Velocity, va2 = 10.5 ft/s
<u>Explanation:</u>
From the figure:
Length of the cable = Sa + 2Sb = l
∴ vₐ = -2vb
Applying the principle of Impulse and momentum in x-direction
![mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2](https://tex.z-dn.net/?f=mv_x_1%20%2B%20%5Cint%5Climits%5Et_t%20%7BF_x%7D%20%5C%2C%20dt%20%3D%20mv_x_2)
Limit is t1 to t2
-(1)
Applying the principle of Impulse and momentum in y-direction
![mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2](https://tex.z-dn.net/?f=mv_y_1%20%2B%20%5Cint%5Climits%5Et_t%20%7BF_y%7D%20%5C%2C%20dt%20%3D%20mv_y_2)
Limit is t1 to t2
-(2)
Solving equation (1) and (2), we obtain
T = 1.6lb
va2 = 10.5 ft/s
Answer:
Force of attraction is 2,46*10^-8N![F= G*(m1*m2/r^2)\\where \\\\G=6,67*10^-11 (N*m^2/kg^2)\\m1=42kg\\m2=55kg\\r=2,5m\\therefore\\F=6,67*10^-11*(42*55/2.5^2)\\ F=2,46*10^-8 N](https://tex.z-dn.net/?f=F%3D%20G%2A%28m1%2Am2%2Fr%5E2%29%5C%5Cwhere%20%5C%5C%5C%5CG%3D6%2C67%2A10%5E-11%20%28N%2Am%5E2%2Fkg%5E2%29%5C%5Cm1%3D42kg%5C%5Cm2%3D55kg%5C%5Cr%3D2%2C5m%5C%5Ctherefore%5C%5CF%3D6%2C67%2A10%5E-11%2A%2842%2A55%2F2.5%5E2%29%5C%5C%20%3C%2Fp%3E%3Cp%3EF%3D2%2C46%2A10%5E-8%20N)
Explanation:
Using the Law of Universal Gravitation, proposed by Newton.
Answer:
current in series is 1.33 mA
current in parallel is 5.48 mA
Explanation:
given data
voltage = 8 V
resistors R1 = 2.5 kilo ohms
resistors R2 = 3.5 kilo ohms
to given data
current flow
solution
current flow in series is express as
current = voltage / resistor
put here all value
current = 8 / (2.5 + 3.5)
current = 8 / 6
current = 1.33 mA
and
current flow in parallel is express as
current = voltage / resistor
put here all value
current = 8 / (1/ (1/2.5 + 1/3.5))
current = 8 / 1.45
current = 5.48 mA
The difference in weight is due to the
displacement of water (the buoyancy of water is acting on the athlete thus
giving her smaller weight).<span>
The amount of weight displaced or the amount of buoyant force
is: </span>
Fb = 690 N - 48 N
Fb = 642 N
From newtons law, F = m*g. Using this formula, we
can calculate for the mass of water displaced:
m of water displaced = 642N / 9.8m/s^2
m of water displaced = 65.5 kg
Assuming a water density of 1 kg/L, and using the
formula volume = mass/density:
V of water displaced = 65.5kg / 1kg/L = 65.5 L
The volume of water displaced is equal to the
volume of athlete. Therefore:
V of athlete = 65.5 L
The mass of athlete can also be calculated using,
F = m*g
m of athlete = 690 N/ 9.8m/s^2
m of athlete = 70.41 kg
Knowing the volume and mass of athlete, her
average density is therefore:
average density = 70.41 kg / 65.5 L
<span>average density = 1.07 kg/L = 1.07 g/mL</span>
Answer:
(A) 0.54 kg.m^{2}
(B) 0.0156 N
Explanation:
from the question you would notice that there are some missing details, using search engines you can find similar questions online here 'https://www.chegg.com/homework-help/questions-and-answers/small-ball-mass-120-kg-mounted-one-end-rod-0860-m-long-negligible-mass-system-rotates-hori-q7245149'
here is the complete question:
A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of 2.60 x 10^{-2} N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?.
solution
mass of the ball (m) = 1.5 kg
length of the rod (L) = 0.6 m
angular velocity (ω) = 4900 rpm
air drag (F) = 2.60 x 10^{-2} N = 0.026 N
(take note that values from the original question are used, with the exception of the air drag which was not in the original question)
(A) because the rod is mass less, the rotational inertia of the system is the rotational inertia of the rod about the other end, hence rotational inertia =
where m = mass of ball and L = length of rod
=
= 0.54 kg.m^{2}
(B) The torque that must be applied to keep the ball in motion at constant speed = FLsin90
= 0.026 x 0.6 x sin 90 = 0.0156 N