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kirill [66]
3 years ago
13

What is an astronomical unit (AU)?

Physics
1 answer:
avanturin [10]3 years ago
4 0

The AU ... Astronomical Unit ... used to be defined as the average distance between the Sun and Earth during the year.

Now it's defined as 149,597,870,700 meters exactly.

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SONAR was developed during World War II, to find submarines. Submarines were sneaking up on ships underwater and releasing torpe
Reika [66]
As long as you describe the chemical reactions within the torpedo and explain how torpedoes advanced modern warfare.
3 0
4 years ago
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What happens to particle spacing when temperature increase, What we call this process?
Slav-nsk [51]

Answer:

The atoms start vibrating really fast, making the space between them larger. It's an increase in Kinetic Energy.

6 0
4 years ago
Which of the balls will exert the smallest force on object A?Why
olganol [36]
The 1kg ball would exert the smallest force.

As force = mass x gravity, this means that the smaller the mass (kg), the lesser the force.

When the mass is lighter (1kg):

Force = mass x gravity
Force = 1 x 9.8
Force = 9.8N

Compared to when the mass is heavier (10kg)

Force = mass x gravity
Force = 10 x 9.8
Force = 98N

Where this proves that the lighter the mass, the smaller the force exerted.
6 0
3 years ago
How is the mass and speed of a particle related to its kinetic energy and temperature?
Brut [27]
Higher the mass will cause more friction which will make its temperature higher and it would also need more kinetic energy to move at a higher speed<span />
6 0
3 years ago
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where
const2013 [10]

Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:  

W = ^{x_{2}}_{x_{1}}\int F_{x} dx

Where:

x₁: is the lower limit = 0 m    

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction =  (4.5 + 13.7x - 1.5x²)N

W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx  

W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}  

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}    

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}      

W = 29.85 J

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!                                

6 0
3 years ago
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