Wind and amplitude creates the waves in an ordinary ocean
<span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
Answer:
is the time taken by the car to accelerate the desired range of the speed from zero at full power.
Explanation:
Given:
Range of speed during which constant power is supplied to the wheels by the car is 
.
- Initial velocity of the car,
- final velocity of the car during the test,
- Time taken to accelerate form zero to 32 mph at full power,
- initial velocity of the car,
- final desired velocity of the car,
Now the acceleration of the car:
Now using the equation of motion:
is the time taken by the car to accelerate the desired range of the speed from zero at full power.
I'm not too sure but I think it's nuclear decay
<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>
