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Scorpion4ik [409]
3 years ago
12

How does a plane and parachute manipulate air resistance differently?

Physics
1 answer:
dimaraw [331]3 years ago
3 0
A planes wings are slightly tilted catching the air as it passes, keeping it afloat
a parachute has a cupping effect where it can only take so much, so it slows down your descendence
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A 103 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 93 m below
siniylev [52]

Answer:

125.83672 seconds

Explanation:

P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)

m = Mass of professor = 103 kg

g = Acceleration due to gravity = 9.8 m/s²

h = Height of professor = 93 m

Work done would be equal to the potential energy

W=mgh\\\Rightarrow W=103\times 9.8\times 93\\\Rightarrow W=93874.2\ J

Power is given by

P=\frac{W}{t}\\\Rightarrow t=\frac{W}{P}\\\Rightarrow t=\frac{93874.2}{746}\\\Rightarrow t=125.83672\ seconds

The time taken by the horse to pull the professor is 125.83672 seconds

6 0
3 years ago
What does an atomic number represent in an atom?
Bezzdna [24]

The number of protons in an atom is known as the atomic number

6 0
2 years ago
Read 2 more answers
A regulation basketball has a 32 cm diameter
Rudik [331]

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

3 0
3 years ago
Andy took a bus and then walked from his home to downtown. For the first 1.6 hour, the bus drove at an average speed of 15 km/h
Yuliya22 [10]

<em>1.6x15=24 </em>

<em> </em>

<em>0.4x4.5=1.8 </em>

<em> </em>

<em>24+1.8=25.8 </em>

<em> </em>

<em>1.6+0.4=2 </em>

<em> </em>

<em>25.8/2=12.9</em>

4 0
3 years ago
Read 2 more answers
The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha
belka [17]

Given that,

Energy H=2.7\times10^{31}\ W

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

H=Ae\sigma T^4

A=\dfrac{H}{e\sigma T^4}

4\pi R^2=\dfrac{H}{e\sigma T^4}

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}

R=5.0\times10^{10}\ m

(b). Given that,

Radiates energy H=2.1\times10^{23}\ W

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}

R=5.42\times10^{6}\ m

Hence, (a). The radius of the star is 5.0\times10^{10}\ m

(b). The radius of the star is 5.42\times10^{6}\ m

8 0
3 years ago
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