Answer:
141.78 ft
Explanation:
When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.
Calculating the acceleration using one of Newton's equations of motion:
Note: The negative sign denotes deceleration.
When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2
Hence, we can find the minimum stopping distance using:
The minimum stopping distance is 141.78 ft.
Answer:
i guess 0.8 miles away
Explanation:
mathematically:-
speed is given in question
Time is given
and we have to find distance
simply by using speed formula (s) = d/t
we get answer
Answer:
9.36*10^11 m
Explanation
Orbital velocity v=√{(G*M)/R},
G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,
M = mass of the star
R =distance from the planet to the star.
v=ωR, with ω as the angular velocity and R the radius
ωR=√{(G*M)/R},
ω=2π/T,
T = orbital period of the planet
To get R we write the formula by making R the subject of the equation
(2π/T)*R=√{(G*M)/R}
{(2π/T)*R}²=[√{(G*M)/R}]²,
(4π²/T²)*R²=(G*M)/R,
(4π²/T²)*R³=G*M,
R³=(G*M*T²)/4π²,
R=∛{(G*M*T²)/4π²},
Substitute values
R=9.36*10^11 m
B is the answer you need and i honestly got this question on a middle school test
you must be in different area then me
