Answer:
The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
Explanation:
The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.
When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.
So the centripetal acceleration, a' = 16v²/R.
To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16
a'/a = 16
a' = 16a.
So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
Answer:
r = 2.031 x 10⁶ m = 2031 km
Explanation:
In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:
Centripetal Force = Gravitational Force
mv²/r = GmM/r²
v² = GM/r
r = GM/v²
where,
r = radial distance = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Planet = 3.52 x 10¹³ kg
v = tangential speed = 0.034 m/s
Therefore,
r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²
<u>r = 2.031 x 10⁶ m = 2031 km</u>
There must be a centripetal force to move the object move in a curve path.
Answer:
The motion of the block is downwards with acceleration 1.7 m/s^2.
Explanation:
First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.
Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.
Now, let’s investigate the free-body diagram of the block.
Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,
As for the static friction, we will consider the angle 31.8, but just before the block starts the move.
Answer:
u=36.8m/s
Explanation:
because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations
u^2=v^2-2ā*s. where:
u^2 stands for intial velocity
v^2 stands for final velocity
since the cougar skidded to a complete stop the final velocity is zero.
u^2=v^2-2ā*s
u^2=(0)^2 -2(-2.87 m/s^2)*236 m
u^2=0+5.74m/s^2* 236m
u^2=1354.64m^2/s^2
u=√1354.64m^2/s^2
u=36.8m/s (approximate value)
when ever the acceleration is constant you can use one of the following equation to find the required value.
1. v = u + at. (no s)
2. s= 1/2(u+v)t. (no ā)
3. s=ut + 1/2at^2. ( no v)
4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)
