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:
the distance between the particles and also the amount of electric charge they carry.
Explanation:
yup
Answer:
Some examples of vegetative propagation are farmers creating repeated crops of apples, corn, mangoes or avocados through asexual plant reproduction rather than planting seeds. Vegetative propagation can be accomplished from side-shoots, slips, stems and sections of tubers, bulbs or rhizomes.
Explanation:
Answer:
Part a)

Part b)

Part c)

Part d)
Net force on a closed loop in uniform magnetic field is always ZERO

Explanation:
As we know that force on a current carrying wire is given as

now we have
Part a)
current in side 166 cm and magnetic field is parallel
so we have

here we know that L and B is parallel to each other so

Part b)
For 68.1 cm length wire we have

here we know that


so we have


Part c)
For 151 cm length wire we have

here we know that


so we have


Part d)
Net force on a closed loop in uniform magnetic field is always ZERO

Answer:
each resistor is 540 Ω
Explanation:
Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance
defined by the formula:

Therefore, R/3 is the equivalent resistance of the initial circuit.
In the second circuit, two of the resistors are in parallel, so they are equivalent to:

and when this is combined with the third resistor in series, the equivalent resistance (
) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

The problem states that the difference between the equivalent resistances in both circuits is given by:

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:
