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Degger [83]
2 years ago
9

3. You are flying 2586 miles from San Francisco to New York. An hour into the flight, you are 600

Physics
2 answers:
Andrei [34K]2 years ago
6 0

Answer:

268.22 m/s

Explanation:

An hour into the flight, you are 600 miles from San Francisco.

This sentence says that speed is 600 mi/h.

Now we need to convert it into m/s.

1 mi = 1609.34 m

1 h = 60 min = 60 min *60 s/1min = 3600 s

600 mi/h * 1609.34 m/1 mi * 1h/ 3600s= 600*1609.34/3600 m/s= =268.22 m/s

Citrus2011 [14]2 years ago
3 0

Answer:

268.2 m/s (1dp)

Explanation:

600 miles = 965,606m (1mile = 1609.34m)

965606/60 = 16,0943.43 ..... (metres traveled every minute)

16,0943.43/60 = 268.2238 .....

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What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and th
saveliy_v [14]

Answer:

The concentration of mole evil at oxygen on that day is 0.00858 mol/L

Explanation:

Here, we want to calculate the concentration of molecular oxygen

The pressure on that day is 1.0 atm

Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm

Now let’s calculate the concentration;

From Ideal gas law;

PV = nRT

This can be written as;

P/RT = n/V

The term n/V refers to concentration;

Let’s make substitutions now;

P = pressure = 0.21 atm

R = molar gas constant = 0.0821 L•atm/mol•k

T = temperature = 25 = 25 + 273.15 = 298.15 K

Substituting these values, we have;

n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L

6 0
2 years ago
A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
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From the geometry, the normal component of Fp can be calculated: 
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The total normal force Fn then is: 
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= 1721.63 N</span>

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
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> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

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2 years ago
Which pair of sentences is describing the same velocity? A car is parked. A car is moving in circles. A bus drives 40 miles per
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The bus and the truck have the same velocity.

Also, Valerie and Owen have the same velocity.
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A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th
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Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s<span>².</span>
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A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

7 0
2 years ago
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