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mezya [45]
3 years ago
13

A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below t

he interface (Fig. E12.33). The density of the oil is 790 kg>m3 . (a) What is the gauge pressure at the upper face of the block
Physics
1 answer:
bezimeni [28]3 years ago
4 0

Answer:

the gauge pressure at the upper face of the block is 116 Pa

Explanation:

Given the data in the question;

A cubical block of wood, 10.0 cm on a side.

height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m

density ρ = 790 kg/m³

Using expression for the gauged pressure;

p-p₀ = ρgh

where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.

we know that, acceleration due to gravity g = 9.8 m/s²

so we substitute

p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m

= 116.13 ≈ 116 Pa

Therefore, the gauge pressure at the upper face of the block is 116 Pa

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During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima
vova2212 [387]

Answer:

The value  F  =  0.1396 \ N

Explanation:

From the question we are told that

   The volume blood  ejected is  V  =  65 \ cm^3  = 65*10^{-6} \  m^3

    The velocity of the blood ejected is  v  = 103 \  cm/s  = \frac{103}{100} = 1.03 \ m/s

    The density of blood is  \rho = 1060 \  kg/m^3

     The heart beat is R = 59 \  bpm(beats \  per \  minute) = \frac{59}{60}= 0.9833\ bps

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      F  =  2 * \rho  *  V  *  R *  v

=>    F  =  2 * 1060  *  65*10^{-6}  *  0.9833 *  1.03

=>    F  =  0.1396 \ N

8 0
3 years ago
If a ball is 10m high with what velocity will it fall?
Semmy [17]

14m/s

Explanation:

Given parameters:

Height of the ball = 10m

Unknown:

Velocity of fall or final velocity = ?

Solution:

We are going to use the appropriate equation of motion to solve this problem.

The object is falling with respect to gravity.

  V² = U² + 2gH

where V is the final velocity

            U is the initial velocity

             g is the acceleration due to gravity 9.8m/s²

             H is the height of fall

The initial velocity here is zero and

      V² = 2 x 9.8 x 10 = 196

       V = 14m/s

learn more:

Motion problems brainly.com/question/5248528

#learnwithBrainly

6 0
3 years ago
Hurry Please !!!!!<br><br> Study the diagram<br> Point C identifies the____<br> of the wave
Nata [24]

Answer: Trough

Explanation: The point labeled C in the wave diagram above is the TROUGH of the wave motion. The trough of a wave motion identifies or signifies the point of least or minimum Displacement by measuring the downward Displacement of the wave. The point A is the CREST which is the opposite of the trough, signifying the point of maximum or upward Displacement of the wave cycle.

Point B is the wave amplitude which signifies the maximum extent of vibration from the equilibrium position of a wave. The point labeled D refers to the wavength of the wave motion which is the distance between successive crest or troughs of a wave motion.

4 0
3 years ago
What 2 things need to be known about an object in order to determine its kinetic energy?
NikAS [45]
I believe the answer is the mass of the object and the speed at which it is moving. 
7 0
3 years ago
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
2 years ago
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