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3 years ago

14

Which layer of the atom sphere contacts the surface of earth

1 answer:

Keith_Richards [23]3 years ago

4 0

The troposphere is the layer of the atmosphere which contacts the surface of the Earth. This layer is <span>0 to 12 km (0 to 7 miles).

Faith xoxo</span>

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I AN TIMED, HURRY!!

kramer

Answer:

C)0.59 N

Explanation:

By the Achemides principle we know that the buoyancy force on an object which is immersed in a incompressible fluid at rest is equal to the weight of the fluid displaced. So all we have to do is to find the mass of 60ml.

Density = mass / volume

Mass = density × volume

= 1 g/cm³× 60 cm³

{ as 1ml=1cm³}

= 60 g

So force equals to,

Force = weight = mass × gravitational acceleration

= 0.06×9.8 = 0.59 N

6 0

4 years ago

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5. A man in a car is listening to the radio. The radio station is broadcasting at a frequency of 85 MHz from two radio transmitt

nasty-shy [4]

Answer:

Speed = 0.00392 m/s

Explanation:

Solution:

Frequency of the radio = 85 MHz

If we have the frequency, we can calculate the wavelength of the radio wave.

As we know,

Frequency = speed of light/wavelength

wavelength = c/f

c = speed of light = 3 x $10^{8}$ m/s

So,

Wavelength = 3 x $10^{8}$ m/s / 85 x $10^{6}$ Hz

Wavelength = 3.5294 m

Man gets disturbed reception at t = 15 min

t = 15 x 60 = 900 s

t = 900 s

Speed = distance/time

Here, distance is wavelength. So,

Speed = 3.5294 m / 900 s

Speed = 0.00392 m/s

Hence, the man's car is going with speed of 0.00392 m/s

3 0

3 years ago

Which best explains why the shape of a liquid can change? (1 point) a The particles have just enough energy to move past each ot

FrozenT [24]

Answer:

The particles have just enough energy to move past each other.

4 0

3 years ago

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The x-axis of a trajectory represents its _____.

sineoko [7]

The x-acis of a trajectory represents its C

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4 years ago

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A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

7 0

3 years ago