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Bond [772]
3 years ago
11

The ionosphere lies between the mesosphere and exosphere is it true or false

Physics
1 answer:
Helga [31]3 years ago
6 0

False because the Ionsophere lies between the Mesosphere and the Theromsphere. If can can you give me brainliest :o ?

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A body is accelerated continuously. What is the form of the graph
Andrej [43]

Answer:

If a body is constantly accelerating the graph would be one-quadrant

4 0
3 years ago
When devising a model, scientists can only use the information available during their lifetime. This means that the current mode
11111nata11111 [884]

no, it not useless. we still learn Bohr's model in HS n dats almost 200 yr old! while there may be new models, previous one is good for explaining the basics. it is also useful to learn previous model n see how our understanding improves over time.


6 0
3 years ago
Read 2 more answers
(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface
Keith_Richards [23]

Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}

The expression of escape velocity is given by

v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s

The escape speed is 1777.92 m/s

3 0
3 years ago
Read 2 more answers
3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number
Rina8888 [55]

ANSWER

\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

F=\frac{GMm}{r^2}

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\  \\ F=4.67*10^{-9}\text{ }N \end{gathered}

That is the answer.

6 0
1 year ago
An object undergoes two successive displacements:
Gre4nikov [31]

The magnitude of the net displacement is 95.3 m

Explanation:

To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.

1st displacement is:

d_1=79 m at 16.9^{\circ}

So its components are

d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m

2nd displacement is:

d_2=16.7 m at 31.1^{\circ}

So its components are

d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m

Therefore, the x- and y-components of the net displacement are:

d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m

Therefore, the magnitude of the final displacement is:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

8 0
3 years ago
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