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____ [38]
3 years ago
8

What is the density of a block of wood with a mass of 120 g and a volume of 200 cm

Physics
1 answer:
creativ13 [48]3 years ago
7 0
D=m/V therefore the answer is 120/200 or 0.6
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12) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a flow rate is 1.00 L
masha68 [24]

Answer:

The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]

Explanation:

We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.

For this case, we don't use the elevation difference and therefore those terms can be cancelled.

When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.

5 0
4 years ago
Which of the following accurately shows how to calculate the weight of a 20kg object
forsale [732]
The answer to your question is "20kgx9.8m/s" because weight is the force an object is exerting on another object, and the formula used to calculate force is <em>Force = Mass * Acceleration</em>.
4 0
3 years ago
Read 2 more answers
Suppose a large truck on a hydraulic lift exerts 7,000 N of force on a piston with an area of 0.4 m2. The piston can only suppor
uysha [10]

Pressure is the amount of force per unit area. In formula it is,

P = F ÷ A

P = 7000 N ÷ 0.4 m2

P = 17,500 N/m2

The amount of pressure the truck exerts on the piston is 17,500 N/m2

7 0
3 years ago
A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su
labwork [276]

Answer:

a

 \theta _2  = 13^o

b

 \theta _1  =32.94^o

c

 \theta_c  =  53.05^o    

Explanation:

From the question we are told that

    The angle of incidence is  \theta_1 =  10^o

    The refractive index of water is  n_1 = 1.3

  Generally Snell's law is mathematically represented as

          n_1 sin(\theta_1) =  n_2 sin(\theta_ 2)

Here n_2 is the refractive index of air with value  n_2 =  1

         \theta_2  is the angle of refraction

So  

        \theta _2  =  sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]

=>     \theta _2  =  sin^{-1}[\frac{1.3 * sin(10)}{1} ]

=>     \theta _2  = 13^o

Given that the angle should not be greater than \theta _2 =45^o  then the angle of incidence will be

       \theta _1  =  sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]

=>     \theta _1  =  sin^{-1}[\frac{1 * sin(45)}{1.3} ]

=>     \theta _1  =32.94^o

Generally for critical angle is mathematically represented as

        \theta_c  =  sin^{-1}[\frac{n_2}{n_1} ]

=>     \theta_c  =  sin^{-1}[\frac{1}{1.3} ]  

=>     \theta_c  =  53.05^o            

4 0
3 years ago
Para que sirve la caida libre
masya89 [10]
Eh? i don’t understand huhu
4 0
3 years ago
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