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ddd [48]
3 years ago
9

A panda is flying at 3 m/s using a hang-glider. If the panda’s mass is 120 kg and the hang glider’s mass is 50 kg, what is the p

anda’s speed immediately after it lets go of the hang glider? Explain.
Physics
1 answer:
WINSTONCH [101]3 years ago
7 0

Answer:

hmmmmm

Explanation:

(B)

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A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0
3 years ago
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Thorium ____ has 90 protons and 137 neutrons.
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Answer:

Thorium 227 (also known as Radioactinium)

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If a proton is released at the equator and falls toward earth under the influence of gravity, the magnetic force on the proton w
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Positive positive side naturally because protons always move to positive side and electron always move to negative side
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what is the process by which overlying pressure from rocks and soil reduces the size or volume of sediments.
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Read more on Brainly.com - brainly.com/question/14295948#readmore

Explanation:

6 0
4 years ago
2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to ma
Marizza181 [45]

Answer:

6.54 × 10⁻⁵ Pa-s

Explanation:

Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m

Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

So,  μ = F/A ÷ u/y

substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m

μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s

μ = 6.54 × 10⁻⁵ Ns/m²

μ = 6.54 × 10⁻⁵ Pa-s

5 0
3 years ago
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