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prohojiy [21]
3 years ago
15

Is air made of matter ?​evidence and reasoning.

Physics
1 answer:
Fiesta28 [93]3 years ago
4 0
Yes, air is matter because it is a gas composed mostly of nitrogen, oxygen, and argon, and gases do have mass because they’re molecules bouncing around a contained volume (here, the atmosphere)
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What is a bond between a positive and a negative ion charge?
klasskru [66]

Answer:

Ionic Bond

Explanation:

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3 years ago
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how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5
Alexandra [31]

Answer: a) 12857.1 m/s/s b) 578.6 N

Explanation:

Impulse = change in momentum

Ft = mV2 - mV1

V = AT, 45 / .0035 = 12857.1 m/s/s

(b) .045 x 12857.1 = 578.6 N

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3 years ago
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1. How does the structure of a hard-boiled egg compare to the crust, mantle, and core structure of Earth? What do the egg's laye
lutik1710 [3]

Answer:

2. Earth layers compare in egg layer it is hard on the outside

Explanation:

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3 years ago
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A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0
3 years ago
A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what
pychu [463]
By definition we have that the final speed is:
 Vf² = Vo² + 2 * a * d
 Where,
 Vo: Final speed
 a: acceleration
 d: distance.
 We cleared this expression the acceleration:
 a = (Vf²-Vo²) / (2 * d)
 Substituting the values:
 a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
 a = -77268 mi / h ^ 2
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
 First you must make a free body diagram and see the acceleration of the car:
 g = 32.2 feet / sec ^ 2
 a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
 a = -31.48 feet / sec ^ 2
 A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
 A = -22.07 feet / sec ^ 2
 Clearing the braking distance:
 Vf² = Vo² + 2 * a * d
 d = (Vf²-Vo²) / (2 * a)
 Substituting the values:
 d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
 d = 175.44 feet
 answer:
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
5 0
3 years ago
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