Answer:
2000 ohms
Explanation:
Resisters in series just add.
Rt = R1 + R2 + R3
R1 = 650 ohm
R2 = 350 ohm
R3 = 1000 ohm
Rt = 650 + 350 + 1000
Rt = 2000 ohms.
You need to set their position functions equal to one another and so for the time t when that is true. That is when the tiger and the deer are in the same place meaning the tiger catches the dear
Xdear= 2t+15 deer position function.
(I integrated the velocity function )
To get the Tigers position function you must integrate the acceleration twice. This becomes
Xtiger=t^2
Now t^2=2t+15
Time t is when the tiger catches the deer
t^2-2t-15=0
(t-5)(t+3)=0 factored
t=5s is the answer you use (t=-3 is a meaningless solution)
<span>Our equation 1 would be
m*v=M*V1+m*V2
v=V1+V2
v-V1=V2
the equation 2 would look like this
</span>V^2=V1^2+V2^2
V^2-V1^2=V2^2
(V-V1)*(V+V1)=V2^2Dividing with the 1
V+V1=V2
It’s 600 x 2.5 which is 1500. So the plane flies about 1500km in 2.5 hrs.
Answer:
![\mu_s \geq 0.27](https://tex.z-dn.net/?f=%5Cmu_s%20%5Cgeq%200.27)
Explanation:
The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:
![f_s=\frac{mv^{2}}{R}](https://tex.z-dn.net/?f=f_s%3D%5Cfrac%7Bmv%5E%7B2%7D%7D%7BR%7D)
But we know that:
![f_s\leq \mu_s N](https://tex.z-dn.net/?f=f_s%5Cleq%20%5Cmu_s%20N)
And the normal force is given by the sum of the forces in the vertical direction:
![N-mg=0 \implies N=mg](https://tex.z-dn.net/?f=N-mg%3D0%20%5Cimplies%20N%3Dmg)
Finally, we have:
![f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27](https://tex.z-dn.net/?f=f_s%3D%5Cfrac%7Bmv%5E%7B2%7D%7D%7BR%7D%20%20%5Cleq%20%5Cmu_s%20mg%5C%5C%5C%5C%5Cimplies%20%5Cmu_s%5Cgeq%20%5Cfrac%7Bv%5E%7B2%7D%7D%7BgR%7D%20%20%5C%5C%5C%5C%5Cmu_s%5Cgeq%20%5Cfrac%7B%286%5Cfrac%7Bm%7D%7Bs%7D%29%20%5E%7B2%7D%7D%7B%289.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%2813.5m%29%20%7D%5C%5C%5C%5C%5Cmu_s%5Cgeq0.27)
So, the minimum value for the coefficient of friction is 0.27.