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3 years ago

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A charged particle is injected at 211 m/s into a 0.0633‑T uniform magnetic magnetic field perpendicularly to the field. The diam

eter of its orbit is measured and found to be 0.0389 m. What is the charge–to–mass ratio of this particle?

1 answer:

lakkis [162]3 years ago

7 0

Answer:

$85.68*10^3C/kg$

Explanation:

For this problem we need the concept about Force in a Magnetic field,

For definition we know that

$F=m\frac{v^2}{r}$

Where v is the velocity, m the mass and r the radius or distance between the two points.

We know as well that

$F = qvB$

where q is the charge of a proton

v the velocity and B the magnetic field, then matching the two equation,

$qvB=m\frac{v^2}{r}$

Re-arrange for q/m (charge to mass ratio)

$\frac{q}{m} = \frac{v}{Br}$

Our values are,

$v=211m/s$

$B= 0.0633T$

$r=0.0389m$

Substituting,

$\frac{q}{m} = \frac{211}{0.0633*0.0389}$

$\frac{q}{m} = 85689.8 C/kg = 85.68*10^3C/kg$

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DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

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with $\theta=41.9^{\circ}$

- The frictional force, whose magnitude is

$F_f = \mu mg$

where $\mu=0.33$ , m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

$F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N$

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3 years ago

A car accelerates from rest at 3.00 m/second squared What is the velocity after 5 secs ? What is the displacement after 5 second

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Explanation:

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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

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Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

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aev [14]

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

$F_y = F sin \theta$

$F_y = 245\times sin 55^0$

$F_y =200.69$

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

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g A thin uniform film of oil that can be varied in thickness covers a sheet of glass of refractive index 1.52. The refractive in

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Answer:

The right solution is "84.09 nm".

Explanation:

The given values are:

Refractive index of glass,

= 1.52

Refractive index of oil (n),

= 1.64

Wavelength (λ),

= 555 nm

Now,

The thickness of the film (t) will be:

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= $\frac{555}{4\times 1.65}$

= $\frac{555}{6.6}$

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