An object is in circular motion. How will the object behave if the centripetal force is removed

1 answer:

3 0

The object will sail away in a straight line ... continuing in the same direction it was going when the centripetal force stopped.

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Vector a s is 2.80 cm long and is 60.0° above the x-axis in the first quadrant. vector b s is 1.90 cm long and is 60.0° below th

zysi [14]

Refer to the diagram shown below.

Vector a is 2.8 cm long and 60° above the x-axis in the first quadrant. Therefore it s represented by
$\vec{a} = 2.8 (cos60^{o} \vec{i}+sin60^{o}\vec{j})$

Vector b is 1.9 cm long and 60° below the x-axis in the fourth quadrant. Therefore it is represented by
$\vec{b}=1.9(cos60^{o}\vec{i}-sin60^{o}\vec{j})$

The vectors may be written as
a = 2.8*(cos60°, sin60°) = (1.445, 2.5028)
b = 1.9*(cos60°, -sin60°) = (0.950, -1.6454)

Part (a)
a + b = (1.445+0.950, 2.5028-1.6454)
= (2.395, 0.8574)
The magnitude of this vector is
√[2.395² + 0.8574² ] = 2.5438 cm
Its direction is
tan⁻¹ (0.8574/2.395) = 19.7° above the x-axis, in the first quadrant.
The resultant vector is shown in the figure below.

Part (b)
Similarly obtain
a - b = (0.495, 4.1485)
The magnitude is 4.1777.
The direction is 83° above the x-axis, in the 1st quadrant.
The resultant vector is shown in the figure below.

Part (c)
Similarly, obtain
b - a = (-0.495, -4.1483)
The magnitude is 4.1777.
The direction is 83° below the negative x-axis, in the 3rd quadrant.
The resultant vector is shown in the figure below.

5 0

3 years ago

To win the game, a place kicker must kick a football from 30m from the goal, and the ball must clear the crossbar which is 3.05

Mashcka [7]

Answer:

Explanation:

Ignore air resistance

The horizontal velocity is

vx =19cos37.2 = 15.134 m/s

The ball travels 30 m in a time of

30 / 15.134 = 1.982 s

In that time, the ball will be at an altitude of

s = vy₀t + ½at²

s = 19sin37.2(1.982) + ½(-9.8)1.982² = 3.5137... m

so the ball clears the bar by

h = 3.5137 - 3.05 = 0.4637... = 0.46 m

7 0

3 years ago

Can anyone help me with any parts in question number 1: parallel circuit?

Leno4ka [110]

Question #1:

a). The sketch is attached to this answer.

b). The equivalent resistance of 30Ω and 50Ω in parallel is

1 / (1/30 + 1/50) = 18.75 Ω

c). I = V/R = (100/30) = (3 and 1/3) Amperes

d). Follow the wires, and you see that the 50Ω resistor is
connected directly to the battery, and so is the voltmeter.
So the voltage across the 50Ω resistor, and the reading
on the voltmeter, is 100 volts.

e). I = V/R
Through the 30Ω resistor: I = 3-1/3 A
Through the 50Ω resistor: I = 2 A

f). In the parallel circuit, both resistors are connected
directly to the battery. So neither resistor even knows
that the other one is there. Each resistor sees 100 volts,
and the current through each resistor is 100/R, just as if
it were the only resistor in the circuit.

4 0

3 years ago

What is the technical name for the animal kingdom

enyata [817]

The community off animals

6 0

3 years ago

- What forces caused the rocket to no longer 'stay at rest' (its INERTIA)?

horsena [70]

Answer:

Newtons first law

Explanation:

Objects stay at rest until a force acts on it causing it to move from its resting position

6 0

3 years ago