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3 years ago

Name:

Physics

1 answer:

aliya0001 [1]3 years ago

4 0

Answer:

Kinetic energy is 1000 J.

Explanation:

Given:

Mass of the bicycle, $m=80\textrm{ kg}$

Distance traveled, $d=100\textrm{ m}$

Time of travel, $t=20\textrm{ s}$

Unknowns:

Speed of the bicycler, $v$ .

Kinetic energy of the bicycler.

Equation:

Speed of the bicycler is, $v=\frac{d}{t}=\frac{100}{20}=5\textrm{ m/s}$

Therefore, the kinetic energy of the bicycler is given as,

$KE=\frac{1}{2}mv^{2}$

Substitute:

Plug in 5 m/s for $v$ and 80 kg for $m$ . Solve for KE.

Solve:

$KE=\frac{1}{2}\times 80\times (5)^{2}\\KE=40\times 25=1000\textrm{ J}$

Therefore, his kinetic energy is 1000 J.

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Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released qu

Kruka [31]

Answer: a) 1766 sec. b) 55.5 MJ c) 13.9 MW d) -12,944 Nm

Explanation:

a) The torque and the angular acceleration, are related by the following expression, which resembles very much to the Newton's 2nd Law for point masses:

ζ = I . γ, where ζ=external torque, I = rotational inertia and γ = angular acceleration.

We also know that a flywheel is a solid disk, so the rotational inertia for this type of body is equal to MR² / 2.

By definition, angular acceleration is the rate of change of angular velocity with time, so we can write the following:

γ = ωf -ω₀ / t

Assuming that the flywhel starts from rest, we know that ω₀ = 0, and ωf = 12,000 rpm.

As all the units are given in SI units, it is advisable to convert the rpm to rad/sec, as follows:

12,000 rpm = 12,000 rev. (2π/rev) . (1min/60 sec) = 400 π rad/sec

Returning to the original equation, we have:

ζ = MR² / 2 . (ωf/ t)

Replacing by the values, and solving for t, we have:

t = 250 Kg. (0.75)² m² . 400 π / 2. 50 Nm = 1,766 sec.

b) Due to the flywheel is just rotating, all the stored energy is rotational kinetic energy, which can be written as follows:

K = 1/2 I ωf² = 1/2 (MR²/2) ωf² = 1/4. 250 Kg. (0.75)² m². (400π)²

K= 55.5 MJ

c) Power is defined as energy delivered in a given time.

The energy delivered, is just the half of the originally stored value, i.e. , 55.5 MJ /2, equal to 27.75 MJ.

Dividing this value by 2.0 sec, we have the average power delivered to the machine, that we found to be equal to 27.75 MJ / 2s = 13. 9 MW

d) Using the same relationship than in a), we can write the following:

ζ = I. γ

I remains the same (as the flywheel is the same), so the only unknown is the angular acceleration.

Angular acceleration, by definition, is as follows:

γ = ωf - ω₀ / t

We know the value of ω₀, as it is the top speed value that we have already got,i.e., 400 π rad/sec.

We don't know the value for ωf, but we know the value of the rotational kinetic energy after 2.0 secs, which is equal to the half of the one we obtained in step b).

So, we can write the following:

Kf = 1/2 I ωf² = 1/2 (1/2 I ω₀²) ⇒ 1/ 2 ωf² = 1/4 ω₀² ⇒ωf = ω₀/√2

Replacing in the expression for angular acceleration:

γ = (ω₀/√2 - ω₀) / t = -0.29. 400. π/ 2 rad/sec²= -184.1 rad/sec²

Finally, we can get the torque as follows:

ζ = (250 kg. (0.75)² m² /2) . 184.1 rad/sec² = -12,944 Nm

6 0

4 years ago

On story ghost boy can you guys tell me something common and uncommon about them

Savatey [412]

Answer:

Sarah has no emotions

Explanation:

7 0

2 years ago

A cube icebox of side 3cm has a thickness of 5.0cm. If 4.0 kg of ice is put in the box estimate the amount of ice remaining afte

qaws [65]

Answer:

The amount of solid ice remaining after 6 hours is approximately 3.68664 kg

Explanation:

The given parameters are;

The side length of the cube box, s = 3(0) cm = 0.3 m

The thickness of the cube box, d = 5.0 cm = 0.05 m

The mass of ice in the box, m = 4.0 kg

The outside temperature of the cube box, T₁ = 45°C

The temperature of the melting ice inside the box, T₂ = 0°C

The latent heat of fusion of ice, $L_f$ = 3.35 × 10⁵ J/K/hr/kg

The surface area of the box, A = 6·s² 6 × (0.3 m)² = 0.54 m²

The coefficient of thermal conductivity, K = 0.01 J/s·m⁻¹·K⁻¹

For thermal equilibrium, we have;

The heat supplied by the surrounding = The heat gained by the ice

The heat supplied by the surrounding, Q = K·A·ΔT·t/d

Where;

ΔT = T₁ - T₂ = 45° C - 0° C = 45° C

ΔT = 45° C

Q = K·A·ΔT·t/d = 0.01 × 0.54 × 45 × 6× 60×60/0.05 = 104976

∴ The heat supplied by the surrounding, Q = 104976 J

The heat gained by the ice = $L_f$ × $m_{melted \ ice}$ =3.35 × 10⁵ J/kg × $m_{melted \ ice}$

Therefore, from Q = $L_f$ × $m_{melted \ ice}$ , we have;

Q = 104976 J = $L_f$ × $m_{melted \ ice}$ = 3.35 × 10⁵ J/kg × $m_{melted \ ice}$

104976 J = 3.35 × 10⁵ J/kg × $m_{melted \ ice}$

$m_{melted \ ice}$ = 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg

The mass of melted ice, $m_{melted \ ice}$ ≈ 0.31336 kg

∴ The amount of solid ice remaining after 6 hours, $m_{ice}$ = m - $m_{melted \ ice}$

Which gives;

$m_{ice}$ = m - $m_{melted \ ice}$ = 4.0 kg - 0.31336 kg ≈ 3.68664 kg

The amount of solid ice remaining after 6 hours, $m_{ice}$ ≈ 3.68664 kg.

8 0

3 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$