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AfilCa [17]
2 years ago
15

The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horiz

ontal direction, determine the Cartesian representation of v→ relative to the (iˆ, jˆ) component system.
Physics
1 answer:
kotegsom [21]2 years ago
3 0

Answer:

\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

Explanation:

The x- and y- components of the velocity vector can be written as following:

\vec{v}_x = ||\vec{v}||\cos(\theta)\^i

\vec{v}_y = ||\vec{v}||\sin(\theta)\^j

Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

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Answer: Acceleration will have 2 components, vertical and horizontal.

Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.

Net-horizontal acceleration will  be equal to the horizontal component of the applied acceleration.

Explanation:

Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.

<u>For vertical acceleration:</u>

If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.

In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.

<u>For horizontal acceleration:</u>

This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.

But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.

8 0
3 years ago
Read 2 more answers
Answer these questions plz​
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2 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull
DiKsa [7]

Answer:

F_{net} = 232.8 N

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

F_x = F_1 + F_2cos45 + F_3cos45

here we know that

F_1 = 77.3 N

F_2 = 61.7 N

F_3 = 147 N

F_x = 77.3 + 61.7 cos45 + 147 cos45

F_x = 224.9 N

Similarly in Y direction we will have

F_y = F_3 sin45 - F_2 sin45

F_y = (147 - 61.7)sin45

F_y = 60.3 N

Now the net force on the donkey is given as

F_{net} = \sqrt{F_x^2 + F_y^2}

F_{net} = \sqrt{224.9^2 + 60.3^2}

F_{net} = 232.8 N

Now direction of force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{60.3}{224.9}

\theta = 15^o towards right so it is -15 degree

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Aloiza [94]

Answer:

yes

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