A 3.63.kgkg chihuahua charges at a speed of 3.3m/s3.3m/s. What is the magnitude of the average force needed to bring the chihuah
2 answers:
Answer:
The magnitude of the average force needed to bring the chihuahua to a stop is 23.96 N
Explanation:
Newton's second law states that the acceleration of an object is inversely proportional to the mass of the object. The greater the mass of an object, the less its acceleration will be if a given net force is applied to it. In other words, mass is a property of objects that opposes acceleration when a force is applied.
In this way, Newton's second law says that the acceleration of a body is proportional to the resultant of forces on the acting and inversely proportional to its mass. This is expressed by the mathematical relationship:
F = m * a
where:
F = Force [N]
m = Mass [kg]
a = Acceleration []
You know that 3.63 kg chihuahua charges at a speed of 3.3m/s and you need to bring the chihuahua to a stop in 0.50s.
A motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies by the same amount in each unit of time, so that the acceleration is constant. In this case the uniformly varied rectilinear motion is fulfilled. Being:
velocity (final) = velocity (initial) + acceleration * time
vf = vi + a * t
the acceleration is:
In this case:
- vf= 0
because the chihuahua stops.
- vi= 3.3
- t=0.50 s
Replacing:
Solving:
a= -6.6
Then the magnitude of the average force can be calculated as:
F=m*a
F=3.63 kg* (-6.6 )
Solving:
<u><em>F= - 23.96 N</em></u>
<u><em>The magnitude of the average force needed to bring the chihuahua to a stop is 23.96 N</em></u>
<u><em></em></u>
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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
The bullet travels a horizontal distance of 276.5 m
The bullet is shot forward with a horizontal velocity . It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>
The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.
The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.
Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,
Substitute 0 m/s for , 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.
The horizontal distance traveled by the bullet is given by,
Substitute 500 m/s for and 0.5530s for t.
The bullet travels a distance of 276.5 m.
Answer:
Explanation:
The equation for Power is
P = Work/time to do work and the equation for work is
Work = FΔx
We first need to find the amount of work done, then we can find the power it took to do that work.
W = 2000(9.8)(28) so
W = 550,000 N*m
Now we fill that into the power equation:
gives us
P = 18000 Watts. But we need kW, so we divide by 1000 to get
P = 18 kW of power.