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TEA [102]
3 years ago
15

The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

inate y reaches 150mm, find the v and a vectors for P. Hint: for a circle with origin shown x2+y2=r2. The radius of the slot is 300 mm.
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
4 0

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

P_x=rcos\theta

P_y=rsin\theta

where \thetais angle made by horizontal bar with x axis

Velocity at y=150 mm

150=300sin\theta

thus \theta =30^{\circ}

position ofP_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}

P_x=259.80 mm

P=259.80\hat{i}+150\hat{j}

Velocity at this instant

u_x=-rsin\theta =300\times sin30=-150 mm/s

u_y=rcos\theta =300\times cos30=259.80 mm/s

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Answer:

Explanation:

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3 years ago
How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2
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A transverse wave on a long horizontal rope with a
frosja888 [35]

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and  speed by the equation

f=\frac{v}{\lambda}

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f is the frequency

v is the speed of the wave

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For the wave in this problem,

v = 2 m/s

\lambda=8 m

So the frequency is

f=\frac{2}{8}=0.25 Hz

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

T=\frac{1}{f}=\frac{1}{0.25}=4 s

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

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2 years ago
A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons
torisob [31]

Answer:

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

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a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

6 0
2 years ago
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