The runner's acceleration during this time interval is 10
<u>Given the following data:</u>
- Initial velocity, U = 0 m/s (since the sprinter is starting from rest).
- Final velocity, V = 10.0 m/s
To calculate the runner's acceleration during this time interval, we would use the first equation of motion;
Mathematically, the first equation of motion is calculated by using the formula;
<u>Where:</u>
- U is the initial velocity.
- t is the time measured in seconds.
Substituting the given parameters into the formula, we have;
Therefore, the runner's acceleration during this time interval is 10
Read more: brainly.com/question/8898885
Answer:
well at the rate he goes he would be at 2.5km
Answer:
43850.49112 kg
Explanation:
M = Mass of moon =
r = Distance from Moon's center =
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
m = Mass of black hole
F = Force = 7870 N
From the universal force of gravity we have
The mass of the mini black hole is 43850.49112 kg
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
1 / p = 0.107375
p = 93.13 cm
Answer:
"we both attract each other with the same force but we know that attraction between two bodies depends upon their mass, greater the mass of two bodies is the force of attraction between them"(got this off the internet).