If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero. So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".
If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.
And if there are four ropes holding it up, then the sum of
the four tensions is 100N. If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.
Answer:
16613 m/s
Explanation:
Given that
mass of the fly, m = 0.55 g = 0.55*10^-3 kg
Kinetic Energy of the fly, E = 7.6*10^4 J
Speed of the fly, v = ? m/s
We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.
The Kinetic Energy, KE of any object is represented by the formula
KE = 1/2 * m * v²
If we substitute the values in the relation, we have,
7.6*10^4 = 1/2 * 0.55*10^-3 * v²
v² = (15.2*10^4) / 0.55*10^-3
v² = 2.76*10^8
v = √2.76*10^8
v = 16613 m/s
Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J
Answer:
0
Explanation:
Forces with equal magnitudes and opposite directions cancel each other out, so the net force is 0.
Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
Answer:
C
Explanation:
horizintal speed stays same
only vertical speed changes
so H speed will stay 30 m/s