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pickupchik [31]

3 years ago

8

In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other e

nd rests on a concrete pillar. When a clown (mass 79.0 kg ) steps out halfway onto the board, the board tilts so the rope end is 30∘ from the horizontal and the rope stays vertical. Calculate the force exerted by the rope on the board with the clown on it.

1 answer:

tester [92]3 years ago

4 0

Answer:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N

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Sivilculture isis the art and science of managing forests for desired outcomes.

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3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

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Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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2 years ago

A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate

qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

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radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

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3 years ago

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The transfer of energy that occurs when a force is applied over a distance is WORK.

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3 years ago

Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for ea

NARA [144]

It's not possible to answer the question exactly the way it's written.
That's because we don't know anything about the direction they
drive at any time during the trip.

You see, "velocity" is not just a word that you use for 'speed' when
you want to sound smart and technical, like this question is doing.
"Velocity" is a quantity that's made up of speed AND THE DIRECTION
of the motion. If you don't know the direction of the motion, then you
CAN'T tell the velocity, only the speed.

Here are the average speeds that Lori's family drove on each leg
of their trip:

Speed = (distance covered) / (time to cover the distance) .

Leg-A:
Speed = 15km/10min = 1.5 km/min

Leg-B:
Speed = 20km/15min = (1 and 1/3) km/min

Leg-C
Speed = 24km/12min = 2 km/min

Leg-D:
Speed = 36km/9min = 4 km/min

Leg-E:
Speed = 14km/14min = 1 km/min

From lowest speed to highest speed, they line up like this:

[Leg-E] ==> [Leg-B] ==> [Leg-A] ==> [Leg-C] ==> [Leg-D]
1.0 . . . . . . . . 1.3 . . . . . . . 1.5 . . . . . . . 2.0 . . . . . . . 4.0 . . . . km/minute

Whoever drove Leg-D should have been roundly chastised
and then abandoned by the rest of the family.  36 km in 9 minutes
(4 km per minute) is just about 149 miles per hour !

4 0

2 years ago

Read 2 more answers