A

1) A pendulum is configured to have a period of 2 seconds.

vazorg [7]

"The" (and any subsequent words) was ignored because we limit queries to 32 words.

3 0

2 years ago

According to the concept of punctuated equilibrium, _____.

STatiana [176]

<span>According to the concept of punctuated equilibrium, </span>new species evolve suddenly over relatively short periods of time (a few hundred to a thousand years), followed by longer periods in which little genetic change occurs. Hope this helps. Have a nice day.

7 0

2 years ago

A 5.00 L air sample at a temperature of -50 °C has a pressure of 107 kPa. What will be the new pressure if the temperature is ra

Serjik [45]

Its simple use formuila ,
PV=nRT
n,R is constant as the both have same moles.
so,
(p1v1)/T1 = (p2v2)/T2
so, 128.53338kpa

4 0

2 years ago

A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the

k0ka [10]

<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>

8 0

2 years ago

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

Pie

Answer:

Part a)

$v = 7407.1 m/s$

Part b)

$v_{rel} = 1.05 \times 10^4 m/s$

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

$v = \sqrt{\frac{GM}{R+h}}$

here we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.37 \times 10^6 m$

$h = 900 km = 9.0 \times 10^5 m$

now we have

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}$

$v = 7407.1 m/s$

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

$v_{rel} = \sqrt{2} v$

$v_{rel} = 1.05 \times 10^4 m/s$

6 0

2 years ago