Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + 
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
DescriptionThe Quartering Acts were two or more Acts of British Parliament requiring local governments of the American colonies to provide the British soldiers with housing and food. Each of the Quartering Acts was an amendment to the Mutiny Act and required annual renewal by Parliament.
Hope this helps
Answer:
a = 9.86 m/s²
Explanation:
given,
distance between the centers of wheel = 156 cm
center of mass of motorcycle including rider = 77.5 cm
horizontal acceleration of motor cycle = ?
now,
The moment created by the wheels must equal the moment created by gravity.
take moment about wheel as it touches the ground, here we will take horizontal distance between them.
then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.
now equating both the moment
m g d = F h
d is the horizontal distance
h is the vertical distance
m g d = m a h
term of mass get eliminated
g d = a h
so,
a = 9.86 m/s²
Answer:
20.96 m/s
Explanation:
Using the equations of motion
y = uᵧt + gt²/2
Since the puck slides off horizontally,
uᵧ = vertical component of the initial velocity of the puck = 0 m/s
y = vertical height of the platform = 2 m
g = 9.8 m/s²
t = time of flight of the puck = ?
2 = (0)(t) + 9.8 t²/2
4.9t² = 2
t = 0.639 s
For the horizontal component of the motion
x = uₓt + gt²/2
x = horizontal distance covered by the puck
uₓ = horizontal component of the initial velocity = 20 m/s
g = 0 m/s² as there's no acceleration component in the x-direction
t = 0.639 s
x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m
For the final velocity, we'll calculate the horizontal and vertical components
vₓ² = uₓ² + 2gx
g = 0 m/s²
vₓ = uₓ = 20 m/s
Vertical component
vᵧ² = uᵧ² + 2gy
vᵧ² = 0 + 2×9.8×2
vᵧ = 6.26 m/s
vₓ = 20 m/s, vᵧ = 6.26 m/s
Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s
Answer:
The free-body diagram of the cannonball is found in the attachment below
<em>Note The question is incomplete. The complete question is as follows:</em>
<em>A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglected.</em>
<em>Draw the free-body diagram of the cannonball.</em>
Explanation:
Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
In order to construct free-body diagrams, it is important to know the various types of forces acting on the object in that situation. Then, the direction in which each of the forces is acting is determined. Finally the given object is drawn using any given representation, usually a box, and the direction of action of the forces are represented using arrows.
In the given situation of a cannonball which has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction., the forces acting on it are:
F = force exerted by the cannon acting in the direction of angle of projection
Fdrag = drag force. The drag force acts in a direction opposite to the force exerted by the cannon
Fw = weight of the cannonball acting in a downward direction
The free body diagram is as shown in the attachment below.
