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2 years ago

Are outer planets gaseous

Physics

1 answer:

bija089 [108]2 years ago

6 0

Answer:

Yes

Explanation:

The four outer planets are all gas giants made primarily of hydrogen and helium. They have thick gaseous outer layers and liquid interiors.

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What is the gravitational acceleration close to the surface of a planet with a mass of 9ME and radius of 3RE, where ME and RE ar

Papessa [141]

Answer:

9.78 m/s²

Explanation:

To solve this, we use the gravitational formula

g = GM/r², where

g = acceleration due to gravity

G = gravitational constant

M = mass of the planet

r = radius of the planet

From the question, we got that the mass of the planet is

M = 9ME, where ME = 5.95*10^24

M = 9 * 5.95*10^24

M = 5.355*10^25 kg

Also, the Radius of the planet, R = 3RE, where RE = 6.37*10^6

R = 3 * 6.37*10^6

R = 1.911*10^7 m

On applying the values of both R and M to the equation, we get

g = GM/r²

g = (6.67*10^-11 * 5.355*10^25) / (1.911*10^7)²

g = 3.57*10^15/3.65*10^14

g = 9.78 m/s²

Therefore, the acceleration due to gravity on the planet is 9.78 m/s²

Please vote brainliest if it helped you <3

5 0

3 years ago

Determine the image distance and image height for a 5.00 cm tall object placed 30.0 cm from a double convex lens with a focal le

vlabodo [156]

Answer:

The image distance is 30 cm

image height = - 5 cm

Explanation:

The formula for calculating the image distance is expressed as

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

From the information given,

u = 30

f = 15

By substituting these values into the formula,

1/15 = 1/30 + 1/v

1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30

Taking the reciprocal of both sides,

v = 30

The image distance is 30 cm

magnification = image height/object height = - v/u

Given that object height = 5 cm, then

image height/5 = - 30/30 = - 1

image height = - 5 * 1

image height = - 5 cm

8 0

1 year ago

A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a

3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec

Explanation:

We have given acceleration of the car $a_1=5.9m/sec^2$

Acceleration of the stock car $a_2=3.6m/sec^2$

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So $s_1=s_2$

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So $\frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6$

After solving t = 3.56 sec

(b) From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m$

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0

3 years ago

What role does team collaboration play in successfully planning a mission to Mars?

mihalych1998 [28]

Answer:it helps get everyone in the same page

Explanation:

This is important so everyone knows what they should be doing

4 0

2 years ago

Describe the general relationship between the weight of adult men and the average number of Calories needed daily to maintain a

Alexeev081 [22]

Answer:

The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking.

The amount of calories contained in the food we eat every day must represent the amount of calories eliminated by the body in that time to have a steady weight.

Explanation:

The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking. If total quantity of calories in the food we consume every day is higher that the average number of calories needed daily, then weight increases by fat accumulation.

3 0

2 years ago