Let the key is free falling, therefore from equation of motion
.
Take initial velocity, u=0, so
.
![h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }](https://tex.z-dn.net/?f=h%20%3D%200%5Ctimes%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dg%20t%5E2%3D%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20%5C%5C%5C%20t%20%3D%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%20%7D)
As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula
![d= v \times t](https://tex.z-dn.net/?f=d%3D%20v%20%5Ctimes%20t)
From above substituting t,
.
Now substituting all the given values and g = 9.8 m/s^2, we get
.
Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.
Answer:
mass is 6.97 pounds
Explanation:
given data
volume = 3.1 liters
density = 1.020 g/ml = 1.02 kg/l
to find out
How many pounds of blood plasma
solution
we know mass formula from density that is
density = mass / volume
so
mass = density × volume ...............1
so put all value to get mass
mass = 1.02 × 3.1
mass = 3.162 kg
mass = 3.162 × 2.205 pounds
so mass is 6.97 pounds
Answer: 7.53 μC
Explanation: In order to explain this problem we have to use the gaussian law so we have:
∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0
Q inside= 0 = q+ Qinner surface=0
Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement
![-h= vsinA\times t-gt^2/2](https://tex.z-dn.net/?f=-h%3D%20vsinA%5Ctimes%20t-gt%5E2%2F2)
putting values h=15 m, v=0.8
............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get
![-15 = 0.8\times40/0.6 - 4.9t^2](https://tex.z-dn.net/?f=-15%20%3D%200.8%5Ctimes40%2F0.6%20-%204.9t%5E2)
solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = ![-H =v sinA t - gt^2/2](https://tex.z-dn.net/?f=-H%20%3Dv%20sinA%20t%20-%20gt%5E2%2F2)
⇒ ![4.9t^2 - 8.9\times0.8t - 100 = 0](https://tex.z-dn.net/?f=4.9t%5E2%20-%208.9%5Ctimes0.8t%20-%20100%20%3D%200)
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m
Explanation:
PE= mgh
6 J= (3m) (9.81 m/s2) (mass)
mass=( 6)/(3×9.81)
mass= 0.20 Kg