Answer:
Question 1)
a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.
![v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2](https://tex.z-dn.net/?f=v_1%20%3D%20v_0%20%2B%20at%20%5C%5C4%20%3D%202%20%2B%204a%5C%5Ca%20%3D%200.5~ft%2Fs%5E2)
This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,
![v = \omega R\\a = \alpha R](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20R%5C%5Ca%20%3D%20%5Calpha%20R)
where α is the angular acceleration.
In order to continue this question, the radius of the drums should be given.
Let us denote the radius of the drums as R, the angular acceleration of drum B is
α = 0.5/R.
b) The distance travelled by the drums can be found by the following kinematics formula:
![v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft](https://tex.z-dn.net/?f=v_1%5E2%20%3D%20v_0%5E2%20%2B%202ax%5C%5C4%5E2%20%3D%202%5E2%20%2B%202%280.5%29x%5C%5Cx%20%3D%2012%20ft)
One revolution is equal to the circumference of the drum. So, total number of revolutions is
![x / (2\pi R) = 6/(\pi R)](https://tex.z-dn.net/?f=x%20%2F%20%282%5Cpi%20R%29%20%3D%206%2F%28%5Cpi%20R%29)
Question 2)
a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:
![a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bv_%7Bfuel%7D%7D%7Bm%7D%5Cfrac%7Bdm%7D%7Bdt%7D%20%3D%20-%5Cfrac%7B13000%7D%7B2600%7D25%20%3D%20125~ft%2Fs%5E2)
b) ![a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7Bv_%7Bfuel%7D%7D%7Bm%7D%5Cfrac%7Bdm%7D%7Bdt%7D%20%3D%20-%20%5Cfrac%7B13000%7D%7B400%7D25%20%3D%20812.5~ft%2Fs%5E2)
<h2>
Power is 11 W</h2>
Explanation:
Power = Work ÷ Time
Work = Force x Displacement
Force = 22 N
Displacement = 3 m
Time = 6 seconds
Substituting
Work = Force x Displacement
Work = 22 x 3 = 66 J
Power = Work ÷ Time
Power = 66 ÷ 6
Power = 11 W
Power is 11 W
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.
Answer:
a) 25.5°(south of east)
b) 119 s
c) 238 m
Explanation:
solution:
we have river speed
=2 m/s
velocity of motorboat relative to water is
=4.2 m/s
so speed will be:
a)
=
+![v_{m/r}](https://tex.z-dn.net/?f=v_%7Bm%2Fr%7D)
solving graphically
![v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}](https://tex.z-dn.net/?f=v_%7Bm%7D%3D%5Csqrt%7Bv%5E2_%7Br%7D%2Bv%5E2_%7Bm%2Fr%7D%7D)
=4.7 m/s
Ф=![tan^{-1} (\frac{v_{r}}{v_{m/r}} )](https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20%28%5Cfrac%7Bv_%7Br%7D%7D%7Bv_%7Bm%2Fr%7D%7D%20%29)
=25.5°(south of east)
b) time to cross the river: t=
=
=119 s
c) d=
=(2)(119)=238 m
note :
pic is attached
Answer:
![F_2 = 29.54 N](https://tex.z-dn.net/?f=F_2%20%3D%2029.54%20N)
Explanation:
As we know that the combination is maintained at rest position
So we will take net torque on the system to be ZERO
so we know that
![\tau = \vec r \times \vec F](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cvec%20r%20%5Ctimes%20%5Cvec%20F)
here we will have
![\vec r_1 \times F_1 = \vec r_2 \times F_2](https://tex.z-dn.net/?f=%5Cvec%20r_1%20%5Ctimes%20F_1%20%3D%20%5Cvec%20r_2%20%5Ctimes%20F_2)
so we have
![13 \times 50 = 22 \times F_2](https://tex.z-dn.net/?f=13%20%5Ctimes%2050%20%3D%2022%20%5Ctimes%20F_2)
so we have
![F_2 = \frac{13 \times 50}{22}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7B13%20%5Ctimes%2050%7D%7B22%7D)
![F_2 = 29.54 N](https://tex.z-dn.net/?f=F_2%20%3D%2029.54%20N)