The element is iridium and it has 77 electrons
The equation for force is F=ma. Because we have the value of mass (0.42 kg) and the acceleration (14.8 m/s^2), simply plug them into the equation for force to get

The answer is 6.22 N because newtons are the unit used to measure force.
Answer: 1608.39 J
Explanation: Given that the
mass M = 42kg
U = 11.5m/s
V = 3.33m/s
how much work did friction do
Work done = Force × distance
Work done = Ma × distance
But acceleration a = V/t
Work done = M × V/t × d
Work done = M × V × d/t
Where d/t = velocity
Therefore,
Work done = M × U × V
Work done = 42 × 11.5 × 3.33
Work done = 1608.39 J
Answer : Cat will reach tree first.
Explanation :
It is given that,
Total distance between dog and tree is 30 yards. The cat has 10 yards lead. So, the distance between cat and tree is 20 yards.
Speed of cat is 6 Yards/ sec.
Speed of dog is 8 Yards/sec.
speed of cat is, 


Now, speed of dog is, 


It is clear that, cat is taking less time as compared with dog. So, the cat will reach tree first.
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as
C = B × log₂(1 + S/N)
where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.
Since the given SN ratio is in decibels, we must first express it as a ratio with no units as
SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000
Now that we have S/N, we can solve for its capacity (in bits per second) as
C = 4000 × log₂(1 + 1000)
C = 39868.91 bps
Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.
Answer: 40 kbps