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nignag [31]
3 years ago
11

A cart is propelled over an xy plane with acceleration compo- nents ax 4.0 m/s2 and ay 2.0 m/s2. Its initial velocity has com- p

onents v0x 8.0 m/s and v0y 12 m/s. In unit-vector notation, what is the velocity of the cart when it reaches its greatest y coordinate?
Physics
1 answer:
Tasya [4]3 years ago
4 0

Answer:

(-16.0\ \hat i+0\ \hat j)\ \rm m/s.

Explanation:

<u>Given:</u>

  • x-component of acceleration of the cart, \rm a_x = 4.0\ m/s^2.
  • y-component of acceleration of the cart, \rm a_y = 2.0\ m/s^2.
  • x-component of initial velocity of the cart, \rm v_{0x} = 8.0\ m/s.
  • y-component of initial velocity of the cart, \rm v_{0y} = 12\ m/s.

Let the x and y components of the final velocity of the cart when the cart reaches its greatest y coordinate be \rm v_x\ and \ v_y respectively.

When the cart reaches its greatest y-coordinate then the y component of its final velocity should be zero, \rm v_y =0.\\

Let the cart reaches its greatest y-coordinate at time t, then using the following equation:

\rm v_y = v_{0y}+a_yt\\0=12+2\times t\\\Righttarrow t=\dfrac{-12}{2}=-6\ s.

At this time, the x component of the velocity of the particle is given by

\rm v_x = v_{0x}+a_xt=8.0+4.0\times (-6)=8.0-24.0=-16.0\ m/s.

Thus, the velocity of the cart when it reaches its greatest y coordinate, in unit-vector notation, is given by

\vec v = v_x\hat i+v_y\hat j=(-16.0\ \hat i+0\ \hat j)\ \rm m/s.

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Complete Question

The complete question is shown on the first uploaded image

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The value is |v| = 6.93

Explanation:

From the question we are told that

    The initial point is (x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )

    The  terminal point is  (x_2 , y_2 , z_2) = (-5 , 11, 8 )

Generally the magnitude of the vector is mathematically represented as

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g Initially, the motorcycle travels along a straight road with a speed of 35 m/s (this is almost 80 mph). The maximum decelerati
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Given:

Initial speed of the motorcycle (u) = 35 m/s

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Maximum deceleration of the motorcycle (a) = -1.2 m/s²

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By substituting values in the equation, we get:

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4 0
3 years ago
If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial accelerat
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Answer:

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Explanation:

Given Data

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Mass of electron Me=9.109×10⁻³¹ kg

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Charge of electron qe= -e = -1.602×10⁻¹⁹C

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Solution

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From Newtons Second Law of motion

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