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katrin2010 [14]
3 years ago
15

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

jectile (of the same mass) causes the pendulum to swing twice as high, h2 = 5.2 cm. The second projectile was how many times faster than the first?
Physics
1 answer:
Kipish [7]3 years ago
8 0

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

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Answer:

39.05°

Explanation:

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Mathematically,

dsin\theta=n\lambda

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Given that, d is 84 cm, n is 2, and the wavelength can be calculated as,

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Here, c is 343 m/s and f is 1300 Hz,

Therefore,

\lambda=\frac{343 m/s}{1300 Hz}\\\lambda=0.264m

Recall diffraction equation in term of sin\theta.

sin\theta=\frac{n\lambda}{d}

Put all the variables.

sin\theta=\frac{2\times 0.264 m}{84 cm}\\sin\theta=\frac{2\times 0.264 m}{0.84 m}\\\theta=39.05^{\circ}

Therefore, it is the required angle between the first 2 order of diffraction.

3 0
3 years ago
What are the four components that can be derived from a unit of blood?
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Calculate the air pressure in the pressurized tank, if h1 = 0. 18 m, h2 = 0. 2m and h3 = 0. 25m. The density of the mercury, wat
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The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

<h3 /><h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

Pressure is found as the product of the density,acceleraton due to gravity and the height.

P₁=ρ₁gh₁

P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m

P₁=24014.88 N/m²

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P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m

P₂=196.2 N/m²

P₃=ρ₃gh₃

P₃=850 kg/m³×9.81 (m/s²)×0.25

P₃=2084.625  N/m²

Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

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2 years ago
Infuse 1000 mL of NS. The infusion set delivers 15 gtt/mL. Give 300 mL bolus over 20 minutes and then infuse at the hourly rate
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Answer:

rate of infusion is 900 mL/hr

Explanation:

given data

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time t= 20 min

rate = 60 mL/hr

to find out

rate of infusion

solution

we know here we give 300 mL infuse in 20 min

so here for 20 min

rate of infusion is express as

rate of infusion = I2 / t

rate of infusion = 300 / 20

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so rate of infusion is 900 mL/hr

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