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tigry1 [53]
3 years ago
10

A 20-cm long spring is attached to the wall. When pulled horizontally with a force of 100N, the spring stretches to a length of

22cm. What is the value of the constant? A.) The same spring is used in a tug of war. Two people pull on the ends, each with a force of 100N. How long is the stretched string? B.) The same spring is now suspended from a hook and a 10.2kg block is attached to the bottom end. How long is the stretched spring?
Physics
1 answer:
suter [353]3 years ago
3 0

Answer:

Explanation:

Part 0

All the spring moves is 2 cm

x = 2 cm * [1 m / 100 cm ]

x = 0.020 meters

F = k*d

100N = k * 0.02 m

100 N / 0.02 = k

5000 N / m

Part A

The spring feels a force of 100 N - - 100N = 200 N because each person is pulling in the opposite direction.

F = k * x

200N = 5000 N/m * d

200 / 5000 = d

d = 0.04 meters.

Part B

10.2 kg must be converted to a force as experienced here on earth.

F = m * g

g = 9.81

m = 10.2

F = 10.2 * 9.81

F = 100.06 N

F = k * d

100.06 = 5000 * d

d = 100.06 / 5000

d = 0.02 meters.

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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.360 mm wide. The diffraction pattern is observed
Bumek [7]

Answer:

a) 0.0130 m

b') w' = =6.46*10^{-3] m

Explanation:

given data:

\lambda of light = 633 nm

width of siit a =0.360 mm

distance from screen = 3.75 m

a) the first minima is located at

sin\theta = \frac{\lambda}{a}

              == \frac{633 *10^{-9}}{.360*10^{-3}}

           \theta = 0.100

y_1 = dtan\theta_1 = 3.75*tan(0.100) = 6.54 *10^{-3} m

with of central fringe  = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m

b)

width of the first bright fringe on either side of the central one = w' = y_2 -y_1

calculation for y_2

sin\theta = 2\frac{\lambda}{a}

              = = 2*\frac{633 *10^{-9}}{.360*10^{-3}}

             \theta  = 2*0.100 = 0.200

y_2 = dtan\theta_1 = 3.75*tan(0.200) =0.0130 m

w' = 0.0130  -6.54 *10^{-3}

w' = =6.46*10^{-3] m

6 0
3 years ago
Use the following photoelectric graph to answer the following question:
8090 [49]

Answer:

2.9 E14 Hz

Explanation:

As we know by Einstein's equation that energy incident on the photo sensitive surface will be used by the surface to eject electron out of the surface with some kinetic energy.

This is given by

E = \phi + KE

now the threshold frequency is the minimum frequency of the incident photons due to which electrons are ejected out with minimum kinetic energy or least kinetic energy.

So here when KE = 0 in the graph then corresponding to that position the frequency will be given as threshold frequency

so here from graph when KE = 0

f = 2.9 E14 Hz

8 0
3 years ago
In which era did the universe’s clouds start to condense and the universe became transparent for the first time?
alukav5142 [94]

Answer:

The Universe became transparent to the light left over from the Big Bang when it was roughly 380,000 years old

Explanation:

8 0
3 years ago
A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the
wel

Answers:

B.) F cos\theta=F_{n}

C.) F sin\theta=F_{g} \pm F_{f}

Explanation:

The image attached shows the way the force F is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

-F_{n}+F cos\theta=0 (1)

Where:

F_{n} is the Normal force

F is the magnitude of the force exerted on the block

\theta is the angle

Net Force in Y:

F sin\theta \pm F_{f}-F_{g}=0 (2)

Where:

F_{f} is the Friction force (it is expresed with the \pm sign because this force may be up or down, we cannot know because the block is at rest)

F_{g} is the gravity force

Rewrittin (1):

F cos\theta=F_{n} (3) This is according to option B

Rewritting (2):

F sin\theta=F_{g}\pm F_{f} (3) This is according to option C

3 0
3 years ago
How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?
Nady [450]

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations.  We are going to use vf = vi + at.  Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.  

5 0
3 years ago
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