To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,
Here
n = Number of node
T = Tension
= Linear density
L = Length
Replacing the values in the frequency and value of n is one for fundamental overtone
Similarly plug in 2 for n for first overtone and determine the value of frequency
Similarly plug in 3 for n for first overtone and determine the value of frequency
<h2>
Law 1:</h2><h3>An object already in motion stays in motion, unless acted upon by a force.</h3><h3 /><h2>Law 2:</h2><h3>
</h3><h3>f = forces on an object</h3><h3>m = mass of that object</h3><h3>a = acceleration of that object</h3><h3 /><h2>Law 3:</h2><h3>Everything has an equal and opposite reaction.</h3><h3 /><h3>Hope this helps!</h3>
Answer:
Explanation:
Two straight wires
Have current in opposite direction
i1=i2=i=2Amps
Distance between two wires
r=5mm=0.005m
Length of one wire is ∞
Length of second wire is 0.3m
Force between the wire,
The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by
F/l = μoi1i2/2πr
F/l=μoi²/2πr
μo=4π×10^-7 H/m
The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.
F/l = μoi1i2/2πr
F/0.3=4π×10^-7×2²/2π•0.005
F/0.3=1.6×10^-4
Cross multiply
F=1.6×10^-4×0.3
F=4.8×10^-5N
Answer:
The answer is C)The force of gravity from Earth acting on the spacecraft decreased because the distance from Earth increased.
Explanation:
Gravity, a force, is dependent on the mass of the object exerting the gravity and the distance of an outside object from that object. The larger the object, the more gravity it will exert on an outside object. This force decreases as you move away from the object, but it will always still exist and never be equal to 0.
