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Ksivusya [100]
3 years ago
11

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi

ned by the equation d = 144 – 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. For which interval of time is Greg’s car moving in the air?
0 -3 t>3
0
Physics
1 answer:
Talja [164]3 years ago
3 0
The solution to the problem is as follows:

 <span>Average = 80 
So Sum = 80 * 5 = 400 
Mode = 88, so two results are 88 (if three results were 88, then the median would be 88). 
Three results are 81, 88, and 88. 
That leaves 143. We could still have one 81 score, so that leaves the lowest score as 62. 

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 - 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground? 
d = 144 - 16t2 
0 = 144 - 16t2 
16t^2=144 
t^2=9 
t=3</span>
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You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be
Alex

Answer:

The amplitude  is  A =  90.2 \ m

Explanation:

From the question we are told that

    The frequency of when sound is approaching observer is   f = 392 Hz

     The frequency as the move away from observer  is  f_ a =  330 \ Hz

    The time between the pitch are t =  10 \ s

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

       T =  2 t

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

          T = \frac{2 \pi}{w}

Where  w is the angular velocity

=>   \frac{2 \pi}{w}  =  2 * 10

=>   w =  0.314 \ rad/sec

Now using Doppler Effect,

   The source of the sound is approaching the observer

The

          f = f_o (\frac{v}{v- wA} )

         392  = f_o (\frac{v}{v- wA} )

Where A is the amplitude

    So when the source is moving away from the observer

         f_a =  f_o (\frac{v}{v+ wA} )  

        330  =  f_o (\frac{v}{v+ wA} )  

Here  f_o is the fundamental frequency

Dividing the both equation  we have

           \frac{392}{330}  =  \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}

           1.1878  = \frac{v+wA}{v-wA}

         1.1878 v -  1.1878 wA = v+wA

        1.1878 v = 2.1878 wA

=>     A =  \frac{(0.1878 * (330))}{(2.1878)* (0.314)}

         A =  90.2 \ m

7 0
3 years ago
Atom X contains seven protons and seven neutrons. atom Z contains seven protons and eight neutrons. which of the following state
stiks02 [169]

Answer:

The correct answer is option A: they are isotopes.

Explanation:

From atom X we know that the number of protons is 7 and the number of neutrons is 7 and from atom Z we know that the number of protons is 7 and the number of neutrons is 8.

Since the number of protons of atom X and atom Z is the same, we have that atom X and atom Z is the same element. The difference in the number of neutrons tells us that atom X and atom Z are isotopes. Remember that an isotope is one element that has atoms with different numbers of neutrons.            

The mass number is given by:              

A = p + n

Where <em>n</em> is the number of neutrons and <em>p </em>is the number of protons.

For atom X and atom Z we have:

A_{x} = 7 + 7 = 14

A_{z} = 7 + 8 = 15

Hence, they have a different mass number.  

We know that the element with 7 protons is nitrogen. The first isotope is ^{14}_{7}N and the second isotope is ^{15}_{7}N. Both isotopes are stables (they are not radioactive).  

Therefore, the correct answer is option A: they are isotopes.

I hope it helps you!                                                          

3 0
3 years ago
A Porsche challenges a Honda to a 400 m race. Because the Porsche's acceleration of 3.4 m/s2 is larger than the Honda's 3.0 m/s2
padilas [110]

Answer:

Winner wins by 0.969 s

Explanation:

For the Porche:

Given:

Displacement of Porsche s = 400 m

Acceleration of Porsche a = 3.4 m/s^2

From Newton's second equation of motion,

s = ut + (1/2) a t^2 (u = 0 as the car was initially at rest)

Substituting the values into the equation, we have

t^2 = (2 * 400) / 3.4

= 235.29 / 3.4

t = 15.33 s

For the Honda:

Displacement of Honda = 310 m

Acceleration of Honda = 3 m/s^2

Applying Newton's second equation of motion

s = ut + (1/2) a t^2 (u = 0 for same reason)

Substituting the values into the equation, we obtain

t^2 = (2 * 310) / 3

= 620 / 3

t = 14.37 s

Hence

The winner (honda) wins by a time interval of = 15.33 - 14.37    

=0.969 s

8 0
3 years ago
Weight on planet Mars ​
Sunny_sXe [5.5K]

Mars: 0.38

weight = mass x surface gravity

multiplying your weight on Earth by the number above will give you your weight on the surface of Mars

If you weigh 150 pounds (68 kg.) on Earth, you would weigh 57 lbs. (26 kg.) on Mars

5 0
3 years ago
Read 2 more answers
A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C
Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

d_{ab} = distance traveled from station A to station B

t_{ab} = time of travel between station A to station B

we know that

Time = \frac{distance}{speed}

t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}

Total distance traveled is given as

d = d_{ab} + d_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }

Given that :

d_{ab} = 4 d_{bc}

So

v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}

2)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

t_{ab} = time of travel between station A to station B

d_{ab} = distance traveled from station A to station B

we know that

distance = (speed) (time)

d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = time of travel for train from station B to station C

we know that

distance = (speed) (time)

d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}

Total distance traveled is given as

d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}

Given that :

t_{ab} = 3 t_{bc}

So

v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}

4 0
3 years ago
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