Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.
Answer:
200 km/hr
Explanation:
Since he goes 80km per hour, multiply this by 2.5 or two and a half hours.
80 x 2.5 = 200 km/hr.
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
Where
Angular acceleration
Angular velocity
t = Time
Our values are
Replacing at the previous equation we have that the angular velocity is
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
Therefore the angular acceleration of a point on the outer edge of the tires is 
I am not so sure about this it is too difficult
Answer:
c: long and thin resistor.
Explanation:
The resistance of a resistor is given by:
R = ρ*L/A
where:
R = resistance
ρ = resistivity (depends on the material)
L = length of the material
A = cross-sectional area of the material
We can see that the length is on the numerator, which means that if we increase the length, then the resistance is increased.
We also can see that the cross-sectional area is on the denominator, then if we increase the area (for example, with a ticker resistor) the resistance decreases.
Then if we want to maximize the resistance, we need to have a long and thin resistor, so the correct answer is c.
