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igomit [66]
3 years ago
11

Us

Physics
1 answer:
ElenaW [278]3 years ago
3 0

The answer to your question is law.

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Given 4.8 moles of a gas at 37 degrees Celsius and at 792 torr, what is the volume of the gas? (The ideal gas constant is 0.0821
Anastaziya [24]

Answer:

1.17 x 10^2 L

Explanation:

We can find the volume of the gas by using the ideal gas law:

pV=nRT

where we have:

p=792 Torr \cdot \frac{1 atm}{760 Torr} = 1.04 atm is the pressure

V is the volume

n = 4.8 mol is the number of moles

R = 0.0821 L · atm/mol · K is the ideal gas constant

T=37^{\circ}+273 =310 K is the temperature

Solving the equation for V, we find the volume

V=\frac{nRT}{p}=\frac{(4.8 mol)(0.0821 L atm/mol K)(310 K)}{1.04 atm}=117.5 L = 1.17\cdot 10^2 L

4 0
3 years ago
8) A racecar accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the
Vlada [557]

8)Acceleration of the car is 11.17 m/s².

9) Acceleration of skater is 0.57 m/s².

10) The deceleration is 1.8 m/s².

11) The average acceleration of the sprinter is 2 m/s².

12) The velocity of the stroller after 4.75 minutes is 171 m/s.

13) The skier will be moving as 11 m/s speed.

14) 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

Answer:

Explanation:

8) Acceleration of the car is determined by finding the ratio of change in velocity to the time taken for the change in velocity. As here the initial velocity is 18.5 m/s and the final velocity is 46.1 m/s, then in 2.47 seconds, the acceleration is

Acceleration = (46.1-18.5)/2.47 = 11.17 m/s².

Acceleration of the car is 11.17 m/s².

9)Similarly, the acceleration of the skater with initial velocity zero to final velocity as 5.7 m/s in 10 seconds is

Acceleration = (5.7-0)/10=0.57 m/s².

So acceleration of skater is 0.57 m/s².

10) In this case, the shuttle bus is stopping so its speed is decreasing from high value to low value. This kind of acceleration related to the decrease in velocity is termed as deceleration as the value of acceleration will be coming as negative.

Acceleration = (0-9)/5=-1.8 m/s².

So the deceleration is 1.8 m/s².

11) Average acceleration of sprinter = (Final velocity-Initial velocity)/Time

Average acceleration = (7.5-5)/1.25=2 m/s².

So the average acceleration of the sprinter is 2 m/s².

12)Since the acceleration of the stroller is given as 0.6 m/s². And the initial velocity is given as zero. So for the time of 4.75 minutes, the velocity will be equal to the final velocity.

As per equations of motion,

v = u +at

As u =0, a = 0.6 m/s² and t = 4.75 ×60 s = 285 s

So v = 0.6×285 = 171 m/s

Thus, the velocity of the stroller after 4.75 minutes is 171 m/s.

13) Similarly, in this case, u = 0, a = 2.2 m/s² and t = 5 s

Then v = u+at=0+(2.2×5)=11 m/s

So the skier will be moving as 11 m/s speed.

14) Here a = 50 m/s² and v = 28 m/s with u = 0 so time taken to reach the speed of 28 m/s is

v = u +at = 0+ (50 t)

28 = 50 t

t = 28/50 = 0.56 s

So 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

4 0
4 years ago
Calculate the percentage increase in speed of the cyclist when the power output changes from 200 W to 300 W.
Rom4ik [11]

Answer:

<em>50%</em>

Explanation:

Given

Initial power = 200W

Final power = 300W

Increment = 300 - 200 = 100W

percentage increase = increment/initial power * 100

percentage increase = 100/200 * 100%

percentage increase = 0.5 * 100

percentage increase = 50%

<em>Hence the percentage increase in speed is 50%</em>

6 0
3 years ago
In a lab, the mass of object a is 52 kg. object a weighs
aksik [14]
If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
F=52*9.8=509.6
therefore the weight of the object is 509.6N
5 0
3 years ago
Read 2 more answers
A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant an
masya89 [10]

Answer:

Explanation:

Given

Wheels are rotating with constant angular velocity let say \omega

Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.

But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.

(a) yes, there will be tangential velocity which is given by

v=r\cdot \omega

where r=radial distance from center

(b)tangential acceleration

there would be no tangential acceleration as velocity is constant

(c)centripetal acceleration

Yes, there will be centripetal acceleration given by

a_c=\omega ^2\times r

                                   

7 0
3 years ago
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