We use the kinematic equations,
(A)
(B)
Here, u is initial velocity, v is final velocity, a is acceleration and t is time.
Given,
,
and
.
Substituting these values in equation (B), we get
.
Therefore from equation (A),

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s
(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)
</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
Answer:
6 hours 15 minutes
Explanation:
On the trip from L.A. to London, the plane travels at 750 mph against a headwind of 50 mph, and that makes the net 700 mph (in aviation speak, 750 is the airspeed, while 700 is the groundspeed). 5000 miles divided by 700 mph results in about 7.14 hours, or about 7 hours and 9 minutes. On the return trip, ASSUMING THE SAME WIND, the plane travels at 750 mph, but this time the wind of 50 mph is a tail wind. So the net (groundspeed) is 800 mph. Traveling 5000 miles at 800 mph only takes 6.25 hours, or 6 hours and 15 minutes.
Outbound flight 7 hours 9 minutes
Return flight 6 hours 15 minutes
Answer:
0.647 nC
Explanation:
The force experienced by a charge due to the presence of an electric field is given by

where
q is the charge
E is the magnitude of the electric field
In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

Therefore, if the electric field magnitude is

Then the charge on each antenna would be
