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ruslelena [56]
2 years ago
14

Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming

that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9m2 .
Physics
1 answer:
Firlakuza [10]2 years ago
8 0

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

tan \ \theta = \dfrac{12}{100} \\ \\  tan \ \theta = 0.12 \\ \\  \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0

The drag force D = Wsinθ

\dfrac{1}{2}C_v \rho AV^2 = W sin \theta

where;

\rho = 1.23 \ kg/m^3

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}

C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}

C_D =0.9378

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madreJ [45]

Answer:

\huge\boxed{m = 70 \ kg}

Explanation:

<u>Given Data:</u>

Momentum = P = 700 kg m/s

Velocity = v = 10 m/s

<u>Required:</u>

Mass = m = ?

<u>Formula:</u>

P = mv

<u>Solution:</u>

m = P / v

m = 700 / 10

m = 70 kg

\rule[225]{225}{2}

Hope this helped!

<h3>~AnonymousHelper1807</h3>
5 0
2 years ago
What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J
FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0
2 years ago
José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg


8 0
2 years ago
Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart
Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0
2 years ago
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Damm [24]

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

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First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

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P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

5 0
2 years ago
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