Answer:
The gravitational force is definitely acting downwards towards the ground and this is equal to the weight of the skydiver.
the acceleration a = 7.8 m/s²
Explanation:
Given that :
the mass of the skydiver = 60 kg
Velocity = 50 m/s
Thus; gravitational force is definitely acting downwards towards the ground and this is equal to the weight of the skydiver.
Also; the air resistance is acting upward and the resultant of both forces = mass×acceleration
So;
mg-R = ma
60(9.8) - 120 = 60(a)
588 -120 = 60a
468 = 60a
a = 
a = 7.8 m/s²
Hence, the acceleration a = 7.8 m/s²
Answer:
28.2 m/s
Explanation:
The range of a projectile launched from the ground is given by:
where
v is the initial speed
g = 9.8 m/s^2 is the acceleration of gravity
is the angle at which the projectile is thrown
In this problem we have
d = 81.1 m is the range
is the angle
Solving for v, we find the speed of the projectile:
Answer:
Explanation:
This problem is based on conservation of rotational momentum.
Moment of inertia of rod about its center
= 1/12 m l² , m is mass of the rod and l is its length .
= 1 / 12 x 4.6 x .11²
I = .004638 kg m²
The angular momentum of the bullet about the center of rod = mvr
where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .
5 x 10⁻³ x v sin60 x .11 x .5 where v is velocity of bullet
According to law of conservation of angular momentum
5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and ( I + mr²) is moment of inertia of bullet rod system .
.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12
.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12
.238 x 10⁻³ v = 55.8375 x 10⁻³
.238 v = 55.8375
v = 234.6 m /s
309.5752 m/s.
First, split the "1385 m" in half, because it takes 2 hours to get that far. So this way, we're turning it into mph (miles per an hour). (If m = miles)
Secondly, turn the mph (miles per an hour) into m/s.
Third, just some calculation, and you're done.
