Answer:
19.2m/s
Explanation:
Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,
plug in known varibles,
v=2.4*8
v=19.2m/s
Answer:
ok
Explanation:
the correct answer is C please follow me
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
The conversion for km to inches is:
1km=39370.1in
Now we can solve for 56 km..
56km=39370.1*56
56km=<span> 2204725.6in
Answer=2,204,725.6in</span>
Answer:
The correct answers are
(a) It decreases to 1/3 L
(ii) is (c) It is constant
Explanation:
to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem
The given variables are as follows
Initial volume V1 = 1L
V2 = Unknown
Initial Temperature T1 = 300K
let us assume that the balloon is perfectly elastic
At 300K the balloon is filled and it stretches to maintain 1 atmosphere
at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,
For (i) since we know that the pressure of the balloon is constant
by Charles Law V1/T1 =V2/T2
or V2 = (V1/T1)×T2 =
× 
= 
× L = L/3 hence the correct answer to (i) is 1/3L
