Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

Gravity at a height h above the surface of the moon will be given as:

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

3 years ago

The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of

kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

$P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}$

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

$=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m$

Δh_air=714m

7 0

3 years ago

An element becomes a positively charged ion when it

Anuta_ua [19.1K]

An element becomes a + ion when it...loses electrons

6 0

3 years ago

Read 2 more answers

A car travels for 5.5 h at an average speed of 75 km/h. How far did it travel? T TE

ANEK [815]

Answer:

$\boxed {\boxed {\sf 412.5 \ kilometers }}$

Explanation:

Distance is the product of speed and time.

$d=s*t$

The speed of the car is 75 kilometers per hour. It traveled for 5.5 hours.

$s= 75 \ km/hr \\t= 5.5 \ hr$

Substitute the values into the formula.

$d= 75 \ km/hr * 5.5 \ hr$

Multiply. Note that the hours will cancel each other out.

$d= 75 km * 5.5 \\d= 412.5 \ km$

The car travelled 412.5 kilometers.

5 0

3 years ago

It takes a truck 3.56 seconds to slow down from 112 km/h to 87.4 km/h. What is its average acceleration?

maxonik [38]

Answer:

option B

−1.92 m/s2

Explanation:

Given in the question,

time took by truck to slow down = 3.56 sec

initial speed of truck = 112 km/h

final speed of truck = 87.4 km/h

1 km/h = 0.277778 m/s

112 = 31.1 m/s

87.4 = 24.28 m/s

Formula use to calculate the acceleration

v - u = at

where v is final speed

u is initial speed

a is acceleration

t is time

plug values in the equation

24.28 - 31.1 = a(3.56)

-6.8 = a(3.56)

a = -6.8 / 3.56

a = -1.9 m/s²

3 0

4 years ago