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2 years ago

15

a guage is connected to tank in which the pressure of the fluid is 305kpa above atmospheric. if the absolute presssure of the fl

uid remains unchanged but the gage is in chamber where the air pressure is reduce to a vaccum of 648mmHg what reading in psi will then be observed

2 answers:

Serjik [45]2 years ago

7 0

Answer:

Explanation:

Hydrostatic pressure at any height below the water surface is calculated by P=hdg where h is the height below the open water surface, d is the density of water and g is the acceleration due to gravity. So if you want to calculate gauge pressure at height h then use formula P=hdg+P∘ where P∘ is atmospheric pressure

raketka [301]2 years ago

3 0

It had yk be yk me we don’t see the yk up there so

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On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed

JulsSmile [24]

Answer: 1.25 m

Explanation:

Given

initial velocity $(v_i )=3 m/s$

Final velocity $(v_f)=0.55\times 3=1.65 m/s$

Change in kinetic Energy =work done by Friction

change in Kinetic Energy $=\frac{m}{2}\left ( v_i^2-v_f^2\right )$

work done by friction $=\mu mgL$

$\frac{m}{2}\left ( 3^2-1.65^2\right )=0.25\cdot mg\times L$

$3.135=0.25\times 9.8\times L$

L=1.25

7 0

3 years ago

The swinging pendulum has 10 joules of potential energy at its maximum height at points (1) and (5). If the mass of the pendulum

SVEN [57.7K]

The speed of the pendulum at point 3 is 1.4 m/s

Explanation:

We can solve this problem by using the law of conservation of energy. In fact, the mechanical energy of the pendulum (which is the sum of his potential energy + his kinetic energy) must be conserved. So we can write:

$U_1 +K_1 = U_3 + K_3$

where

$U_1$ is the initial potential energy, at the highest position

$K_1$ is the initial kinetic energy, at the highest position

$U_3$ is the final potential energy, at the lowest position

$K_3$ is the final kinetic energy, at the lowest position

We are told that:

$U_1 = 10 J$ is the potential energy of the pendulum at the maximum height

$K_1 = 0$ (when the pendulum is at maximum height, the speed is zero, so the kinetic energy is zero)

$U_3 = 0$ (potential energy is zero at the lowest position)

Therefore,

$K_3 = U_1 = 10 J$

Kinetic energy can be rewritten as

$K_3 = \frac{1}{2}mv^2$

where

m = 10 kg is the mass of the pendulum

v is its speed at point 3

Solving for v,

$v=\sqrt{\frac{2K_3}{m}}=\sqrt{\frac{2(10)}{10}}=1.4 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

4 0

3 years ago

A 2.0 kg particle moves in a circle of radius 3.1 m. As you look down on the plane of its orbit, the particle is initially movin

Ghella [55]

Answer

given,

L(t) = 10 - 3.5 t

mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

τ = $\dfrac{dL}{dt}$

τ = $\dfrac{d}{dt}(10 - 3.5 t)$

τ = -3.5 N.m

Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.

L decreases the angular acceleration upward. so, net torque is upward.

b) Moment of inertia of the particle

I = m R^2

I = 2 x 3.1²

I = 19.22 kg.m²

L = I ω

ω = $\dfrac{L}{I}$

ω = $\dfrac{10 - 3.5 t}{19.22}$

ω = $0.520 - 0.182 t$

A = 0.52 rad/s B = -0.182 rad/s²

5 0

3 years ago

Read 2 more answers

In grassland regions, rainy seasons and drought seasons determine, in part, the _____. kinds of resident organisms spread of fir

qaws [65]

kinds of resident organisms

6 0

3 years ago

Read 2 more answers

1. A soccer player kicks a soccer ball towards the goal. If she kicks it with a velocity of 24 m/s at an angle of 31 degrees abo

ruslelena [56]

Answer:

<em>The ball will go as high as 8.46 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object launched at a certain height above the ground and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g the acceleration of gravity, then the maximum height hm can be calculated as follows:

$\displaystyle h_m=\frac{v_o^2\cdot \sin^2\theta}{2g}$

The soccer ball is kicked at a speed of vo=24 m/s at an angle of θ=31°. Taking the value of $g=9.8 m/s^2$ , then:

$\displaystyle h_m=\frac{24^2\cdot \sin^2 31^\circ}{2\cdot 9.8}$

$\displaystyle h_m=\frac{576\cdot 0.2653}{19.6}$

$h_m=7.80~m$

The ball will go as high as 8.46 m

4 0

3 years ago