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3 years ago

Joe balances a stationary coin on the the tip of his finger 20 cm from the top of the table. How much work is Joe doing?

Physics

1 answer:

adell [148]3 years ago

7 0

The work done by Joe is 0 J.

<u>Explanation</u>:

When a force is applied to an object, there will be a movement because of the applied force to a certain distance. This transfer of energy when a force is applied to an object that tends to move the object is known as work done.

The energy is transferred from one state to another and the stored energy is equal to the work done.

W = F . D

where F represents the force in newton,

D represents the distance or displacement of an object.

Force = 0 N, D = 20 cm = 0.20 m

W = 0 $\times$ 0.20 = 0 J.

Hence the work done by Joe is 0 J.

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Which of the following keys moves the insertion point to the beginning of data in a cell? (Points : 2)

Korvikt [17]

Insert moves the insertion point to the beginning of data in a cell so the answer is INSERT :)))
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6 0

3 years ago

The bob (weight) at the end of a pendulum has a mass of 0.3 kilograms. The bob is pulled to position B and allowed to swing. It

Ivahew [28]

Answer:

0.147 J

Explanation:

The total energy that has been transformed into thermal energy is equal to the loss of gravitational potential energy between the initial situation (bob at h=0.5 m above the ground) and the final situation (bob back but at h=0.45 m above the ground).

Therefore, we have

$E_{thermal}=\Delta U=mgh_1 - mgh_2 = mg(h_1 -h_2)$

where

m = 0.3 kg is the mass of the bob

g = 9.8 m/s^2

h1 = 0.5 m is the initial height

h2 = 0.45 m is the final height

Substituting, we find the thermal energy

$E_{thermal}=(0.3 kg)(9.8 m/s^2)(0.5 m-0.45 m)=0.147 J$

Therefore, the energy transformed into thermal energy is 0.147 J.

3 0

3 years ago

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

Crazy boy [7]

Answer:

$F = -1.07 *10^{-8} \ N$

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

6 0

4 years ago

A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius

alisha [4.7K]

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

$B=\frac{\mu_oI}{2R}$ (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

$B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}$ (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

$\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A$