Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M
= 1.99 × 10³⁰ kg
Mass of the neutron star
M
= 2( M
)
M
= 2( 1.99 × 10³⁰ kg )
M
= ( 3.98 × 10³⁰ kg )
Radius of neutron star R
= 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω
.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM
= / R
² = mR
ω
²
ω
² = GM
= / R
³
ω
= √(GM
= / R
³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω
= √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω
= √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω
= √ 120831133.3636777
ω
= 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
Answer:
Option C is the correct answer.
Explanation:
Considering vertical motion of ball:-
Initial velocity, u = 2 m/s
Acceleration , a = 9.81 m/s²
Displacement, s = 40 m
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
40 = 2 x t + 0.5 x 9.81 x t²
4.9t² + 2t - 40 = 0
t = 2.66 s or t = -3.06 s
So, time is 2.66 s.
Option C is the correct answer.
<span>At the periphery of a hurricane the air is sinking, and several kilometers above the surface, in the eye, the air is sinking. </span>
To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

Here



Mass inside the orbit in terms of Volume and Density is

Where,
V = Volume
Density
Now considering the volume of the star as a Sphere we have

Replacing at the previous equation we have,

Now replacing the mass at the gravitational acceleration formula we have that


For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

At the same time the general expression for the centripetal acceleration is

Where
is the orbital velocity
Using this expression in the left hand side of the equation we have that



Considering the constant values we have that


As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.
So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density