The approximate acceleration of the car would be <span>3.00
</span>How? We have to use the formula to find the velocity.
v = d/t = 120/6.32
v = 19
Then plug the factor into the acceleration formula to find the acceleration:
a = v/t = 19/6.32
a = 3.00
So, the approximate acceleration of the car would be 3.00
If Resistors are in series= The equivalent is the sum.
E.g R1 and R2 in series, R = R1 + R2.
If in Parallel, equivalent is Product/sum.
E.g If R1 and R2 in parallel, R = (R1*R2)/(R1+R2)
1.) 60 is parallel with 40 and both are then in series with 20.
60//40 = (60*40)/(60+40) = 2400/100 = 24
Now the 24 is in series with the 20
R = 24 + 20 = 44 ohms.
2.) 80 is in series with 40 and both are then in parallel with 40.
Solving the series, R = 80 + 40 =120.
Parallel: 120//40 = (120*40)/(120+40) = 4800/160 = 30
Equivalent Resistance = 30 ohms.
3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.
The 1st parallel = (100*100)/(100+100) = 10000/200 = 50
The 2nd parallel = (50*50)/(50+50) = 2500/100 = 25.
Solving the series = 50 + 25 =75 ohms.
Cheers.
Answer:
Explanation:
Let the highest and lowest score be a and b respectively .
a - b = 10
a = 10 + b
The numbers in descending order are a , 98 , 98 , b .
mean = 100 , so
4 x 100 = a + 98 + 98 + b = 10 + b + 98 + 98 + b = 206 + 2b
400 = 206 + 2b
2b = 194
b = 97
a = 107
So the scores were as follows
107 , 98 , 98 , 97
Answer:
135J
Explanation:
So we know ΔKinetic Energy= ΔWork
Kinetic energy=1/2mv²
So Kf-Ki=ΔK
ΔK=1/2*0.45(25²-5²)=135J
135J=ΔWork
Answer:
103.3 %
Explanation:
For a radioactive isotope, the number of radioactive nuclei left (parent nuclei) after a time t, N(t), is

where
N0 is the initial number of radioactive nuclei
t is the time
is the half-life of the isotope
Here we have

So we find

Which means that the fraction of parent nuclei left after this time is 0.508 (50.8% of the initial value). So the fraction of daugther nuclei at this time is

So the percentage of parent to daughter isotopes is

Which corresponds to 103.3 %.