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Paladinen [302]
3 years ago
9

Unit of speed is derived unit why​

Physics
2 answers:
madam [21]3 years ago
7 0

Answer:

Explanation:

Therefore, the unit of speed is the meter per second, or m/s. The unit meter per second is called a derived unit, meaning that it is derived from the seven SI base units.

Archy [21]3 years ago
7 0

Answer:

Derived SI Units

Explanation

In physics there are many quantities that cannot be expressed by a single base unit. ... Therefore, the unit of speed is the meter per second, or m/s. The unit meter per second is called a derived unit, meaning that it is derived from the seven SI base units.

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What is the acceleration of an object with mass of 42.6 kg when an unbalanced force of 112 N is applied to it
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The answer is 2.63m/s^2! You use the formula F=ma, 112 = 42.6(a), a= 2.63m/s^2.
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3 years ago
Water moves through a constricted pipe in steady, ideal flow. At the
Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: 1.8\cdot 10^{-3} m^3/s

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2

where

p_1=1.75\cdot 10^4 Pa is the pressure in the lower section of the tube

h_1 = 0 is the heigth of the lower section

\rho=1000 kg/m^3 is the density of water

g=9.8 m/s^2 is the acceleration of gravity

v_1 is the speed of the water in the lower pipe

p_2 is the pressure in the higher section

h_2 = 0.250 m is the height in the higher pipe

v_2 is hte speed in the higher section

We can re-write the equation as

v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho} (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-section of the lower pipe, with

r_1 = 3.00 cm =0.03 m is the radius of the lower pipe (half the diameter)

A_2 = \pi r_2^2 is the cross-section of the higher pipe, with

r_2 = 1.50 cm = 0.015 m (radius of the higher pipe)

So we get

r_1^2 v_1 = r_2^2 v_2

And so

v_2 = \frac{r_1^2}{r_2^2}v_1 (2)

Substituting into (1), we find the speed in the lower section:

v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s

B)

Now we can use equation (2) to find the speed in the lower section:

v_2 = \frac{r_1^2}{r_2^2}v_1

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s

C)

The volume flow rate of the water passing through the pipe is given by

V=Av

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume  flow rate is constant, so

r_1=0.03 cm

v_1=0.638 m/s

Therefore, the volume flow rate is

V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

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3 years ago
A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50
loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

 U = \dfrac{1}{2}kx^2

 U = \dfrac{1}{2}\times 4.5 \times 0.13^2

        U = 0.038 J

at x = 0,the only energy will be kinetic .

 \dfrac{1}{2}mv^2=0.038

 \dfrac{1}{2}\times 0.23 \times v^2=0.038

         v² = 0.3304

         v = 0.575 m/s

displacement of the glider

      using conservation of energy

 \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

 x =v\sqrt{\dfrac{m}{k}}

 x =3\times \sqrt{\dfrac{0.23}{4.5}}

        x = 0.678 m

8 0
3 years ago
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