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3 years ago

6

Two objects are moving at equal speed along a level, frictionless surface. the second object has twice the mass of the first obj

ect. they both slide up the same frictionless inclined plane. which object rises to a greater height?

1 answer:

Zarrin [17]3 years ago

5 0

first object gors highest

second object goes lowest.

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The density of helium gas at 0.0 ◦C is 0.16 kg/m3 kg/m3. The temperature is then raised to 100.6 ◦C, but the pressure is kept co

Svetach [21]

Answer:

0.11694 kg/m³

Explanation:

T₁ = 0.0 °C = 273.15 K

ρ₁ = Density = 0.16 kg/m³

T₂ = 100.6 °C = 373.72 °C

PV = mRT

Realtion with density

$Pm=\rho RT\\\Rightarrow \rho =\frac{Pm}{RT}\\\Rightarrow \rho T= \frac{Pm}{R}$

P, m and R are constants, so

$\rho_1 T_1=\rho_2 T_2\\\Rightarrow \rho_2=\frac{\rho_1 T_1}{T_2}\\\Rightarrow \rho_2=\frac{0.16\times 273.15}{373.72}\\\Rightarrow \rho_2=0.11694\ kg/m^3$

∴ New density of the gas is 0.11694 kg/m³

6 0

4 years ago

Please help only if you answer all 4

Oduvanchick [21]

Answer:

i think ice burg for 2nd page for penguin sorry if its wrong!

Explanation:

5 0

3 years ago

Read 2 more answers

In the bedroom of your house you would like to run a 240 W toward, a 30 W alarm clock, a 720 W tanning bed, and four 60 W lights

erica [24]

Answer: current = 10.25A

Explanation:

Toward = 240w

Alarm clock = 30w

Tanning bed = 720w

4 60 watts bulb = 60 * 4 = 240w

Total power = 240 + 30 + 720 + 240 = 1230w

Voltage = 120v

Recall that Power = current * voltage

1230 = current * 120

current = 1230/120 = 10.25A

3 0

3 years ago

The speedometer on your car reads 35 mph. What is this measuring? A. average speed B. instantaneous speed C. velocity D. acceler

Elena-2011 [213]

B. instantaneous speed

The speedometer of a car show instantaneous speed.
Instantaneous speed is the motion rate of a car (in this example) at a particular moment or instant in time. It can be calculated by dividing a portion or a part of the total distance (of the trip) by time traveled.

6 0

3 years ago

What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit

Tresset [83]

The highest frequency sound to which the machine can be adjusted is :

4179.33 Hz

<u>Given data :</u>

Pressure = 10 Pa

Speed of sound = 344 m/s

Displacement altitude = 10⁻⁶ m

<h3>Determine the highest frequency sound ( f ) </h3>

applying the formula below

Pmax = $B(\frac{2\pi f}{v}) A$ --- ( 1 )

Therefore :

f = ( Pmax * V ) / $2\pi \beta A$

= ( 10 * 344 ) / 2 $\pi$ * 1.31 * 10⁵ * 10⁻⁶

= 4179.33 Hz

Hence we can conclude that The highest frequency sound to which the machine can be adjusted is : 4179.33 Hz .

Learn more about Frequency : brainly.com/question/25650657

<u><em>Attached below is the missing part of the question </em></u>

<em>A loud factory machine produces sound having a displacement amplitude in air of 1.00 μm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.31×105 Pa. The speed of sound in air is 344 m/s. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?</em>

4 0

2 years ago