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Olegator [25]
2 years ago
12

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

h bus drives a constant 55 mph, find the rate at which the distance between the buses is changing when they are 13mi apart, heading toward each other.

Physics
1 answer:
Oksanka [162]2 years ago
4 0

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h#The speed toward each other.

\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

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klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

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3 years ago
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Red shift of distant galaxies
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Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?​
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Answer:

\huge\star{\underline{\mathtt{\blue{Answer}}}}\huge\star...

<h2>PE=<em>work done</em></h2><h2><em>m</em><em>gh</em><em>=</em><em>2</em><em>0</em><em>×</em><em>1</em><em>0</em><em>×</em><em>2</em><em>0</em><em>.</em><em>.</em></h2>

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2 years ago
This equation is used to calculate the properties of a gas under nonideal conditions.
goblinko [34]

Answer:

Van der Waal's equation

Explanation:

The Van der Waal's equation is use to calculate the properties of a gas under nonideal or real gases conditions.

(P + \frac{an^2}{V^2})(V-nb) = nRT .

Here P, V ,T ,n and R have usual meaning as in the ideal gas equation

that is PV=nRT

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3 years ago
Light enters an equilateral prism with an incident angle of 35° to the normal of the surface. Calculate the angle at which the
julia-pushkina [17]

Answer:

65.9°

Explanation:

When light goes through air to glass

angle of incidence, i = 35°

refractive index, n = 1.5

Let r be the angle of refraction

Use Snell's law

n=\frac{Sini}{Sinr}

1.5=\frac{Sin35}{Sinr}

Sin r = 0.382

r = 22.5°

Now the ray is incident on the glass surface.

A = r + r'

Where, r' be the angle of incidence at other surface

r' = 60° - 22.5° = 37.5°

Now use Snell's law at other surface

\frac{1}{n}=\frac{Sinr'}{Sini'}

Where, i' be the angle at which the light exit from other surface.

\frac{1}{1.5}=\frac{Sin37.5'}{Sini'}

Sin i' = 0.913

i' = 65.9°

4 0
3 years ago
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