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2 years ago

A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?

1 answer:

6 0

Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J

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The conductor is initially electrically neutral, and then a charge q is placed at the center of the hollow space. Suppose the co

Anvisha [2.4K]

The total charge on the interior of the conductor is zero.

The total charge on the exterior of the conductor is 8q.

<h3>Total charge on the interior</h3>

Due to large number of electrons available for conduction in a conductor, most of the electrons moves to surface leaving zero net charge inside the conductor.

Thus, the total charge on the interior of the conductor is zero.

<h3>Total charge on the exterior</h3>

The total charge on the exterior of the conductor is calculated as follows;

Q = q + 7q = 8q

Thus, the total charge on the exterior of the conductor is 8q.

Learn more about net charge on interior and exterior of conductors here: brainly.com/question/14653264

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2 years ago

Urgent! Based on the diagram below, what color will each pigmented paper appear to be to an observer?

Nastasia [14]

Answer:

Example A will appear green, while Example B will appear greenish-blue.

Explanation:

The color of an object depend on which part of the visible light it reflects towards the observer. Visible light is made up of seven colors: Violet, Indigo, Blue, Green, Yellow, Orange, and Red (VIBGYOR). If all the colors will be reflected object will appear white. If all the colors are absorbed the object appears black. In example A, only green color is being reflected so it will appear <em>Green</em>.

In example B, green and blue are being reflected so the object will appear a mix of green and blue. This color is cyan (greenish blue).

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2 years ago

Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro

Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

$F \ \alpha \ \frac{1}{D^{2}}$ for a distance D

If we move them so that D is doubled:

$\frac{1}{2^{2}.D^{2} }$ = $\frac{1}{4} \eq \frac{1}{.D^{2} } \eq$

Then the force they experience is one fourth of the original.

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2 years ago

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

$261.3=0.3\times 9.8\times h$

$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

4 0

3 years ago

According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the

slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

$A=\dfrac{\pi}{4}\times d^2$

Put the value into the formula

$A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2$

$A=1.96\times10^{-9}\ m^2$

We need to calculate the magnitude of the force

Using formula of force

$F=\sigma A$

Put the value into the formula

$F=196\times10^{6}\times1.96\times10^{-9}$

$F=0.38416\ N$

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

$strain=\dfrac{\Delta l}{l_{0}}$

$\Delta l=strain\times l_{0}$

Put the value into the formula

$\Delta l=0.380\times l_{0}$

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

$l=l_{0}+\Delta l$

Put the value into the formula

$I=l_{0}+0.380\times l_{0}$

$l=1.38l_{0}$

$l_{0}=\dfrac{l}{1.38}$

$l_{0}=\dfrac{12\times10^{-2}}{1.38}$

$l_{0}=0.0869\ m$

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0

3 years ago