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2 years ago

7

Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of dis

placement and distance are exactly the same?

1 answer:

Sedaia [141]2 years ago

6 0

Answer: Please find the answer in the explanation

Explanation:

Under what circumstances does distance traveled equal magnitude of displacement?

When a body's motion is linear in one direction. Or a body moving in a straight line without turning back.

What is the only case in which magnitude of displacement and distance are exactly the same?

When the body is moving in a straight line with without changing direction or without turning back.

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An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the

ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, $E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J$

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.

The magnetic field in north direction, $B_y=19911.5\ nT$

The magnetic field in west direction, $B_x=-3257.1\ nT$

The magnetic field in vertical direction, $B_z=48381.8 \ nT$

Magnetic field, $B=(-3257.1i+19911.5j+48381.8k)\ nT$

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

$E_k=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E_k}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}$

$v=5.92\times 10^7 i\ m/s$ (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

$F=q(v\times B)$

$F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})$

Since, $i\times j=k\ \\j\times k=i\\k\times i=j$

And, $i\times i=j\times j=k\times k=0$

$F=1.6\times 10^{-19}\times [1178 k-2864.20j]$

$|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}$

$F=4.95\times 10^{-16}\ N$

(b) Let a is the acceleration of the electron. It can be calculated as :

$a=\dfrac{F}{m}$

$a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}$

$a=5.43\times 10^{14}\ m/s^2$

Hence, this is the required solution.

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2 years ago

A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm . Both

kakasveta [241]

Answer:

The right response will be "450 volts".

Explanation:

The given values are:

R1 = 4.00 cm

R2 = 6.00 cm

q1 = +6.00 nC

q2 = −9.00 nC

As we know,

The potential difference between the two shell's difference will be:

⇒ $\Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]$

$=K[\frac{q1}{R2}-\frac{q1}{R1} ]$

On substituting the values, we get

$=(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]$

Δ $=450 \ volts$

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A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h

Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = $\frac{dr}{dt} X 12 = 4 \frac{ft}{sec} X 12 sec = 48 ft$

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi (48)(4)$

dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

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Why, on a sunny day, it is normally hot inside a greenhouse

Dahasolnce [82]

Because of the greenhouse gases inside the greenhouse, and the gases trap the heat from the sun so the plant don't freeze, hope this helps

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