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koban [17]
3 years ago
11

Electricity & Magnetism

Physics
1 answer:
ASHA 777 [7]3 years ago
7 0
The answer is A. Sound Energy
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What elements have similar behavior?
ki77a [65]

The elements which have similar behavior are Barium, strontium and beryllium.

7 0
3 years ago
Two moles of helium gas initially at 438 K and 0.44 atm are compressed isothermally to 1.61 atm. Find the final volume of the ga
docker41 [41]

Answer:

44.64335 L

Explanation:

R = Gas constant = 8.314 J/mol K = 0.08205 L atm/mol K

P = Pressure

V = Volume

T = Temperature = 438 K

1 denotes initial

2 denotes final

From ideal gas law we have

PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2\times 0.08205\times 438}{0.44}\\\Rightarrow V=163.35409\ L

So,

P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{0.44\times 163.35409}{1.61}\\\Rightarrow V_2=44.64335\ L

The volume of Helium is 44.64335 L

5 0
3 years ago
A wooden cylinder of length L and cross-sectional area A is partially submerged in a liquid with the axis of the cylinder orient
SCORPION-xisa [38]

Answer:

F=\rho_LAdg

Explanation:

The buoyant force F is equal to the weight of the displaced fluid. The weight of the displaced fluid is W=m_dg, where m_d is the mass of the displaced fluid. The mass of the displaced fluid is m_d=\rho_LV_d, where \rho_L is the density of the fluid and V_d is the displaced volume, which is equal to the submerged volume of the cilinder V_d=V_s=Ad.

Putting all together we have:

F=W=m_dg=\rho_LV_dg=\rho_LAdg

7 0
3 years ago
Why do yall not hate us country people?
sveticcg [70]

Answer:

Because we don't?

Explanation:

4 0
3 years ago
A 3.0 kg object is loaded into a toy spring gun, and the spring has a spring constant 750N/m. The object compresses the spring b
kiruha [24]

Answer:

Explanation:

mass of object, m = 3 kg

spring constant, K = 750 n/m

compression, x = 8 cm = 0.08 m

angle of gun, θ = 30°

(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.

(b) Let v be th evelocity of ball at the tim eof launch.

by using the conservation of energy

1/2 Kx² = 1/2 mv²

750 x 0.08 x 0.08 = 3 x v²

v = 1.265 m/s

By use of the formula of maximum height

h = \frac{v^{2}Sin^{2}\theta}{2g}

h = \frac{1.265^{2}Sin^{2}30}{2\times 9.8}

h = 0.02 m

h = 2 cm  

4 0
3 years ago
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