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dimaraw [331]
3 years ago
11

(only answer if you know the answer or I'll report) Solve it w the steps tysm​

Physics
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

 v_f = 20 m / s

Explanation:

For this exercise let's use the relationship between momentum and moment

        I = Δp

        F t = m v_f - m v₀

as the body starts from rest v₀ = 0

        F t = m v_f

        v_f = \frac{F t}{m}

let's calculate

        v_f = 4 2 / 0.4

         v_f = 20 m / s

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Black_prince [1.1K]

Answer:

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3 years ago
Un automovil transita por una curva en forma de U y recorre una distancia de 400m en 30s sin embargo su posición final está a so
pochemuha

Answer:

Definimos:

Rapidez media es igual al cociente entre la distancia recorrida y el tiempo que se tarda en recorrer esa distancia.

En este caso la distancia recorrida es 400m, y el tiempo que se tarda es 30s, entonces la rapidez media va a ser:

RM = 400m/30s = 13.33 m/s

La velocidad media por otro lado, es igual al cociente entre el desplazamiento y el tiempo necesario para desplazarse.

El desplazamiento es igual a la distancia entre la posición final y la posición inicial, que en este caso eso 40m, y el tiempo necesario sigue siendo 30s, entonces la velocidad media va a ser:

VM = 40m/30s = 1.33 m/s

8 0
3 years ago
Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha
Ksivusya [100]

Answer:

A_c=87.73*10^{21}m/s

Explanation:

From the question we are told that

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T=1.5 \times 10^{-16}

Generally the equation for velocity is mathematically given as

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V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

Generally the equation for Centripetal acceleration is mathematically given as

A_c=\frac{V^2}{r}

A_c=(\frac{20.944*10^5)}{r5*10^{-11}}

A_c=87.73*10^{21}m/s

8 0
3 years ago
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

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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
Sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motionstays in motion with the sam
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Answer:

D

Explanation:

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Ps- The object will stay moving in the same speed and direction.

5 0
3 years ago
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