Ask question

Ask question

All categories

Maksim231197 [3]

3 years ago

12

Assuming that 70 percent of the Earth’s surface is covered with water at average depth of 0.95 mi, estimate the mass of the wate

r on Earth. One mile is approximately 1.609 km and the radius of the earth is 6.37 × 10^6 m. Answer in units of kg. Your answer must be within ± 20.0%

1 answer:

Alla [95]3 years ago

8 0

Answer:

The mass of the water on earth is $5.4537\times10^{20}\ kg$

Explanation:

Given that,

Average depth h= 0.95 mi

$h=0.95\times1.609$

$h =1.528\ km$

$h=1.528\times10^{3}\ m$

Radius of earth $r= 6.37\times10^{6}\ m$

Density = 1000 kg/m³

We need to calculate the area of surface

Using formula of area

$A =4\pi r^2$

Put the value into the formula

$A=4\pi\times(6.37\times10^{6})^2$

$A=5.099\times10^{14}\ m^2$

We need to calculate the volume of earth

$V = Area\times height$

Put the value into the formula

$V=5.099\times10^{14}\times1.528\times10^{3}$

$V=7.791\times10^{17}\ m^3$

Now, 70 % volume of the total volume

$V= 7.791\times10^{17}\times\dfrac{70}{100}$

$V=5.4537\times10^{17}\ m^3$

We need to calculate the mass of the water on earth

Using formula of density

$\rho = \dfrac{m}{V}$

$m = \rho\times V$

Put the value into the formula

$m=1000\times5.4537\times10^{17}$

$m =5.4537\times10^{20}\ kg$

Hence, The mass of the water on earth is $5.4537\times10^{20}\ kg$

You might be interested in

If an object is thrown upward with an initial velocity of 128 ft/second, then its height after t seconds is given by the follow

IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

$h = 128t-32t^2$ ....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

$\dfrac{dh}{dt}=0$

From equation (I)

$\dfrac{dh}{dt}=128-64t$

$128-64t=0$

$t=\dfrac{128}{64}$

$t=2\ sec$

Now, for maximum height

Put the value of t in equation (I)

$h =128\times2-32\times4$

$h=128\ ft$

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

$128t-32t^2=0$

$t(128-32t)=0$

$128=32t$

$t=4\ sec$

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0

3 years ago

A train brakes from 25 m/s to rest in 30 sec. What is its deceleration?

givi [52]

Answer:

a= -0.83m\s^2

Explanation:

a = v \ t

a = -25 \ 30 = -0.833 m\s^2

the object is slowing down 0.83 meter every second

8 0

2 years ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

$\tau = I * \alpha (1)$

Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
τ = F*r (2)
For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
Replacing (2) and (3) in (1), we can solve for α, as follows:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
Replacing this value and (7) in (8), we get:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

b)

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
Replacing this value and (4), in (9), we get:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

$a_{c} = \omega^{2} * r (11)$

Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
Replacing this value and (7) in (11) we get:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$

6 0

2 years ago

Which best describes what occurs when an object takes in a wave as the wave hits it?.

statuscvo [17]

Answer:

Absorption

Explanation:

7 0

1 year ago

What are two ways an engineer can build a car in order for it to accelerate faster

Ket [755]

Explanation:

Take F=ma

a = F/m

For a higher, F higher or m lower

Means higher horse power for engine or lower mass for the car

4 0

2 years ago