The initial velocity is
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.
![\frac{dv}{dt} =10-0.8v \\\\ \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \, \frac{dv}{10-0.8v} = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\ \frac{0.8v -10}{3.2} =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D10-0.8v%20%5C%5C%5C%5C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%3Ddt%20%5C%5C%5C%5C%20%5Cint_%7B16.5%7D%5E%7Bv%7D%20%5C%2C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%20%20%3D%20%5Cint_%7B0%7D%5E%7Bt%7D%20dt%20%5C%5C%5C%5C%20-%20%5Cfrac%7B1%7D%7B0.8%7D%20%5Bln%2810-0.8v%29%5D_%7B16.5%7D%5E%7Bv%7D%3Dt%20%5C%5C%5C%5C%20ln%20%5Cfrac%7B10-0.8v%7D%7B-3.2%7D%3D-0.8t%20%5C%5C%5C%5C%20%20%5Cfrac%7B0.8v%20-10%7D%7B3.2%7D%20%20%3De%5E%7B-0.8t%7D%20%5C%5C%5C%5C%200.8v%20%3D%2010%20%2B%203.2e%5E%7B-0.8t%7D%20%5C%5C%5C%5C%20v%3D12.5%2B4e%5E%7B-0.08t%7D)
The velocity function is
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
The depth of the lake is
![d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft](https://tex.z-dn.net/?f=d%3D%5Cint_%7B0%7D%5E%7B5.7%7D%20%5C%2C%20%2812.5%2B4e%5E%7B-0.8t%7D%29dt%20%5C%5C%5C%5C%20%3D%2012.5%285.7%29%2B%20%5Cfrac%7B4%7D%7B%28-0.8%29%7D%5Be%5E%7B-0.8t%7D%5D_%7B0%7D%5E%7B5.7%7D%20%5C%5C%5C%5C%20%3D71.25-5%280.0105-1%29%20%3D76.198%20%5C%2C%20ft)
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
Answer:
The force of repulsion between the two balls will increase
Explanation:
The electrostatic force between two charged objects is given by
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the distance between the centres of the two objects
We see that the magnitude of the force is directly proportional to the charges on the two objects. in this problem, we have two positively charged balls (so, there is a force of repulsion between them, since like charges repel each other and unlike charges attract each other), and the positive charge in one of them is slowly increased: this means that either 
or 
in the formula is increasing, and so the magnitude of the force is increasing.
Answer: 4.27 *10^6 N/C
Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:
E=k*x/(a^2+x^2)^3/2 where a is the ring radius and x the distance to the point measured from the center of the ring.
Replacing the data we have:
E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2
then
E=4.27 * 10^6 N/C
Answer:
A. the magnitude of the velocity at which the two players move together immediately after the collision is 7.9m/s
B. The direction of this velocity is due north as the linebacker since he has obviously has more momentum
Explanation:
This problem bothers on the inelastic collision
Given data
Mass of linebacker m1= 110kg
Mass of halfbacker m2= 85kg
Velocity of linebacker v1= 8.8m/s
Velocity of halfbacker v2= 7.2m/s
Applying the principle of conservation of momentum for inelastic collision we have
m1v1 +m2v2= (m1+m2)v
Where v is the common velocity after impact
Substituting our data into the expression we have
110*8.5+85*7.2= (110+85)v
935+612=195v
1547=195v
v=1547/195
v=7.9m/s
Momentum of linebacker after impact = 110*7.9= 869Ns
Momentum of halfbacker after impact = 85*7.9= 671.5Ns
the direction after impact is due north since the linebacker has greater momentum
