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2 years ago

In general, how did the water pressure in the tank change when mass was added to the fluid?

Physics

1 answer:

MissTica2 years ago

3 0

Answer:

As the height increases the pressure must increase.

Explanation:

When we add masses to the fluid, the amount of fluid in the tank increases, therefore its height increases and the pressure is described by the expression

P = ρ g h

where rho is constant for a given fluid and h is the height measured from the surface of the fluid.

As the height increases the pressure must increase.

You might be interested in

uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0

olganol [36]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

3 0

3 years ago

b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

$= moment\ of\ inertia \times angular\ acceleration$

$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0

3 years ago

An internal resistance of 5 ohm and a battery of 15 ohm is connected to a resistance of 20 ohm calculate the electric current

Allushta [10]

Answer:

.6 A

Explanation:

Battery 15 VOLTS

V = IR

V / R = I

15 / ( 5+20) = .6 amps

5 0

2 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle

Neporo4naja [7]

Answer:

f= 4,186 10² Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El momento de inercia de indican que giran un eje que pasa por enronqueces

I= ½ M R2

reduzcamos las cantidades al sistema SI

R= 1,4 cm = 0,014 m

M= 430 g = 0,430 kg

substituimos

w= √ (2 k/M R2)

calculemos

w = RA ( 2 370 / (0,430 0,014 2)

w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

w =2pi f

f= w/2π

f= 2,963 10³/ (2π)

f= 4,186 10² Hz

5 0

3 years ago