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A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc

blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0

3 years ago

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

#SPJ4

3 0

2 years ago

A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m

Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

Horizontal displacement = 40.5 m

Time

$t=\frac{40.5}{28.64}=1.41s$

Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

Time to reach the goal posts 40.5 m away = 1.41 seconds

Acceleration = -9.81m/s²

Substituting in s = ut + 0.5at²

s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

Height of throw = 1.4 m

Height traveled by ball = 2.83 m

Total height = 2.83 + 1.4 = 4.23 m

When the ball goes to first base it will be 4.23 m high.

8 0

3 years ago

A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

$v^2=u^2+2as$

$0^2=14.5^2+2\times a\times 18.25$

$a=-5.76m/sec^2$

We know that acceleration is given by

$a=\mu g$

$5.76=\mu\times 9.81$

$\mu =0.587$

So coefficient of friction will be 0.587

6 0

3 years ago

A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr

jasenka [17]

Answer: The original temperature was

$T_{1}=126.51K$

Explanation:

Let's put the information in mathematical form:

$V_{1}=12.7m^{3}$

$T_{1}=?$

$V_{2}=32.5m^{3}$

$T_{2}=323K$

$P_{1}=P_{2}=3atm$

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

$PV=nRT$

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking $R=0.08205746\frac{atm.l}{mol.K}$ , we have:

$n=\frac{P_{2}V_{2}}{RT_{2}}$

⇒ $n=3.67mol$

Now:

$T_{1}=\frac{P_{1}V_{1}}{nR}$

⇒ $T_{1}=126.51K$

7 0

2 years ago

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