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4vir4ik [10]
2 years ago
11

An airplane cruises at 850 km/h relative to the air. It is flying from Denver, Colorado, due west to Reno, Nevada, a distance of

1200 km, and will then return. There is a steady 90km/h wind blowing to the east.
Physics
1 answer:
skad [1K]2 years ago
8 0

Answer:

difference in flight time= 0.3023 hour

Explanation:

The question is incomplete, but I found it in your textbook.

Spped of aircraft = 850 km/h

Opposing speed of wind = 90km/h

Hence, the net speed when it's travelling west = 850 - 90 = 760 km/hr

The distance covered = 1200km

time taken = distance/ time = 1200/ 760 = 1.5789 hours

When coming back, the speed of the wind is complementary to the speed of the aircraft so

net speed when it's coming back = 850 +90 = 940 km/hr

time taken in this instance = 1200/ 940 = 1.2765 hours

Hence, the difference in flight time= 1.5789 - 1.2765 = 0.3023 hour

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The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies
algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0
2 years ago
When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a
Oxana [17]

Answer:

5 mg, 5\cdot 10^{-3}g

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

m=5\cdot 10^{-3}g

If  we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

1 g : 1000 mg = 0.005 g : x

and we find

x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg

4 0
2 years ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For maximum emf, \sin\omega t=1

So,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

So, the maximum emf generated in the loop is 5.32 V.

3 0
2 years ago
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s
Anastaziya [24]

Answer:

the correct answer is c     v₁> 12.5 m / s

Explanation:

This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.

            v² = v₀² + 2 a₁ x

as part of rest v₀ = 0

           a₁ = v² / 2x

           a₁ = 25² / (2 120)

           a₁ = 2.6 m / s²

now we can find the velocity for the distance x₂ = 60 m

           v₁² = 0 + 2 a1 x₂

           v₁ = Ra (2 2,6 60)

           v₁ = 17.7 m / s

these the speed at 60 m

we see that the correct answer is c     v₁> 12.5 m / s

5 0
2 years ago
A 28-kg particle exerts a gravitational force of 8.3 x 10^-9 N on a particle of mass m, which is 3.2 m away. What is m? A) 140 k
xxTIMURxx [149]

Answer:

Mass of another particle, m = 46 kg  

Explanation:

it is given that,

Mass of first particle, m₁ = 28 kg

Gravitational force, F=8.3\times 10^{-9}\ N

Distance between the particles, d = 3.2 m

We need to find the mass m of another particle. It is given by the formula as follows :

F=G\dfrac{m_1m}{d^2}

m=\dfrac{Fd^2}{Gm_1}

m=\dfrac{8.3\times 10^{-9}\ N\times (3.2\ m)^2}{6.67\times 10^{-11}\times 28\ kg}

m = 45.5 kg

or

m = 46 kg

So, the correct option is (d) "46 kg". Hence, this is the required solution.

6 0
2 years ago
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