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3 years ago

11

The area of the velocity time graph gives displacement of the body true or false?

2 answers:

mote1985 [20]3 years ago

8 0

The area of the velocity time graph gives displacement of the body true or false?

The answer this is true.

zaharov [31]3 years ago

6 0

the correct one is the first option

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The electromagnetic wave that CT scans are based on is called a

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X-ray)
because Electromagnetic waves are in act

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3 years ago

a light ray traveling in air enters a second medium and its speed slows to 1.71 × 10^8 meters per second. what is the absolute i

givi [52]

The absolute refractive index is equal to the speed of light of the wave in air divided by the speed of light in the second medium. This means that it is equal to 3 x10^8 / 1.71 x10^8. This means the answer is 1.75

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3 years ago

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If 5 complete oscillations of a sound wave pass through a point in 0.5 s and the speed of sound was recorded to be 10 m/s, then

LekaFEV [45]

Answer:

λ = 2.5m

Explanation:

Given the following :

Speed of sound (v) = 10m/s

If 5 oscillations pass through a point in 0.5seconds;

Time taken (period) for 1 oscillation is :

Number of oscillations / total time taken

5 / 0.5 = 0.25 seconds

Wavelength, period and Velocity are related by the formula:

v = λ / T

λ = v * T

λ = 10 * 0.25

λ = 2.5 m

3 0

3 years ago

A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t

viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

$mv+MV=(m+M)v'$

Here , $v'$ is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

$0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s$

Hence , this is the required solution .

3 0

3 years ago

Calculate the pressure exerted by 11.1 moles of neon gas in a volume of 5.45 L at 25°C using (a) the ideal gas equation and (b)

Ilia_Sergeevich [38]

Answer:

49.82414 atm

50.74675 atm

Explanation:

P = Pressure

V = Volume = 5.45 L

R = Gas constant = 0.08205 L atm/mol K

T = Temperature = 25°C

a = 0.211 atm L²/mol²

b = 0.0171 atm L²/mol²

From ideal gas law we have

$PV=nRT\\\Rightarrow P=\dfrac{nRT}{V}\\\Rightarrow P=\dfrac{11.1\times 0.08205(273.15+25)}{5.45}\\\Rightarrow P=49.82414\ atm$

The pressure is 49.82414 atm

From Van der Waals equation we have

$\left(P+\frac{an^2}{V^2}\right)\left(v-nb\right)=nRT\\\Rightarrow P=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\\Rightarrow P=\dfrac{11.1\times 0.08205\times (273.15+25)}{5.45-(11.1\times 0.0171)}-\dfrac{0.211\times 11.1^2}{5.45^2}\\\Rightarrow P=50.74675\ atm$

The pressure is 50.74675 atm

3 0

3 years ago