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xeze [42]
3 years ago
13

A 20-cm-long spring is attached to a wall. When pulled horizontally with a force of 100 N, the spring stretches to a length of 2

2 cm. The same spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end. How long is the stretched spring in cm? (Do not include unit in answer)
Physics
1 answer:
Delvig [45]3 years ago
7 0

Answer:

the length of stretched spring  in cm is 22

Explanation:

given information:

spring length, x1 = 20 cm = 0.2 m

force, F = 100 N

the length of spring streches, x2 = 22 cm = 0.22 m

According to Hooke's law

F = - kΔx

k = F/*=(x2-x1)

  = 100/(0.22 - 0.20)

  = 5000 N/m

if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end

m = 10.2 kg

W = m g

    = 10.2 x 9.8

    = 99.96 N

F = - k Δx

Δx = F / k

     = 99.96 / 5000

     = 0.02

Δx = x2- x1

x2 = Δx + x1

    = 0.20 + 0.02

    = 0.22 m

     = 22 cm

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9966 [12]
The force equation can easily prove this.  F=ma.  This states that the force on an object is equal to mass times acceleration.  If the mass stays the same and the velocity of the cars increases than that means there is a larger force.  This is because in both cases the cars are stopping in almost an instant and the times of the crashes are theoretically identical.  Acceleration is the change in velocity over time.  If the velocity is higher with the same amount of time than that means there is a higher acceleration.  If you plug a higher acceleration into the force equation then you wind up with a higher force and in turn a more damaging collision.
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8 0
3 years ago
Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
zlopas [31]

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

8 0
3 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
4 years ago
A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0
3 years ago
The driver sees that the road is empty and accelerates at 1.0 m/s2 for 5.0 s. what can you determine about the truck's motion us
vekshin1

kinematic equation

v=u+at

v-u=at

v-u = 1x5

the driver will have increased speed by 5 m/s. actual speeds unknown

7 0
3 years ago
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