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serious [3.7K]
3 years ago
9

As a motorcycle takes a sharp turn, the type of motion that occurs is called _______________ motion.

Physics
1 answer:
Darya [45]3 years ago
4 0

Answer:

circular motion

Explanation:

As a motorcycle takes a sharp turn, the type of motion that occurs is called circular motion.

Circular motion is a movement of an object along a circular path. As this motorcycle makes the sharp turn, it is acted upon by a centripetal force which directs the motorcycle towards the center.

Therefore, circular motion is the correct answer to the question.

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Considerando que você comece a caminhar em velocidade constante, inicialmente a 350 m de um ponto referencial escolhido. Você ca
ExtremeBDS [4]

Answer:

a. S(t)=350−1t

Explanation:

To determine the equation of motion you take into account the general form of motion with constant velocity:

S(t)=S_o+vt    ( 1 )

So is the initial position from a specific reference frame. In this case is 350 m.

v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.

You replace the values of So and v in the equation ( 1 ) and you obtain:

S(t)=350-(1m/s)t

Hence, the answer is:

a. S(t)=350−1t

- - - - - - - - - - - - - - - - - - - - -

Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:

             (1)

Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.

v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.

Você substitui os valores de So ev na equação (1) e obtém:

Portanto, a resposta é:

uma. S (t) = 350-1t, movimento retrógrado

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Answer:

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2 years ago
The magnitude of the electric force between two protons is 2.30 10-26 n. how far apart are they?
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The electric force between two charge objects is calculated through the Coulomb's law.
                               F = kq₁q₂/d²
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                           2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m. 
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The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th
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