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2 years ago

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A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13

N, what is the strength of the magnetic field? (Express your answer using two significant figures.)

1 answer:

tia_tia [17]2 years ago

8 0

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

$F = qvB$

here we have

$B = \frac{F}{qv}$

here we know that

$F = 4.8 \times 10^{-13} N$

$q = 1.6 \times 10^{-19} C$

$v = 4 \times 10^6 m/s$

now from above equation we have

$B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}$

$B = 0.75 T$

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The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress

PolarNik [594]

Answer:

T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42-ksi = 42 x 6.89 = 289.58 MPa

Torque = T =?

<u>Solution:</u>

σ = (P x r) / T

T = (P x r) / σ

T = (386.11 x 0.02) / 289.58

T = 0.03 Nm.

7 0

2 years ago

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl

Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0

2 years ago

A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di

Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

$F . t = m(v_f - v_i)$

so here we know that object is at rest initially so we have

$80 (12) = 80( v_f - 0)$

$v_f = 12 m/s$

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

$F = 80 - \frac{40}{3} t$

now again by same equation we have

$\int F .dt = m(v_f - v_i)$

$\int (80 - (40/3)t) dt = 80(v_f - 12)$

$80 t - \frac{40t^2}{6} = 80(v_f - 12)$

put t = 6 s

$480 - 240 = 80(v_f - 12)$

$v_f = 12 + 3$

$v_f = 15 m/s$

6 0

2 years ago

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the

Diano4ka-milaya [45]

Answer:

<em>The object could fall from six times the original height and still be safe</em>

Explanation:

<u>Free Falling</u>

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

$U=m.g.h$

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

$\displaystyle K=\frac{1}{2}m.v^2$

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

$\displaystyle m.g.h=\frac{1}{2}m.v^2$

Solving for h

$\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}$

If the drop had occurred in the Moon, then

$\displaystyle h_M=\frac{v_M^2}{2g_M}$

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

$\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}$

$\displaystyle h_M=6\frac{v^2}{2g}=6h$

This means the object could fall from six times the original height and still be safe

6 0

2 years ago

A long coaxial cable (Fig. 2.26) carries a uniform volume charge density rho on the inner cylinder (radius a), and a uniform sur

Yuki888 [10]

Answer:

a) E = ρ / e0

b) E = ρ*a / (e0 * r)

c) E = 0

Explanation:

Because of the geometry, the electric field lines will all have a radial direction.

Using Gauss law

$Q/e0 = \int \int E * dA$

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.

We can make this cylinder of different radii to evaluate the electric field at different points.

Then:

A = 2*π*r (area of cylinder per unit of length)

Q/e0 = 2*π*r*E

E = Q / (2*π*e0*r)

Where Q is the charge contained inside the cylinder.

Inside the cable core:

There is a uniform charge density ρ

Q(r) = ρ * 2*π*r

Then

E = ρ * 2*π*r / (2*π*e0*r)

E = ρ / e0 (electric field is constant inside the charged cylinder.

Between ther inner cilinder and the tube:

Q = ρ * 2*π*a

E = ρ * 2*π*a / (2*π*e0*r)

E = ρ*a / (e0 * r)

Outside the tube, the charges of the core cancel each other.

E=0

4 0

2 years ago