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beks73 [17]
2 years ago
13

1. (3 points) Convert the given distances to kilometers using the

Physics
1 answer:
Lena [83]2 years ago
4 0
That’s a lot ........
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How to find the density of air?
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You have to divide the pressure exerted by the air into two partial pressures: of the dry air and of the water vapor. Combining these two values gives you the parameter.
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3 years ago
What is the speed of sound in air that has a temperature of 19.9 celcius
klasskru [66]

Answer4.4 :

Explanation:

7 0
3 years ago
A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
enyata [817]

Answer:

W = 19.8 J

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

F = 14 lbf

F = 6.35 \times 9.8 N

F = 62.3 N

stretch in the spring is given as

x = 4 in = 0.1016 m

now we will have

F = kx

62.3 = k(0.1016)

k = 613.125 N/m

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

W = \frac{1}{2}kx^2

W = \frac{1}{2}(613.125)(0.254)^2

W = 19.8 J

7 0
3 years ago
an object at rest has no ________ energy, but it may have ________ energy resulting from its location or structure.
77julia77 [94]

Answer: first blank: kinetic

Second blank: potential

Explanation:

Hope this helps and please consider choosing me for Brainiest!

5 0
2 years ago
A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. Th
ipn [44]

Answer:

a) F_{fric} = 692 N

b) F_{applied} = 932 N

Explanation:

a)

According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object  in all directions.

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

F_{applied} = 800 N

Mass = m = 90 kg

acceleration = a = 1.2 m/s²

F_{fric} = ?

800 - F_{fric} = (90)(1.2)

F_{fric} = 692 N

b)

According to newton's second law of motion,

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

If we assume the same friction and acceleration between player's feet and ground as calculated in part a

F_{fric} = 692 N

acceleration = a = 1.2 m/s²

We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.

Mass = M = m₁ + m₂ = 110 kg + 90 kg

= 200 kg

F_{applied} = ?

F_{applied} - 692 N = (200)(1.2)

F_{applied} = 692 + 240

F_{applied} = 932 N

7 0
3 years ago
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