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FinnZ [79.3K]
3 years ago
7

PART ONE

Physics
1 answer:
stepladder [879]3 years ago
5 0

Answer:

1. A1, B2, C3

2. 47.1°

Explanation:

Sum of forces in the x direction:

∑Fₓ = ma

f − Fᵥᵥ = 0

f = Fᵥᵥ

Sum of forces in the y direction:

∑Fᵧ = ma

N − W = 0

N = W

Sum of moments about the base of the ladder:

∑τ = Iα

Fᵥᵥ h − W (b/2) = 0

Fᵥᵥ h = ½ W b

Fᵥᵥ (l sin θ) = ½ W (l cos θ)

l Fᵥᵥ sin θ = ½ l W cos θ

The correct set of equations is A1, B2, C3.

At the smallest angle θ, f = Nμ.  Substituting into the first equation, we get:

Nμ = Fᵥᵥ

Substituting the second equation into this equation, we get:

Wμ = Fᵥᵥ

Substituting this into the third equation, we get:

l (Wμ) sin θ = ½ l W cos θ

μ sin θ = ½ cos θ

tan θ = 1 / (2μ)

θ = atan(1 / (2μ))

θ = atan(1 / (2 × 0.464))

θ ≈ 47.1°

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Mandarinka [93]

Answer:

F=9.09\times 10^{-9} N

Explanation:

We are given that

Mass of spherical shell,m_1=1900 kg

Mass=m_2=1.80 kg

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

F=G\frac{m_1m_2}{r^2}

G=6.67\times 10^{-11} Nm^2/kg^2

Using the formula

F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}

F=9.09\times 10^{-9} N

Hence,the gravitational force =F=9.09\times 10^{-9} N

6 0
3 years ago
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los A
WARRIOR [948]

Answer:

a.Beth

b.2232 s

Explanation:

We are given that

Distance,d=400 mi

Speed of Alan,v=45 mph

Speed of Beth,v'=55 mph

a.Time =\frac{distance}{speed}

Using the formula

Time taken by Alan=\frac{400}{45}=8.89 hr

Time taken by Beth=\frac{400}{55}=7.27hr

Alan will reach San Francisco at 4:53 PM

Beth will reach San Francisco at 4:16 PM

Beth will reach before Alan.

b.Difference between time=8.89-7.27=1.62 hr

t=1.62 hr

1.62-1=0.62 hr

0.62 hr=0.62\times 60\times 60=2232 s

Hence, Beth has to wait 2232 s for Alan to arrive .

6 0
3 years ago
In the real world, How does conduction, convection and radiation work? (Examples needed) (WILL MARK BRAINLIST)
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3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

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}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
Read 2 more answers
A car tire is filled with air at pressure 325000 pa at 303 K. When the tire cools,its pressure reduces to 299000 Pa. What is the
max2010maxim [7]

Answer:

new temperature of the tire will be 278.76 K

Explanation:

when the temperature increases, the particles will have greater kinetic energy and also the pressure will be increase for the gas particles.

so when the temperature increases, pressure  will also increase  and vice versa

 T is directly proportional to P

T1  = initial temperature= 303 k

P1= Initial pressure =  325000 pa

T2= Final temperature=  ?

P2= Final pressure =   299000 pa

mathematically

P1/T1= P2/T2

T2= P2 x T1/P1

T2 = 299000 x 303/ 325000=  278.76 k

6 0
3 years ago
Read 2 more answers
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