300 000 000 m/s in km/s
1000 m = 1km
300 000 000 m/s = 3 * 10⁸ m/s = 3 * 10⁵ *10³ m/s
= 3 * 10⁵ *10³ m/s
= 3 * 10⁵ km/s
Speed = 3 * 10⁵ km/s
Answer:
Option B. 8.1
Explanation:
From the question given above, the following data were obtained:
Angle θ = 71°
Hypothenus = 25
Adjacent = x
Thus, we can obtain the x component of the vector by using the cosine ratio as illustrated below:
Cos θ = Adjacent /Hypothenus
Cos 71 = x/25
Cross multiply
x = 25 × Cos 71
x = 25 × 0.3256
x = 8.1
Therefore, the x component of the vector is 8.1
Answer:
10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)
Explanation:
The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.
The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.
We use the first equation of motion for a free-falling body to obtain v as follows;
v = u + gt....................(1)
where g is acceleration due to gravity taken as 9.8m/s/s
It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.
To obtain t, we use the second equation of motion as stated;

Given; h = 6.10m.
since u = 0 for the vertical motion; equation (2) can be written as follows;

substituting;

Putting this value of t in equation (1) we obtain the following;
v = 0 + 9.8*1.12
v = 10.93m/s
Answer:
23.5 mV
Explanation:
number of turn coil 'N' =22
radius 'r' =3.00 cm=>
0.03m
resistance = 1.00 Ω
B= 0.0100t + 0.0400t²
Time 't'= 4.60s
Note that Area'A' = πr²
The magnitude of induced EMF is given by,
lƩl =ΔφB/Δt = N (dB/dt)A
=N[d/dt (0.0100t + 0.0400 t²)A
=22(0.0100 + 0.0800(4.60))[π(0.03)²]
=0.0235
=23.5 mV
Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV
Because the position depends on the amount of time that has passed.