<span>When an object moves in a circle, the acceleration points toward the center of the circle. This acceleration is called centripetal acceleration.
We can use a simple equation to find centripetal acceleration.
a = v^2 / r
We can use this same equation to find the speed of the car.
v^2 = a * r
v = sqrt { a * r }
v = sqrt{ (1.50)(9.80 m/s^2)(11.0 m) }
v = 12.7 m/s
The speed of the roller coaster is 12.7 m/s</span>
Answer:
This is due to impulse
Explanation:
Impulse equal to mΔv and FΔt
You can set these equal as mΔv = FΔt
When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.
Answer:
Explanation:
The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth 
, where 
is the radius of the Earth.
We have then:
And then we use the centripetal acceleration formula:
Answer:
Greater than
Explanation:
Here, angular momentum is conserved.
When the cloud shrinks under the right conditions, a star may be formed.
Thus, Diameter of clouds are much higher than a star.
Moment of inertia of cloud is greater than the star's inertial.
so, angular velocity of the star would be greater than angular velocity of the rotating gas.
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
