Answer:
The distance of fly travel is 115.06 m.
Explanation:
Given that,
Distance = 115 mm
Speed = 1.10 m/s
Speed of fly = 2.20 m/s
We need to calculate the relative speed
Using formula of relative speed
![v=v_{1}+v_{2}](https://tex.z-dn.net/?f=v%3Dv_%7B1%7D%2Bv_%7B2%7D)
Put the value into the formula
![v=1.10+1.10](https://tex.z-dn.net/?f=v%3D1.10%2B1.10)
![v=2.20\ m/s](https://tex.z-dn.net/?f=v%3D2.20%5C%20m%2Fs)
We need to calculate the time for the two steamrollers to meet each other
Using formula of time
![t=\dfrac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bd%7D%7Bv%7D)
Put the value into the formula
![t=\dfrac{115}{2.20}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B115%7D%7B2.20%7D)
![t=52.3\ sec](https://tex.z-dn.net/?f=t%3D52.3%5C%20sec)
We need to calculate the distance of fly travel
Using formula of distance
![d=vt](https://tex.z-dn.net/?f=d%3Dvt)
Put the value into the formula
![d=2.20\times52.3](https://tex.z-dn.net/?f=d%3D2.20%5Ctimes52.3)
![d=115.06\ m](https://tex.z-dn.net/?f=d%3D115.06%5C%20m)
Hence, The distance of fly travel is 115.06 m.
<em>The correct option is</em> D. The initial momentum of the blue and green train combined during the collision is 200 kgm/s.
<h3>
Initial momentum of the blue and green train</h3>
Apply the principle of conservation of linear momentum as follows;
Pi = m1v1 + m2v2
where;
- m1 is mass of blue train
- m2 is mass of green train
- v1 is velocity of blue train
- v2 is velocity green train
- Pi is the initial momentum of the two trains
Pi = (50 x 4) + 30(0)
Pi = 200 kgm/s
Thus, the initial momentum of the blue and green train combined is 200 kgm/s.
Learn more about momentum here: brainly.com/question/7538238
#SPJ1
Because we should convert units within the metric system
Answer:19.34 miles
Explanation:
Given
First car travels with 32 mi/h
second travels with 54 mi/h
they leave at 2:00 Pm at 45
distance traveled by them is 0.5 hr
suppose First is traveling towards x axis and another travels 45 to it
Distance traveled by first car is ![32\times 0.5=16 miles](https://tex.z-dn.net/?f=32%5Ctimes%200.5%3D16%20miles)
Position vector of first car after 0.5 hr
![\vec{r_1}=32\times 0.5(\cos (45)\hat{i})](https://tex.z-dn.net/?f=%5Cvec%7Br_1%7D%3D32%5Ctimes%200.5%28%5Ccos%20%2845%29%5Chat%7Bi%7D%29)
Position vector of second car after 0.5 hr
![\vec{r_2}=54\times 0.5(\cos (45)\hat{i}+\sin (45)\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Br_2%7D%3D54%5Ctimes%200.5%28%5Ccos%20%2845%29%5Chat%7Bi%7D%2B%5Csin%20%2845%29%5Chat%7Bj%7D%29)
distance between them
![\vec{r_{21}}=27(\cos (45)\hat{i}+\sin (45)\hat{j})-16(\cos (45)\hat{i})](https://tex.z-dn.net/?f=%5Cvec%7Br_%7B21%7D%7D%3D27%28%5Ccos%20%2845%29%5Chat%7Bi%7D%2B%5Csin%20%2845%29%5Chat%7Bj%7D%29-16%28%5Ccos%20%2845%29%5Chat%7Bi%7D%29)
![\vec{r_{21}}=3.09\hat{i}+27\hat{j}\sin (45)](https://tex.z-dn.net/?f=%5Cvec%7Br_%7B21%7D%7D%3D3.09%5Chat%7Bi%7D%2B27%5Chat%7Bj%7D%5Csin%20%2845%29)
distance between them![|\vec{r_{21}}|=19.34](https://tex.z-dn.net/?f=%7C%5Cvec%7Br_%7B21%7D%7D%7C%3D19.34)
Answer : I = 0.0906 A.
Explanation :
It is given that,
No of electrons flowing in a wire, ![q=3.4\times 10^{19}](https://tex.z-dn.net/?f=q%3D3.4%5Ctimes%2010%5E%7B19%7D)
We know that the charge on an electron is given by, ![q_e=1.6\times 10^{-19}\ C](https://tex.z-dn.net/?f=q_e%3D1.6%5Ctimes%2010%5E%7B-19%7D%5C%20C)
So the total charge becomes:
![q=3.4\times 10^{19}\times 1.6\times 10^{-19}\ C](https://tex.z-dn.net/?f=q%3D3.4%5Ctimes%2010%5E%7B19%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5C%20C)
![q=5.44\ C](https://tex.z-dn.net/?f=q%3D5.44%5C%20C)
We know that the current in wire is defined as, ![I=\dfrac{q}{t}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7Bq%7D%7Bt%7D)
![I=\dfrac{5.44\ C}{60\ s}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B5.44%5C%20C%7D%7B60%5C%20s%7D)
![I=0.0906\ A](https://tex.z-dn.net/?f=I%3D0.0906%5C%20A)
Hence, this is the required solution.