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3 years ago

There is a bell at the top of a tower that is 45 m high. The bell weighs 190 kg. The bell has

Physics

1 answer:

natali 33 [55]3 years ago

6 0

Answer: The bell has 8550 J energy.

given, There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N

i.e., bell is located at the top of tower, h = 45m

weight of the bell, F = 190 N

workdone by the gravitational force = F.hcos180°

[ gravitational force (i.e., weight ) acting downward while body is located 45m above the ground. so, angle between force and h = 180° ]

workdone by the gravitational force = 190 × 45 × (-1)

= -8550 J

we know, potential energy = negative of workdone

= -(-8550 J) = 8550 J

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3 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.

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Answer:

Explanation:

Given:

length of ladder $r_L = 14m$

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position of firefighter $r_F = 3.8m$

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angle of ladder $\alpha = 63$

Unknown:

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force of friction on base of ladder $F_R$

normal force on base of ladder $F_N$

From the free body diagram of the sketch you get 3 equations:

$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$

Solving the equations gives:

$F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}$

$F_R = 238N\\ F_N = 1310N$

$F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3$

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4 years ago

A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor

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Answer:

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Now,

To calculate the output voltage, $V_{o}$ :

WE know that strain is given by:

$\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}$

Thus

$V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}$

Now, substituting the suitable values in the above eqn:

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4 years ago

A ball of 12 kg is attached to a string of 0.8 meter spun at 4

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3 years ago

A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d

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Answer:

The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

(target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

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We are interested in the target movement, so we can solve for that:

(target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

-94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

The target's speed is found by dividing the distance it covers by the time it takes.

√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

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3 years ago