Answer:
(a) 8Ω (b) Ratio = Parra/P8 ohm = 1
Explanation:
Solution
Recall that,
An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω
Now,
(a) How can two 8-Ω speakers be arranged, when one = 4-Ω speaker, and one =12-Ω speaker
The Upper arm is : 8 ohm, 8 ohm
The Lower arm is : 12 ohm, 4 ohm
The Requirement is = (16 x 16)/(16 + 16) = 8 ohm
(b) compare your arrangement power output of with the power output of a single 8-Ω speaker
The Ratio = Parra/P8 ohm = 1
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Answer:
Option E is correct 310N
Explanation:
Given that the force used to push the crate is F = 200N
The force directed 20° below the horizontal
Mass of crate is m = 25kg
Weight of the crate can be determine using
W = mg
g is gravitational constant =9.8m/s²
W = 25×9.8
W = 245 N
Check attachment. For free body diagram and better understanding
Using newton second law along the vertical axis since we want to find the normal force
ΣFy = m•ay
ay = 0, since the body is not moving in the vertical or y direction
N—W—F•Sin20 = 0
N = W+F•Sin20
N = 245+ 200Sin20
N = 245 + 68.4
N = 313.4 N
The normal force is approximately 310 N to the nearest ten
