The correct answer is compound
Answer:
no u tried of the same dam
thing
Explanation:
<span>3.78 m
Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes.
7.2 m/s / 9.81 m/s^2 = 0.77945 s
The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving
d = 1/2 A T^2
d = 1/2 9.81 m/s^2 (0.77945 s)^2
d = 4.905 m/s^2 0.607542 s^2
d = 2.979995 m
So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height.
d = 2.979995 m + 0.8 m = 3.779995 m
Rounding to 2 decimal places gives us 3.78 m</span>
Answer:
58.24 Km/h.
Explanation:
From the question given above, the following data were obtained:
Distance (d) = 495 Km
Time (t) = 8 h 30 mins
Speed (S) =?
Next, we shall express 8 hours 30 mins to hours.
We'll begin by convert 30 mins to hour.
60 mins = 1 h
Therefore,
30 mins = 30 mins × 1 h/ 60 mins
30 mins = 0.5 hour.
Thus,
8 h 30 min = 8 + 0.5 = 8.5 hours
Speed is define as the distance travelled per unit time. Mathematically, it is expressed as:
Speed = Distance /time
With the above formula, we can obtain the speed as shown below:
Distance (d) = 495 Km
Time (t) = 8.5 hour
Speed (S) =?
Speed = Distance /time
Speed = 495 Km / 8.5 hour
Speed = 58.24 Km/h
Thus, the speed is 58.24 Km/h.
Answer:
Explanation:
Let the charge particle have charge equal to +q .
force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )
force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.
F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )
= (Bqv) - j
= - Bqv j
The force will be along - ve y - direction .
If we take charge as negative or - q
force due to electric field will be along - y axis .
magnetic force = F = -q ( v i x B k )
= + Bqv j
magnetic force will be along + y axis
So it is difficult to find out the nature of charge on the particle from this experiment.
