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mario62 [17]
3 years ago
15

Which of these is an example of acceleration? A a bicyclist turning around a corner B a car traveling south with its cruise cont

rol set C a bus traveling north at 25m/s D a helicopter waiting to take off
Physics
1 answer:
Airida [17]3 years ago
4 0

The answer is A.

Acceleration means when you go faster in speed. Since the bicyclist is turning, he has to slow down but when he finishes turning, he is going to pedal harder and gain more speed. This is called Acceleration.

Best of Luck!

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An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle
Dovator [93]

Answer:The velocity of the train is 3.84m/s

Explanation:

According to the Doppler effect, if the source is moving towards you then the apparent frequency of the sound emitted by the source is higher and if the source is moving away from you then the apparent frequency of the sound emitted by the source is smaller.

This is given by:

fo = V +-Vo/ V +-Vo × source

Where fo= observed frequency

V= velocity of sound

Vo= vo it of the observer

fsource= frequency the source

Given:

Observed frequency of the approaching train fo1= 452Hz

The observed frequency of train= fo2= 442Hz

Velocity of sound= 334m/s

Velocity of source=?

Train approaching the observer is given by:

fo1= V/(V - Vs)× source ...eq1

Train passes the student is given by:

fo2= V/(V - Vs)×source ...eq2

Divide eq1 by eq2

452/442 = (343+Vs)/(343 - Vs)

1.02 =(343+Vs)/(343 -Vs)

Cross multiply

1.02(343- Vs) = 343 + Vs

350.76 - 1.02Vs = 343 + Vs

Collecting like terms

350.76 -343= 1.02Vs+ Vs

7.76 = 2.02Vs

Vs= 7.76/2.02

Vs= 3.84m/s

4 0
3 years ago
How many main retrofit strategies help solve the ecological problems with suburbs?
iragen [17]

ANSWER:

5

Explanation:

Because they are elven in numbers

3 0
2 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
STATION 1
jok3333 [9.3K]
Chedda chease snadwhik
8 0
3 years ago
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between
erma4kov [3.2K]

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   COP  =  \frac{T_i}{T_o - T_i}

Here T_i is the inside temperature

while  T_o is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   \frac{T_i}{T_o - T_i} heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         Q \ \alpha \ (T_o - T_i)

=>        Q= k (T_o - T_i)

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   \frac{T_i}{T_o - T_i}  amount of heat

   E  unit of electricity will remove  Q= k (T_o - T_i)

So

      E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }

=>   E = \frac{k}{T_i} (T_o - T_i)^2

given that  \frac{k}{T_i} is constant

    =>  E \  \alpha  \  (T_o - T_i)^2

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  T_i   =  30 ^oC

 and      T_o  =  40 ^oC

Hence  

     E = K (T_o - T_i)^2

Here K stand for a constant

So  

        E = K (40 -  30)^2

=>      E = 100K

Now if  the  T_i   =  20 ^oC

Then

       E = K (40 -  20)^2

=>      E = 400 \ K

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       Q = k (T_o - T_i )^{\frac{1}{2} }

So

       E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }

=>   E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }

Assuming \frac{k}{T_i} is a constant

Then  

     E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

   

4 0
3 years ago
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