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koban [17]
3 years ago
14

8. When a 100 N bag of nails hangs motionless from a single vertical strand of rope, how many newtons of tension are exerted in

the strand? - What if the bag is supported by four vertical strands?
Physics
1 answer:
Svetllana [295]3 years ago
8 0

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero.  So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".

If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.

And if there are four ropes holding it up, then the sum of
the four tensions is 100N.  If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.

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light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur
VashaNatasha [74]

Answer:

The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

Explanation:

Given;

wavelength of light, λ = 600 nm

The minimum thickness of the soap bubble for destructive interference to occur is given by;

t = \frac{\lambda/n}{2}\\\\t = \frac{\lambda}{2n}

where;

n is refractive index of soap film = 1.33

t = \frac{\lambda}{2n} \\\\t = \frac{600*10^{-9}}{2(1.33)}\\\\t = 2.2556 *10^{-7} \ m\\\\t =  225.56 *10^{-9} \ m\\\\t = 225.56 \ nm

Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

4 0
3 years ago
a roller coaster has a speed of 16 m/s at a height 28 meters above the ground what is its speed at the top of the next hill whic
Tju [1.3M]

a una velocidad de

22 m/s, quien lo golpea y devuelve en la misma

dirección con una velocidad de 14 m/s. Si el

tiempo de contacto del balón con la jugadora es

de 0,03 s, ¿con qué fuerza golpeó la jugadora el

balón?

21 Una bala de 0,8 g, está en la recámara de un rifl e

cuando se g

5 0
3 years ago
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0
3 years ago
Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m
inysia [295]

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

      τ = I α

     T r = I α

     T gR = I α

     Alf = T R / I (1)

     T = α I / R

Now let's use Newton's second law in the mass that descends

     W- T = m a

     a = (m g -T) / m

The two accelerations need related

     a = R α

    α = a / R

    a = (m g - α I / R) / m

    R α = g - α I /m R

    α (R + I / mR) = g

    α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

4 0
4 years ago
The tuning fork has a frequency of 426.7 is found to cause resonance in a closed column of air measuring 0.186 (the diameter of
lidiya [134]

Answer:

The velocity of the tuning fork sound is 317.4648 m/s

Explanation:

The given parameters are;

The frequency of the fork, f = 426.7 Hz

The length of the closed air column, L = 0.186 m

The diameter of the tube, d = 0.15 m

At the fundamental frequency in a closed tube, we have;

λ = 4·L

Where;

λ = The wavelength of the wave

L = The length of the tube

From the equation for the velocity of a wave, we have;

v = f·λ

Where;

v = The velocity of the (sound) wave

f = The frequency of the wave = 426.7 Hz

λ = 4·L = 4 × 0.186 m = 0.744 m

∴ v = 426.7 Hz × 0.744 m = 317.4648 m/s

Therefore, the velocity of the sound produced by the tuning fork, v = 317.4648 m/s

3 0
3 years ago
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