A block of mass, m, sits on the ground. A student pulls up on

ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po

Levart [38]

Answer:

$6.046N$

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

$F_n_e_t=F_g+F_t$

$F_t$ is the tension force. $F_g=9.8N/kg$ is the force of gravity.

$F_n_e_t=ma_c=mv^2/r\\$

where $r$ is the rope's radius from the fixed point.

From the net force equation above:

$F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N$

Hence the tension force is 6.046N

8 0

3 years ago

As a way of determining the inductance of a coil used in a research project, a student first connects the coil to a 5.62 V batte

Reptile [31]

Answer:

Its inductance L = 166 mH

Explanation:

Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR

R = V/I = 5.62/0.698 = 8.052 Ω

Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from

V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'

Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω

WE now find the reactance X of the coil from

Z² = X² + R²

X = √(Z² - R²)

= √(97.5² - 8.05²)

= √(9506.25 - 64.8025)

= √9441.4475

= 97.17 Ω

Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.

L = X/2πf

= 97.17/2π(93.1 Hz)

= 97.17 Ω/584.965 rad/s

= 0.166 H

= 166 mH

Its inductance L = 166 mH

5 0

3 years ago

You have 2 objects flying towards you. One of them is havier with lower velocity and the other one is lighter but has higher vel

Lerok [7]

Answer: I would say the object with the Lower velocity because Lighter with Higher velocity makes it heavy, velocity is how heavy something is so the lighter it is the less difficult it will be to catch.

7 0

3 years ago

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.