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diamong [38]
3 years ago
5

A 15m long wire is placed horizontally on the surface of a liquid and is gently pulled up with the force of 60N to keep the wire

in equilibrium.what is the surface tension of the liquid?​
Physics
1 answer:
Ksivusya [100]3 years ago
5 0

Answer:

2 N/m

Explanation:

Given that

The length of a wire, l = 15 m

Force on the wire, F = 60 N

Surface tension on the liquid, S = ?

The formula to solve this problem is given as

F = S * 2 L, where

F = force on the liquid

S = surface tension of the liquid

L= length of the wire

On substituting the values of each, we have

60 = S * (2 * 15)

S = 60 / 30

S = 2 N/m

Thus, we can say then, that the surface tension of the liquid is 2 N/m

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A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in
Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0
3 years ago
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Explain how thermal energy (temperature) affects chemical changes.
Monica [59]
If bonds are broken, the energy is released, and if bonds are formed, energy is absorbed. During conversions from chemical energy to thermal energy, the energy stored in the chemical bonds are released and this energy causes surrounding molecules to move faster thus increasing the thermal energy of a substance.
8 0
3 years ago
A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

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3 years ago
A 1000 kg racecar, which is capable of a top speed of 125 m/s, is sitting in a
k0ka [10]
Sitting = no movement
KE=0
5 0
3 years ago
Pls answer the 15a answer i cant understand it​
Maksim231197 [3]

Materials required for the experiment of limiting force borne by string:-

  1. String balance
  2. weights
  3. light strings
  4. weight hanger
  5. pan for spring balance
  6. Sand

Steps of procedure for for the experiment of limiting force borne by string:-

  1. First we have to tie a light string to the fixed support and then tie the other end with the weight hanger consists of weight.
  2. Add additional weight to the hanger again and again. And continue the same until the string is broken.
  3. Note down the weight (x) where the string is broken.
  4. Suspend spring balance to a support.
  5. Tie the light string at the end of the balance and at the other end suspend the pan for spring balance.
  6. Now place the weights (x-100 grams) in pan.
  7. Observe the reading in the spring balance.
  8. Add a small amount of sand in the pan by observing the readings.
  9. same is to be done till the string is broken.

Learn more about limiting force here:- brainly.com/question/11371672

#SPJ1

6 0
1 year ago
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