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Vladimir79 [104]

2 years ago

10

list 8 countries which have launched satellites into space and also list the name of the satellite which was launched by the cou

ntries

1 answer:

Alja [10]2 years ago

7 0

Launch-capable countries
Order Country Satellite(s)
1 Soviet Union Sputnik 1
2 United States Explorer 1
3 France Astérix
4 Japan Ohsumi
10 more rows

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Which two stimuli did John B. Watson associate in his infamous “Little Albert” experiment? A. a white lab rat and the boy’s moth

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3 years ago

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2 years ago

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3 years ago

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The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35

azamat

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

$g= \frac{GM}{R^2}$

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

$g_{ss} = \frac{GM}{R+h^2}$

$g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}$

$g_{ss} = 8.82m/s^2$

Therefore the gravitational acceleration at 350km is $8.82m/s^2$

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3 years ago

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be

Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

F = $m \times a$

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

F = $m \times a$

= $1090 \times (\frac{21}{16})$

= 1430.625 N

Also, it is known that

$sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}$

$sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}$

$\theta$ = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

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2 years ago