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3 years ago

What forces were exerted during the air resistance experiment?

Physics

1 answer:

zepelin [54]3 years ago

6 0

The forces that were exerted during the air resistance was the gravity.

Explanation:

Air resistance is the frictional power air applies against a moving article. As an article moves, air opposition backs it off. The quicker the article's movement, the more prominent the air obstruction applied against it.

The force acting upon the objects dropped in the air was air resistance and gravity. The force acting upon the objects dropped in the vacuum was gravity only.

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Halleys comet has period of 75.3 years. Using Kepler’s third law, find it’s semimajor axis expressed in astronomical units?

natta225 [31]

Answer: 17.83 AU

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. 

$T^{2}\propto a^{3}$ (1)

Talking in general, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite, comet) orbiting a greater body in space with the size $a$ of its orbit.

However, if $T$ is measured in years, and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes:

$T^{2}=a^{3}$ (2)

This means that now both sides of the equation are equal.

Knowing $T=75.3years$ and isolating $a$ from (2):

$a=\sqrt[3]{T^{2}}=T^{2/3}$ (3)

$a=(75.3years)^{2/3}$ (4)

Finally:

$a=17.83AU$ (5)

4 0

3 years ago

Two poles are connected by a wire that is also connected to the ground. The first pole is 20ft tall and the second pole is 10ft

Natasha2012 [34]

Answer: 31.6ft

Explanation:

Check the attachment for the diagram.

According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|

that is 20ft - 10ft = 10ft

According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.

|AE|^2 + |EC|^2 = |AC|^2

10^2 + 30^2 = |AC|^2

100 + 900 = |AC|^2

|AC| = √1000

|AC| = 31.6ft

Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.

6 0

4 years ago

jarptica [38.1K]

Answer:

5.03

Explanation:

trust me

7 0

3 years ago

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

4 years ago

A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(

Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

$\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}$

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0

3 years ago