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natita [175]
3 years ago
10

At the surface of a certain planet, the gravitational acceleration g has a magnitude of 20.0 m/s^2. A 22.0-kg brass ball is tran

sported to this planet.
What is (a) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?
Physics
1 answer:
Oliga [24]3 years ago
3 0

Answer:

a, 22 kg and 22 kg

b, 215.8 N and 440 N

Explanation:

a

The mass of the ball remains constant and unchanged irrespective of where it has been to, need to go or is going. So, basically the mass of the ball on earth is as the same mass of the ball on the said planet, 22 kg

b

The weight of any object factors in the acceleration due to gravity of the said area(or planet).

W = mg, with m being the mass and g being the acceleration due to gravity.

On earth

W = 22 * 9.81 = 215.8 N

On the said planet,

W = 22 * 20 = 440 N

Do therefore, the weight is 215.8 N on earth and 440 N on the planet

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In a client with lower crossed syndrome, which of the following muscles is lengthened?
jek_recluse [69]

Answer:

Internal Oblique.

Explanation:

Lower crossed syndrome is a condition in which there are strong and weak muscles. So there is an imbalance of muscle strengths. It occurs when some muscles constanly get shortened or lengthened just like in this case internal oblique muscle got lengthened.

6 0
3 years ago
At this rate, what fraction of its mass would io lose in 4. 5 billion years?.
Pie

In 4.5 billion years, the amount of mass lost is 1.41912 * 10¹⁷ kg.

<h3>Equation</h3>

An equation is an expression used to show the relationship between two or more variables and numbers.

Given that:

Io loses about a ton (1000 kilograms) of sulfur dioxide per second to Jupiter's magnetosphere.

In 4.5 billion years, amount of mass lost = 1000 kg/s * (4.5 * 10⁹ * 365 * 60 60 * 24) = 1.41912 * 10¹⁷ kg.

In 4.5 billion years, the amount of mass lost is 1.41912 * 10¹⁷ kg.

Find out more on Equation at: brainly.com/question/13763238

4 0
2 years ago
A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total
worty [1.4K]

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

6 0
3 years ago
27. Improvements in space technology have greatly helped scientists better understand the stars, planets, and other
stellarik [79]

Answer:

B

Explanation:

Technological advancements in astronomy have led to the development of thousands of products such as

new materials, medical devices, and communications satellites.

For example, the production of instruments like telescopes, microscopes, and other measuring devices now enables us to identify things that we couldn't examine with the naked eye.

Similarly, the development of communication satellites has led to the emergence and improvements in the mobile network industry.

8 0
3 years ago
A charge of 5.67 x 10^-18 C is placed 3.5 x 10^-6 m away from another charge of -
Fudgin [204]
+ 1.58 e -15

Please hit thanks button! :)
5 0
2 years ago
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