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Anestetic [448]
3 years ago
15

Find the acceleration of a train whose speed increases from 7m/s to 17 m/s in 120s

Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0
Initial speed of the train = 7 m/s
Final speed of the train = 17 m/s
Change of speed of the train = (17 - 7) m/s
                                               = 10 m/s
Time taken for the change of the speed of the train = 120 s
Then
Acceleration of the train = Change of speed of the train/Time taken for the change of speed
                                       = 10/120 m/s^2
                                       = 1/12 m/s^2
                                       = 0.083 m/s^2
So the acceleration of the train is 1/12 meter per second square or 0.083 m/s^2. I hope this is the answer you were looking for.
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A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string
Darya [45]

Answer:

a

The  speed of  wave is   v_1  = 129.1 \ m/s

b

The new speed of the two waves is v =  129.1 \ m/s

Explanation:

From the question we are told that

    The mass of the string is  m  =  60 \ g  =  60 *10^{-3} \ kg

    The length is  l  =  2.0 \ m

    The tension is  T  = 500 \ N

Now the velocity of the first wave is mathematically represented as

     v_1  = \sqrt{ \frac{T}{\mu} }

Where  \mu is the linear density which is mathematically represented as

      \mu  =  \frac{m}{l}

substituting values    

     \mu  =  \frac{ 60 *10^{-3}}{2.0 }

     \mu  =  0.03\ kg/m

So

   v_1  = \sqrt{ \frac{500}{0.03} }

   v_1  = 129.1 \ m/s

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

     

8 0
3 years ago
at the center of the neutral metal block, what is the direction of the electric field contributed by the charge in and on the me
elena-14-01-66 [18.8K]

The direction of electric field by the charge in and on the metal block will be along the direction line 5 as given in question.

<h3>How to determine electric field direction in a metal block?</h3>

The charge always remain on outer surface of metal and inside the metal block, the net electric field is zero. But due to dipole there is an electric field at the center of metal block i.e. at point R along direction line 1.

Now, to make make the net electric field zero at center, the electric field by the charge in and on the metal block must be equal in magnitude to that of electric field due to dipole at point R and in opposite direction to that of the net electric field at at R due to dipole.

The electric field by the charge in and on the metal block will be making 180° angle to the  electric field due to dipole at point R.

Hence the direction of electric field by the charge in and on the metal block will be along the direction line 5 as given in question.

To know more about electric field, click on brainly.com/question/11509296

#SPJ4

4 0
1 year ago
What is the definition of force according to Newton's first law of motion?​
11111nata11111 [884]
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force
8 0
3 years ago
Read 2 more answers
A quarter-wave monopole radio antenna (also called a Marconi antenna) consists of a long conductor of one quarter the length of
sasho [114]

Answer:

a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m

b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m

Explanation:

(a) Radiowave wavelength= λ = c/f

As we know, Radiowave speed in the air = c = 3 x 10^8 m/s

f = frequency = 604 kHz = 604 x 10^3 Hz

Hence, wavelength = (3x10^8/604x10^3) m

λ = 496.69 m

So the height of the antenna BROADCASTING AT 604 kHz =  λ /4 = (496.69/4) m

= 124.17 m

(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz  (1kHZ = 1000 Hz)

Hence, wavelength =  λ = (3 x 10^8/1710 x 10^3) m

 λ= 175.44 m

So, height of the antenna =  λ /4 = (175.44/4) m

= 43.86 m  

5 0
3 years ago
An astronaut holds a rock 100 m above surface of Planet X. The rock is then thrown upwards with a sleek of 15m/s. The rock reach
Gelneren [198K]

Answer:5 m/s^{2}

Explanation:

This problem is related to vertical motion, and the equation that models it is:

y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2} (1)

Where:

y=0m is the rock's final height

y_{o}=100 m is the rock's initial height

V_{o}=15 m/s is the rock's initial velocity

\theta=90\° is the angle at which the rock was thrown (directly upwards)

t=10 s is the time

g is the acceleration due gravity in Planet X

Isolating g and taking into account sin(90\°)=1 :

g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t) (2)

g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s)) (3)

g=5 m/s^{2} (4) This is the acceleration due gravity in Planet X

5 0
3 years ago
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