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Anastaziya [24]

3 years ago

5

an electrically charged particle with mass m, velocity v, and charge q moves in a circle in magnetic field b. what is the formul

a for the radius of circle r?

1 answer:

kolbaska11 [484]3 years ago

5 0

Answer:

$r=\frac {mv}{qb}$

Explanation:

In this case, since the charged particle moves in circular motion, the centripetal force is equivalent to the magnetic force.

$\frac {mv^{2}}{r}=qvb\\\frac {mv}{r}=qb\\r=\frac {mv}{qb}$

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11. A 6.0‐m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is

Irina-Kira [14]

Answer:

Time will be 19 ms so option (a) is correct option

Explanation:

We have given that mass of wire m = 50 gram = 0.5 kg

Frequency f = 810 Hz

Wavelength = 0.4 m

Velocity is given by

$v=wavelength\times frequency=810\times 0.4=324m/sec$

Amplitude is given as d = 6 m

So time $t=\frac{distance}{velocity}=\frac{6}{324}=0.01851=19ms$

So option (a) is correct option

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3 years ago

When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

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3 years ago

4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

VMariaS [17]

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

$v=\dfrac{d}{t}$

Where. d = distance

t = time

Put the value into the formula

$v=\dfrac{18.288}{2}$

$v=10\ m/s$ during that run

(b). We need to calculate their velocity

Using formula of speed

$v=\dfrac{\Delta d}{\Delta t}$

Put the value into the formula

$v=\dfrac{22.86-36.58}{2}$

$v=-6.86\ m/s$

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

$d=vt$

Put the value into the formula

$d=6.86\times1.3$

$d=8.91\ m$

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

8 0

3 years ago

A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3

marissa [1.9K]

Answer:

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Explanation:

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$\Rightarrow H=\frac{v^2}{2g}+h$

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H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

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$\Rightarrow v^2=\frac{gD^2}{2h}$

thus, height H in terms of D is given by

$H=\frac{D^2}{2h}+h$

$H=\frac{13.9^2}{2\times3}+3$

H=35.20 m

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3 years ago

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Answer:

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Explanation:

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3 years ago