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Len [333]
3 years ago
11

A car traveling east at 46.8 m/s passes a trooper hiding at the roadside. the driver uniformly reduces his speed to 25.0 m/s in

3.70 s. (a) what is the magnitude and direction of the car's acceleration as it slows down? magnitude m/s2 direction (b) how far does the car travel in the 3.70-s time period? m
Physics
1 answer:
Firlakuza [10]3 years ago
8 0
(a). a=5, 89 m/s west.
(b) x=132,82 m.
You might be interested in
The number of kilograms of water in a human body varies directly as the mass of the body. Upper AA 9393​-kg person contains 6262
Pie

Answer:

54 kg

Explanation:

Mass of person = 93 kg = m_p

Mass of water = 62 kg = m_w

Dividing the above two masses we get

\frac{m_p}{m_w}=\frac{93}{62}\\\Rightarrow \frac{m_p}{m_w}=1.5\\\Rightarrow m_p=1.5m_w

Hence, the mass of the person is 1.5 times the mass of the water in them

Now, Mass of person = 81 kg = m_p

m_p=1.5m_w\\\Rightarrow m_w=\frac{1}{1.5} m_p\\\Rightarrow m_w=\frac{1}{1.5} \times 81\\\Rightarrow m_w=54\ kg

So, the mass of water in a person that has mass of 81 kg is 54 kg

3 0
3 years ago
Help me please....................
maksim [4K]
You need to write the question for the solution.
Add the question then only can get the answer
8 0
4 years ago
A force of 50n is applied to an object at an angle of 50 degrees measured relative to the horizontal. Determine the horizontal a
kobusy [5.1K]

Answer:

Fx = 32.14 [N]

Fy  = 38.3 [N]

Explanation:

To solve this problem we must decompose the force vector, for this we will use the angle of 50 degrees measured from the horizontal component.

F = 50 [N]

Fx = 50*cos(50) = 32.14 [N]

Fy = 50*sin(50) = 38.3 [N]

We can verify this result using the Pythagorean theorem.

F = \sqrt{(32.14)^{2}+ (38.3)^{2}} \\F = 50 [N]

3 0
3 years ago
A ruby laser delivers a 16.0-ns pulse of 4.20-MW average power. If the photons have a wavelength of 694.3 nm, how many are conta
stepan [7]

Answer:

The  value is  n  =  2.347 *10^{17} \  photons

Explanation:

From the question we are told that

     The  amount of power delivered is  P  =  4.20 \  M W  =  4.20  *10^{6} \  W

      The  time taken is  t =  16.0ns  =  16.0 *10^{-9} \  s

       The  wavelength is  \lambda  =  694.3 \  nm =  694.3 *10^{-9} \  m

     

Generally the energy delivered is  mathematically represented as

     E  =  P  * t  =  \frac{n  *  h  *  c  }{\lambda }

Where  h is the Planck's constant with value  h  =  6.262  *10^{-34} \  J \cdot  s

           c  is the speed of light with value  c =  3.0*10^{8} \  m/s

     

So  

    4.20 *10^{6}  *  16*10^{-9}=  \frac{n  *  6.626 *10^{-34}  *  3.0*10^{8}  }{694.3 *10^{-9}}

=>    n  =  2.347 *10^{17} \  photons

4 0
3 years ago
Listed below are the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones. Use the given data to const
Fantom [35]

Answer:

The 5-number summary is

1. Median = 0.93 W/kg

2. Lower quartile = 0.69 W/kg

3. Upper quartile = 1.16 W/kg

4. Minimum value = 0.54 W/kg

5. Maximum value = 1.42 W/kg

Explanation:

We are given the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones.

1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57

What is 5-number summary?

A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.

These statistical characteristics are:    

1. Median

2. Lower quartile

3. Upper quartile  

4. Minimum value  

5. Maximum value  

1. Median:

Arrange the data in ascending order

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

(n+1)/2 gives the median value of the data set.

(11 + 1)/2 = 6th position

Therefore, 0.93 W/kg is the median of the data set.

2. Lower quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The lower quartile is the median of the lower half of the data set.

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

The median is 6/2 = 3rd position

Therefore, the lower quartile of the data set is 0.69 W/kg

3. Upper quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The upper quartile is the median of the lower half of the data set.

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The median is 6/2 = 3rd position

Therefore, the upper quartile of the data set is 1.16 W/kg

4. Minimum value:

The minimum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the minimum value of the data set is 0.54 W/kg

5. Maximum value  

The maximum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the maximum value of the data set is 1.42 W/kg

The box plot is illustrated in the attached diagram.

6 0
3 years ago
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