I belive its like 1200 mile per hour ive done he math for it
Scobie will take 10 days to drive around Earth's equator.
To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

<em>Where:</em>
r: is the Earth's radius = 6371 km
Then, the distance is:

Now, if we divide the above distance by the speed of the car we can find the time:

Therefore, Scobie will take 10 days to drive around Earth's equator.
To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults
I hope it helps you!
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
The electric current passing through the bulb would be 3.3A
<u>Explanation:</u>
Given:
Electric charge, q = 800C
Time, t = 4 min
= 4 X 60 sec
= 240 sec
Electric current, I = ?
We know,

On substituting the value we get:

Thus, the electric current passing through the bulb would be 3.3A
Answer:
Explanation:
To solve this, we start by using one of the equations of motion. The very first one, in fact
1
V = U + at.
V = 0 + 0.8 * 3.4 = 2.72 m/s.
2.
V = 0 + 0.8 * 4.3 = 3.44 m/s.
3.
d = ½ * 0.8 * 4.3² + 3.44 * 12.9
d = 7.396 + 44.376
d = 51.77 m.
4.
d = 62 - 51.77 = 10.23 m. = Distance
traveled during deceleration.
a = (V² - Vo²) / 2d.
a = (0² - 3.44²) / 20.46
a = -11.8336 / 20.46 = -0.58 m/s²
5.
t = (V - Vo)/a =(0 - 3.44) / -0.58
t = -3.44/-.58 = 5.93 s
= Stop time.
T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total
time the hare was moving.
6.
d = Vo * t + ½ * a * t² = 62 m.
0 + 0.5 * (23.13)² * a = 61
267.5a = 61
a = 61/267.5
a = 0.23 m/s²