v given t = 2.52 s

v = at + v₀

v = (-3.14 m/s²) (2.52 s) + 27.0 m/s

v = 19.1 m/s

Find: Δx given t = 2.52 s

Δx = v₀ t + ½ at²

Δx = (27.0 m/s) (2.52 s) + ½ (-3.14 m/s²) (2.52 s)²

Δx = 58.1 m

Find: t given v = 0 m/s

v = at + v₀

0 m/s = (-3.14 m/s²) t + 27.0 m/s

t = 8.60 s

Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

Answers

Answer:

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

$Electric FIeld = (Change\: in\: Potential)/(Distance)$

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance= $0.8 * 10^(-2) m$

So, Electric field is

$Electric\: field=(3.8V)/(0.8 * 10^(-2)m )$

$Electric\: field=475 V/m$

So, electric field is 475 Volts per meter.

Note : Also we can say 475 Newtons per coulomb

An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit. Required:
Find the radius and period of the orbit.