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2 years ago

Short Answer

Physics

1 answer:

STatiana [176]2 years ago

4 0

15 degrees because a glass of water won't do anything to a bath tub of 15 degree water

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A certain tuning fork vibrates at a frequency of 215 Hz while each tip of its two prongs has an amplitude of 0.832 mm. (a) What

Marysya12 [62]

Explanation:

It is given that,

Frequency of vibration, f = 215 Hz

Amplitude, A = 0.832 mm

(a) Let T is the period of this motion. It is given by the following relation as :

$T=\dfrac{1}{f}$

$T=\dfrac{1}{215}$

$T=4.65\times 10^{-3}\ s$

(b) Speed of sound in air, v = 343 m/s

It can be given by :

$v=f\times \lambda$

$\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{343\ m/s}{215\ Hz}$

$\lambda=1.59\ m$

Hence, this is the required solution.

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3 years ago

Write the factors of potential energy

IgorLugansk [536]

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2 years ago

If two or more elements are involved in a chemical reaction they will most likely form a?

Vitek1552 [10]

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3 years ago

Read 2 more answers

A block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane. (a) What is its velocity when it reaches t

kramer

Answer:

A.) 8 m/s

B.) 7.0 m

Explanation:

Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.

(a) What is its velocity when it reaches the top of the plane?

Since the plane is frictionless, the final velocity V will be the same as 8 m/s

The velocity will be 8 m/s as it reaches the top of the plane.

(b) How far horizontally does it land after it leaves the plane?

For frictionless plane,

a = gsinø

Acceleration a = 9.8sin28

Acceleration a = 4.6 m/s^2

Using the third equation of motion

V^2 = U^2 - 2as

Substitute the a and the U into the equation. Where V = 0

0 = 8^2 - 2 × 4.6 × S

9.2S = 64

S = 64/9.2

S = 6.956 m

S = 7.0 m

4 0

2 years ago

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

2 years ago