20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.
Here we have to add the two measurements given in the question
The measurement values are given as 1.0090 cm and 0.02 cm.we have to add them on the basis of significant figure rules.
As per the addition rule in terms of significant figures
1-First we have to select the number of significant digits after the decimal point of each quantity.
2-Now we have to remember that during the addition ,the resultant of two quantities will follow the quantity having least number of significant figures after the decimal point.
3-Here we are considering the minimum number of significant figures after the decimal points not the minimum number of significant figures in case of multiplication and division
Now we have to add these two quantities as per the above rule-
1.0090 cm +0.02 cm
=1.0290 cm
Here the result will follow 0.02 which has minimum number of significant figures after the decimal points.
Hence we have to round off the number from 9 of 1.0290
As 9 is greater than 5 ,so he actual result will be 1.03 cm
The correct order they go in is "1-4-2-3" The correct answer is D.
<span>Ball A with a mass of 0.500 kg is moving east at a velocity of 0.800 m/s. It strikes ball B, also of mass 0.500 kg, which is stationary. Ball A glances off B at an angle of 40.0° north of its...Two smart cars depart from the same starting location at the same time and travel different routes to the same destination, arriving at the same time. Explain why the cars travelled different...Given electric flux density D=0.3r^2 ar nc/m^2 in free space a. Find the total charge within the sphere r=3 b. Find the total electric flux leaving the sphere r=4</span>
<span>22.5 newtons.
First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as:
E = 0.5mv^2
where
E = Energy
m = mass
v = velocity
Substituting known values and solving gives:
E = 0.5 3.06 kg (7 m/s)^2
E = 1.53 kg 49 m^2/s^2
E = 74.97 kg*m^2/s^2
Now ignoring air resistance, how much energy should the rock have had?
We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So
3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2
So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is
299.88 J - 74.97 J = 224.91 J
So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division.
224.91 J / 10 m
= 224.91 kg*m^2/s^2 / 10 m
= 22.491 kg*m/s^2
= 22.491 N
Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
