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Afina-wow [57]
3 years ago
7

A sample of helium (He) occupies 8.0 liters at 1 atm and 20.0◦C. What pressure is necessary to change the volume to 1.0 liters a

t 10.0◦C?
Physics
1 answer:
nevsk [136]3 years ago
7 0

Apply the combined gas law

PV/T = const.

P = pressure, V = volume, T = temperature, PV/T must stay constant.

Initial PVT values:

P = 1atm, V = 8.0L, T = 20.0°C = 293.15K

Final PVT values:

P = ?, V = 1.0L, T = 10.0°C = 283.15K

Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:

1(8.0)/293.15 = P(1.0)/283.15

P = 7.7atm

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Stephanie serves a volleyball from a height of 0.80 m and gives it an initial velocity of +7.2 m/s straight up. how high will th
Papessa [141]
<span>3.78 m Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes. 7.2 m/s / 9.81 m/s^2 = 0.77945 s The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving d = 1/2 A T^2 d = 1/2 9.81 m/s^2 (0.77945 s)^2 d = 4.905 m/s^2 0.607542 s^2 d = 2.979995 m So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height. d = 2.979995 m + 0.8 m = 3.779995 m Rounding to 2 decimal places gives us 3.78 m</span>
7 0
3 years ago
What must be true the number of chromosomes in each sex cell
zvonat [6]

Answer:

In humans, each cell normally contains 23 pairs of chromosomes, for a total of 46.

8 0
3 years ago
A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force
MakcuM [25]

The correct answer to the question is : 9375 N.

CALCULATION:

As per the question, the mass of the car  m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = \frac{D}{2}

                                                                  = \frac{200}{2}\ m

                                                                  = 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

                              Force F = \frac{mv^2}{R}

                                            =  \frac{1500\times (25)^2}{100}\ N

                                            = 9375 N.           [ANS}        

The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.


6 0
3 years ago
Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c
Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

7 0
3 years ago
A 795-loop square armature coil with a side of 10. 5 cm rotates at 70. 0 rev/s in a uniform magnetic field of strength 0. 45 t.
Agata [3.3K]

The rms voltage output of the generator is 1.94 × 10⁻ ⁵ V.

RMS is an acronym for root mean squared. An RMS value is more than just the "amount of AC power that causes the same heating impact as an analogous DC power" or something along those lines.

No. of loop = 795

Diameter of the coil = 10.5 cm

Radius of the coil = 5.25 cm

Magnetic Field, B = 0.45 T

Time, t = 70.0 rev/s

              V_{rms} =\frac{NwAB}{\sqrt{2} }

Where,

              N = No. of loop

              A = Area of the coil

              B = Magnetic Field

              V_{rms} = Voltage rms

Area of the coil = πr²

                        = 86.57 cm²

w = 2π/t

   =( 2 × 3.141)/70.0

   = 0.089

V_{rms} =\frac{795*0.089*86.57* 0.45}{\sqrt{2} }\\\\V_{rms} =\frac{2756.36}{\sqrt{2} }\\\\\\V_{rms} =\frac{2756.36}{1.414 }\\\\V_{rms} = 1.94 * 10^-^5 V

Therefore, the rms voltage output of the generator is 1.94 × 10⁻ ⁵ V.

Learn more about rms voltage here:

brainly.com/question/13156072

#SPJ4

8 0
10 months ago
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