Answer:
The length of the sling is increased and the linear velocity of the stone stays the same.
Explanation:
Centripetal acceleration is given by
![a_C = \dfrac{v^2}{r}](https://tex.z-dn.net/?f=a_C%20%3D%20%5Cdfrac%7Bv%5E2%7D%7Br%7D)
where <em>v</em> is the linear velocity and <em>r</em> is the radius of motion.
In terms of the angular velocity, <em>ω</em>
<em />
, since ![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
From the relation,
is directly proportional to the square of the velocity and inversely proportional to the radius.
Hence, an increase in linear velocity results in an increase in the centripetal acceleration if the radius is kept constant. A decrease in the radius also causes an increase of the centripetal acceleration.
1) Impulse: -32,000 kg m/s
The impulse acting on the car is equal to the change in momentum of the car:
![I= \Delta p = m (v-u)](https://tex.z-dn.net/?f=I%3D%20%5CDelta%20p%20%3D%20m%20%28v-u%29)
where in this problem we have
m = 1000 kg is the mass of the car
v = 0 m/s is the final velocity of the car
u = 32 m/s is the initial velocity of the car
Substituting values into the equation, we find
![I=(1000 kg)(0-32 m/s)=-32,000 kg m/s](https://tex.z-dn.net/?f=I%3D%281000%20kg%29%280-32%20m%2Fs%29%3D-32%2C000%20kg%20m%2Fs)
2) -4751 N
The impulse exerted on the car is also equal to the product between the average force, F, and the duration of the collision, t:
![I=Ft](https://tex.z-dn.net/?f=I%3DFt)
where in this situation we know
is the impulse
t = 7 s is the duration of the collision
Solving the formula for F, we find the average force:
![F=\frac{I}{t}=\frac{-32,000 kg m/s}{7 s}=-4,571 N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BI%7D%7Bt%7D%3D%5Cfrac%7B-32%2C000%20kg%20m%2Fs%7D%7B7%20s%7D%3D-4%2C571%20N)
and the negative sign means that the force is in the opposite direction to the motion of the car.
Answer:
Explanation:
a )
Iz( t ) = A + B t²
Iz( t ) = angular velocity
putting dimensional formula
T⁻¹ = A + Bt²
A = T⁻¹
unit of A is rad s⁻¹
BT² = T⁻¹
B = T⁻³
unit of B is rad s⁻³
b )
Iz( t ) = A + B t²
dIz / dt = 2Bt
angular acceleration = 2Bt
at t = 0
angular acceleration = 0
at t = 5
angular acceleration = 2 x 1.6 x 5
= 16 rad / s²
Iz( t ) = A + B t²
dθ / dt = A + B t²
integrating ,
θ = At + B t³ / 3
when t = 0 , θ = 0
when t = 2
θ = At + B t³ / 3
= 2.65 x 2 + 1.6 x 2³ / 3
= 5.3 + 4.27
= 9.57 rad .
Flywheel turns by 9.57 rad during first 2 s .
1. The magnitude of the gravitational force between the Earth and an m is 54.1 N.
2. The magnitude of the gravitational force between the Moon and an m is 1.91 x 10⁻⁴ N.
3. The ratio of the magnitude of the gravitational force between an m on the surface of the Earth due to the Sun to that due to the Moon is 169.6.
<h3>
Gravitational force between Earth and mass, m</h3>
The gravitational force between Earth and mass, m is calculated as follows;
F(Earth) = Gm₁m₂/R²
F(Earth) = (6.67 x 10⁻¹¹ x 5.5 x 5.98 x 10²⁴)/(6,370,000)²
F(Earth) = 54.1 N
<h3>
Gravitational force between Moon and mass, m</h3>
F(moon) = Gm₁m₂/R²
F(moon) = (6.67 x 10⁻¹¹ x 5.5 x 7.36x 10²²)/(3.76 x 10⁸)²
F(moon) = 1.91 x 10⁻⁴ N
<h3>
Gravitational force between Sun and mass, m</h3>
F(sun) = Gm₁m₂/R²
F(sun) = (6.67 x 10⁻¹¹ x 5.5 x 1.99x 10³⁰)/(1.5 x 10¹¹)²
F(sun) = 0.0324 N
<h3>Ratio of F(sun) to F(moon)</h3>
= 0.0324/1.91 x 10⁻⁴
= 169.6
Thus, the magnitude of the gravitational force between the Earth and an m is 54.1 N.
The magnitude of the gravitational force between the Moon and an m is 1.91 x 10⁻⁴ N.
The ratio of the magnitude of the gravitational force between an m on the surface of the Earth due to the Sun to that due to the Moon is 169.6.
Learn more about gravitational force here: brainly.com/question/72250
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