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Marta_Voda [28]
3 years ago
14

A biker is pedaling at a constant speed of 36 km/h. During the last 10 s of the race, he increases his speed with a constant acc

eleration of 0.5 m/s2 . Calculate at which speed he crosses the finish line.
Physics
1 answer:
adell [148]3 years ago
4 0

Answer:

54 km/h

Explanation:

given,

speed of the biker = 36 Km/h

time = 10 s

acceleration = 0.5 m/s²

speed at which it crosses the finish line  = ?

v = 36 x 0.278 = 10 m/s

using equation of motion

v = u + a t

v = 10 + 0.5 x 10

v = 15 m/s

v = 15 x 3.6 = 54 km/hr

speed at which the biker crosses the finish line is equal to 54 km/h

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nalin [4]
A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
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4 0
2 years ago
A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
2 years ago
2. A stone is thrown horizontally at a speed of 6.0 m/s from the top of a cliff 78.4 m high.
Ostrovityanka [42]

Answer:

a) 8 seconds if you are using earth's gravity.

b) 48m if the velocity does not change

c) 9.8m/s

Explanation:

3 0
3 years ago
A liquid is used to make a mercury-type barometer. The barometer is intended for space-faring astronauts. At the surface of the
Anarel [89]

Answer:

Density of liquid = 4730 kg/m³

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Explanation:

Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm

So, ρgh = ρ₁gh₁

ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³

The atmospheric pressure on planet X

P = ρg₁h₃     g₁ = g/4 and h₃ = 725 mm = 0.725 m

on planet X

P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²

6 0
3 years ago
Does this diagram illustrate the second law of thermodynamics? Why or why not
Savatey [412]
<h2>Answer:</h2>

The diagram is not showing the second law of thermodynamics. It is the demonstration of 1st law of thermodynamics.

<h3>Explanation:</h3>

Second law of thermodynamics describes the entropy of the system increase with time, it does not decrease with time. It is constant for ideal systems.

While in first law of thermodynamics, it is stated that the energy of a system can not be lost but it is transferred from one form to other form.

And in this picture, it is shown that the energy released from heat source to cold sink is used in doing work.

Work and heat are forms of energy.

5 0
3 years ago
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