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bixtya [17]
3 years ago
13

A crate is pulled 7 m across a smooth surface. The tension in the rope pulling the crate is 40 N. If the work on the crate is 24

7 J, what is the angle the rope makes with the horizontal
Physics
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

<em>61.9°</em>

Explanation:

The formula for calculating the workdone is expressed as

Workdone = Fdsin theta

F is the force applied on the crate

d is the distance covered

theta is the angle that the rope makes with the horizontal

Given

F = 40N

d = 7m

Workdone = 247J

Substituting into the formula:

247 = 40(7)sin theta

247 = 280sin theta

sin theta = 247/280

sin theta = 0.88214

theta = arcsin(0.88214)

theta = 61.9°

<em>Hence the angle that the rope makes with the horizontal is 61.9°</em>

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(b) Figure 3 shows a wheelbarrow and stones with a total mass of 43 kg. The wheelbarrow is in equilibrium with two of the three
sleet_krkn [62]

An equation showing the relationship between the THREE forces is R -W +F= 0

<h3>What is moment of force?</h3>

The equal and opposite force acting at a point from the axis of rotation is called the moment of force.

M = F x r

Given is a wheelbarrow and stones with a total mass of 43 kg. The wheelbarrow is in equilibrium with two of the three forces acting on it

Taking moment about the wheel, we get

-43x 9.81 x 0.6 + F x 1.5 =0

F = 43 x 0.6/1.5

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Where, R is the normal force, W is the Weight force and F is the applied force.

Thus, this is the equation representing all the three forces acting on wheelbarrow.

Learn more about moment of force.

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3 0
2 years ago
The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the
Ivanshal [37]

Answer:

(a) \alpha = - 1.32\ rev/m^{2}

(b) \theta = 13674\ rev

(c) \alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}

(d) a = 22.458\ m/s^{2}

Solution:

As per the question:

Angular velocity, \omega = 190\ rev/min

Time taken by the wheel to stop, t = 2.4 h = 2.4\times 60 = 144\ min

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

\omega' = \omega + \alpha t

omega' = final angular velocity

\omega = initial angular velocity

\alpha = angular acceleration

Now,

0 = 190 + \alpha \times 144

\alpha = - 1.32\ rev/m^{2}

Now,

(b) The no. of revolutions is given by:

\omega'^{2} = \omega^{2} + 2\alpha \theta

0 = 190^{2} + 2\times (- 1.32) \theta

\theta = 13674\ rev

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}

(d) The radial acceleration is given by:

\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s

Linear acceleration is given by:

a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}

a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}

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What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?
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Answer:

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This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

\frac{dV}{V}=-\frac{1}{RC}dt

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\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt

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ln(\frac{V}{v})=e^{\frac{-t}{RC} }

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