Which of these statements is true?

If you're going 80 mph how long does it take to go 80 miles

Elza [17]

1 mile. Is this a joke lol

6 0

3 years ago

The gravitational field at the Moon in N/kg due to the Earth is approximately (G = 6.67 × 10-11 N m2/kg2, the mass of the Earth

rosijanka [135]

Answer:

F = 2.69 10⁻³ m [ N]

Explanation:

This exercise asks to calculate the gravitational field of the Earth on the lunar surface, let's use the universal gravitation law

F = G m M / r²

where m is the mass of the body, M the mass of the Earth and r the distance between the Earth and the Moon

F = (G M / r²) m

F = (6.67 10⁻¹¹ 5.98 10²⁴ / (3.85 10⁸)² ) m

F = 2.69 10⁻³ m [ N]

This force is directed from the Moon towards the Earth, therefore it reduces the weight of the body

8 0

3 years ago

Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th

dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

Let's write the diffraction equation for a slit

a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

θ = λ / a

In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

θ = 1.22 λ / D

Where D is the diameter of the pupil

Let's apply this equation to our case

θ = 1.22 600 10⁻⁹ / 0.460 10⁻²

θ = 1.591 10⁻² rad

This is the angle separation to solve the two light sources

6 0

3 years ago

A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and th

tamaranim1 [39]

Answer:

<em>The power required by the pump is nearly 230.588 kW</em>

Explanation:

Flow rate of the pump Q = 1 m^3/s

the head flow H = 20 m

specific weight of water γ = 9800 N/m^3

efficiency of the pump η = 85%

First note that specific gravity of water is the product of the density of water and acceleration due to gravity.

γ = ρg

where ρ is density. For water its value is 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

The power to lift this water at this rate will be gotten from the equation

P = ρgQH

but ρg = γ

therefore,

P = γQH

imputing values, we'll have

P = 9800 x 1 x 20 = 196000 W

But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.

we can say that

196000 W is 85% of the power of the pump power needed, therefore

196000 = 85% of $P_{p}$

where $P_{p}$ is the power of the pump needed

85% = 0.85

196000 = 0.85 $P_{p}$

$P_{p}$ = 196000/0.85 = 230588.24 W

<em>Pump power = 230.588 kW</em>

3 0

3 years ago

!!!!!URGENT!!!!!!

Vadim26 [7]

Answer:

The specific heat of air is 100 J/kg°C.

Explanation:

mass, m = 5 kg

Temperature change, T' - T = 37 °C - 22 °C = 15°C

heat, H = 7500 J

let the specific heat of air is c.

The formula of heat is

H = m c (T' - T)

7500 = 5 x c x 15

c = 100 J/kg°C

4 0

3 years ago