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Olegator [25]
1 year 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]1 year 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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A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav
fenix001 [56]

Answer:

The peak current carried by the axon is 5.85 x 10⁻⁸ A

Explanation:

Given;

distance of the field from the axon, r = 1.3 mm

peak magnetic field strength, B = 9 x 10⁻¹² T

To determine the peak current carried by the axon, apply the following equation;

B = \frac{\mu I}{2\pi r}

where;

B is the peak magnetic field

r is the distance of the magnetic field from axon

μ is permeability of free space = 4π x 10⁻⁷

I is the peak current

Re-arrange the equation and solve for "I"

B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

7 0
1 year ago
Max and Jimmy want to jump on a trampoline. Max begins jumping in a steady pattern, making small waves in the trampoline. Jimmy
mylen [45]

Answer:

x_total = (A + B) cos (wt + Ф)

we have the sum of the two waves in a phase movement

Explanation:

In this case we can see that the first boy Max when he enters the trampoline and jumps creates a harmonic movement, with a given frequency. When the second boy Jimmy enters the trampoline and begins to jump he also creates a harmonic movement. If the frequency of the two movements is the same and they are in phase we have a resonant process, where the amplitude of the movement increases significantly.

         Max

               x₁ = A cos (wt + Ф)

         Jimmy

              x₂ = B cos (wt + Ф)

         

total movement

             x_total = (A + B) cos (wt + Ф)

 Therefore we have the sum of the two waves in a phase movement

8 0
1 year ago
A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform
kaheart [24]

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

B=\mu_{0} ni

Put the value into the formula

B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}

B_{c}=0.0000100\ T

B_{c}=10.02\times10^{-6}\ T

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance d=\dfrac{5}{2}

d=2.5\ cm

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

B_{w}=\dfrac{\mu_{0}I}{2\pi d}

Put the value into the formula

B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}

B_{w}=4.0\times10^{-6}\ T

This field is perpendicular to the wire.

The magnitude of magnetic field is

B=\sqrt{B_{c}^2+B_{w}^2}

B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}

B=0.00001078\ T

B=10.78\times10^{-6}\ T

B=10.78\ \mu\ T

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

6 0
1 year ago
Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it
Inessa [10]

Answer:

The average rate of energy transfer to the cooker is 1.80 kW.

Explanation:

Given that,

Pressure of boiled water = 300 kPa

Mass of water = 3 kg

Time = 30 min

Dryness friction of water = 0.5

Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

h_{f}=561.47\ kJ/kg

h_{fg}=2163.8\ kJ/kg

We need to calculate the enthalpy of water at initial state

h_{1}=h_{f}

h_{1}=561.47\ kJ/kg

We need to calculate the enthalpy of water at final state

Using formula of enthalpy

h_{2}=h_{f}+xh_{fg}

Put the value into the formula

h_{2}=561.47+0.5\times2163.8

h_{2}=1643.37\ kJ/kg

We need to calculate the rate of energy transfer to the cooker

Using formula of rate of energy

Q=\dfrac{m(h_{2}-h_{1})}{t}

Put the value into the formula

Q=\dfrac{3\times(1643.37-561.47)}{30\times60}

Q=1.80\ kW

Hence, The average rate of energy transfer to the cooker is 1.80 kW.

3 0
2 years ago
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olga nikolaevna [1]
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5 0
2 years ago
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