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3 years ago

Which layer of the atmosphere is the top layer of the thermosphere?

2 answers:

8 0

The thermosphere is a layer of Earth's atmosphere. The thermosphere is directly above the mesosphere and below the exosphere. It extends from about 90 km (56 miles) to between 500 and 1,000 km (311 to 621 miles) above our planet.

Westkost [7]3 years ago

3 0

The exosphere is the outermost layer<span> of Earth's </span><span>atmosphere.</span>

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A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If th

tensa zangetsu [6.8K]

Answer:

The net charge is $67.89 \mu C$

Solution:

As per the question:

Mass of the plastic bag, m = 12.0 g = $12.0\times 10^{-3}\ kg$

Magnitude of electric field, E = $10^{3}\ N/C$

Angle made by the string, $\theta = 30^{\circ}$

Now,

To calculate the net charge, Q on the ball:

Vertical component of the tension in the string, $T = Tcos\theta$

Horizontal component of the tension in the string, $T = Tsin\theta$

Now,

Balancing the forces in the x-direction:

$Tsin\theta = QE$

$Q = {Tsin\theta}{E}$ (1)

Balancing the forces in the y-direction:

$Tcos\theta = mg$

where

g = acceleration due to gravity = $9.8\ m/s^{2}$

Thus

$T = \frac{mg}{cos\theta }$

$T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N$

Use T = 0.1357 N in eqn (1):

$Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C$

$Q = 67.89\times 10^{- 5}\ C = 67.89\mu C$

7 0

3 years ago

According to Newton's 3rd Law of Motion, Doug, a baseball

sweet-ann [11.9K]

B,
the reaction according to Newton’s third law must be an equal and opposite reaction. The force will be greater in order to bounce off of Doug’s bat

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3 years ago

You did 200 joules of work lifting a 150-newton backpack. How high did you lift the backpack?

kaheart [24]

We know that:
W=Fs
200J=150N*s
s=200J/150N
s=1,33m

8 0

3 years ago

A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is

lbvjy [14]

Answer:

$a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}$

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\$

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}$

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3 years ago

A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk

bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

5 0

3 years ago