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Physics

2 answers:

larisa86 [58]3 years ago

7 0

Waves transfer energy but not matter

Reil [10]3 years ago

4 0

Energy but not matter.

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A weather balloon has a volume of 90.0 l when it is released from sea level. What

frozen [14]

The final atmospheric pressure is $5.19\cdot 10^4 Pa$

Explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

where in our problem we have:

$p_1= 1.03\cdot 10^5 Pa$ is the initial pressure (the atmospheric pressure at sea level)

$V_1 = 90.0 L$ is the initial volume

$p_2$ is the final pressure

$V_2 = 175.0 L$ is the final volume

Solving the equation for p2, we find the final pressure:

$p_2 = \frac{p_1 V_1}{V_2}=\frac{(1.01\cdot 10^5)(90.0)}{175.0}=5.19\cdot 10^4 Pa$

Learn more about ideal gases:

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3 0

3 years ago

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

Anna [14]

Refer to the figure shown below.

g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306