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3 years ago

13

Suppose the earth is shaped as a sphere with radius 4,0004,000 miles and suppose it rotates once every 24 hours. How many miles

along the equator does it rotate each hour? (approximation is acceptable)

1 answer:

Alenkinab [10]3 years ago

8 0

Answer:

1047 miles

Explanation:

The radius of the Earth is

$r = 4000$ (miles)

So its circumference, which is the total length of the equator, is given by

$L=2\pi r= 2\pi(4000)=25133 mi$

Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:

$L' = \frac{25133 mi}{24h}=1047 mi$

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A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)

kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = $\frac{4\pi}{3}\times(0.0991^2-0.0882^2)$

or

volume of sphere = 0.0012 m³

also, volume of half sphere = $\frac{\textup{Total volume}}{\textup{2}}$

or

volume of half sphere = $\frac{\textup{0.0012}}{\textup{2}}$

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density = $\frac{\textup{Mass}}{\textup{Volume}}$

or

Density = $\frac{\textup{0.568}}{\textup{0.0012}}$

or

Density = 474 kg/m³

8 0

2 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

3 0

3 years ago

A car traveling in a straight line has a velocity of 6m/s at some instant. After 6.32s its velocity is 13.2m/s . What is the ave

NeX [460]

Average acceleration = (change in speed) / (time for the change) .

Average acceleration = (13.2 - 6) / (6.32) = 7.2 / 6.32 = about <em>1.139... m/s²</em> .

8 0

3 years ago

A space probe produces a radio signal pulse. If the pulse reaches Earth 12.3 seconds after it is emitted by the probe, what is t

adoni [48]

Answer:

Distance = 3.69 × 10^9 m

The distance from the probe to Earth is 3.69 × 10^9 m

Explanation:

Distance from the probe to the Earth can be derived using the simple motion formula;

Distance = speed × time .....1

Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.

Speed of radio signal = speed of light = 3.0 × 10^8 m/s

time taken to reach the earth = 12.3 seconds

Substituting the values of speed and time into equation 1;

Distance = 3.0 × 10^8 m/s × 12.3 s

Distance = 36.9 × 10^8 m

Distance = 3.69 × 10^9 m

Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s

5 0

2 years ago

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

sp2606 [1]

From the question,
$u = 0m {s}^{ - 1}$
$v = 6m {s}^{ - 1}$

$t = 3s$
$F=12N$

Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

$Ft =m(v-u)$

$12(3.0)=m(6.0- \: 0)$
This implies that

$36.0 = 6m$
$m = \frac{36.0}{6.0}$
$\therefore \: m = 6.0kg$

You can also use the equation of linear motion,
$v = u + at$
$6 = 0 + a(3)$
$6 = 3a$
$a = \frac{6}{3}$

$a = 2 {ms}^{ - 2}$
But
$F=ma$
$12 = m(2)$
$12 = 2m$
$\frac{12}{2} = m$
$\therefore \: m = 6kg$

4 0

3 years ago