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3 years ago

The liquid pressure exerted in one direction only

1 answer:

8 0

The pressure exerted by a liquid on an object increases as we go more deep into the liquid and this pressure is called as hydro static pressure . if we consider a part of the static fluid then all the horizontal forces will cancel out while the vertical forces will add vectorilly and due to which a pressure difference is created . so as we go more deep the pressure increase .

Now pressure is a scalar so it does not depend on direction but when two objects are on the same level with respect to a reference level then the pressure exerted on them by fluid is always the same . hope this helps

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xxMikexx [17]

Answer:

Explanation:

The answer is C

7 0

3 years ago

(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new

Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation

$m \frac{d^{2}x}{dt^{2}} +kx=0$

Definition of parameters

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

$x(t)=c1cos(ωt)+c2Sin(ωt)$

$x(t)=c1cos(5.929t)+c2Sin(5.929t)$

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is

x(t)=0.337sin((5.929t)

6 0

3 years ago

A ball of mass 1.6 kg is attached to the end of a massless string. A circus clown twirls the

erik [133]

Compute the ball's angular speed v :

v = (1 rev) / (2.3 s) • (2π • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s

Use this to find the magnitude of the radial acceleration a :

a = v ²/R

where R is the radius of the circular path. We get

a = v ² / (180 cm) = v ² / (1.8 m) ≈ 13.43 m/s²

The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude

∑ F = m a

where m is the mass of the ball. So, if t denotes the magnitude of the tension force, then

t = (1.6 kg) (13.43 m/s²) ≈ 21 N

7 0

3 years ago

How much current is drawn by a television with a resistance of 30 ω that is connected across a potential difference of 120v?

melomori [17]

By using the Ohms law, we can calculate the current.

Ohms law,

$\frac{V}{I} = R$