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2 years ago

6) What force should be applied to compress a spring

Physics

1 answer:

valina [46]2 years ago

7 0

Answer:

26.6N

Explanation:

Using Hooke’s law

F = kx

Where F = force

K = spring constant

x = displacement

From the question

F= ?

K = 140 N/m

x = 0.19m

Therefore,

F = 140 x 0.19

= 26.6N

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What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?

Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

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5 0

8 months ago

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin

faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance $L=0.450H=0.45\times 10^{-3}H$

Current in the inductor $i=1.90mA=1.90\times 10^{-3}A$

Voltage v = 13 volt

Inductive reactance of the circuit $X_l=\frac{v}{i}$

$X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm$

We know that

$X_l=\omega L=2\pi fL$

$2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10$

f = 2421.127 kHz

6 0

3 years ago

What happens to the speed of light as it passes into a glass block?

Deffense [45]

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➷ The speed would reduce as glass is denser than air.

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7 0

3 years ago

Read 2 more answers

An inductor with an inductance of .5 henrys (H) is to be connected to a 60 Hz circuit. What will the inductive reactance (X L) b

Anika [276]

Answer:

1885.2 ohms

Explanation:

Step one:

given data

L=5H

f=60Hz

Required

The inductive reactance of the inductor

Step two:

Applying the expression

XL= 2πfL

substitute

XL=2*3.142*60*5

XL=1885.2 ohms

4 0

2 years ago

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

= 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

= 3.6 J

Also, we know that

K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

= 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0

3 years ago