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3 years ago

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A particle of mass 7.00 10-8 kg and charge 7.8-µc is traveling due east. it enters perpendicularly a magnetic field whose magnit

ude is 29.0 t. after entering the field, the particle completes one-half of a circle and exits the field traveling due west. how much time does the particle spend traveling in the magnetic field?

1 answer:

sertanlavr [38]3 years ago

6 0

The radial force on the particle is given by the Lorentz force. They don't give you the velocity so you have to solve for it in terms of the radius and get them to cancel.

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For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant

myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4

Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0

3 years ago

Which best describes electromagnetic waves moving from gamma rays to radio waves along the electromagnetic spectrum

kondor19780726 [428]

Answer:

Explanation:

No

9 0

3 years ago

The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em

Vilka [71]

Answer:

$\rho=995.50\ kg.m^{-3}$

$\bar w=9765.887\ N.m^{-3}$

$s=0.9955$

Explanation:

Given:

volume of liquid content in the can, $v_l=0.355\ L=3.55\times 10^{-4}\ L$

mass of filled can, $m_f=0.369\ kg$

weight of empty can, $w_c=0.153\ N$

<u>So, mass of the empty can:</u>

$m_c=\frac{w_c}{g}$

$m_c=\frac{0.153}{9.81}$

$m_c=0.015596\ kg$

<u>Hence the mass of liquid(soda):</u>

$m_l=m_f-m_c$

$m_l=0.369-0.015596$

$m_l=0.3534\ kg$

<u>Therefore the density of liquid soda:</u>

$\rho=\frac{m_l}{v_l}$ (as density is given as mass per unit volume of the substance)

$\rho=\frac{0.3534}{3.55\times 10^{-4}}$

$\rho=995.50\ kg.m^{-3}$

<u>Specific weight of the liquid soda:</u>

$\bar w=\frac{m_l.g}{v_l}=\rho.g$

$\bar w=995.5\times 9.81$

$\bar w=9765.887\ N.m^{-3}$

Specific gravity is the density of the substance to the density of water:

$s=\frac{\rho}{\rho_w}$

where:

$\rho_w=$ density of water

$s=\frac{995.5}{1000}$

$s=0.9955$

3 0

3 years ago

Read 2 more answers

On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m

nika2105 [10]

Answer:

The magnitude of the torque is 263.5 N.

Explanation:

Given that,

Applied force = 31 N

Distance from the axis = 8.5 m

She applies her force perpendicularly to a line drawn from the axis of rotation

So, The angle is 90°

We need to calculate the torque

Using formula of torque

$\tau=Fd\sin\theta$

Where, F = force

d = distance

Put the value into the formula

$\tau=31\times8.5\sin90$

$\tau= 263.5\ N$

Hence, The magnitude of the torque is 263.5 N.

4 0

2 years ago

The relative humidity would be ________% if the actual water vapor in the air was 4 grams per cubic meter, the air's capacity to

AveGali [126]

Answer:

50%

Explanation:

Humidity is the amount water vapor present in the atmosphere.

Relative humidity is defined as the ratio of partial water vapor present in air to the actual water vapor at a particular temperature. It is expressed in percentage and the higher the percentage RH, the more the saturated water vapor present in the atmosphere and vice versa.

It is expressed mathematically as shown;

RH = actual water vapor in air/saturated water vapor × 100%

If the actual water vapor in the air was 4 grams per cubic meter and the air's capacity to hold water vapor was 8 grams per cubic meter

Actual water vapor = 4g/cm³

Air's water capacity (saturated water vapor) = 8g/cm³

RH = 4/8×100

RH = 50%

3 0

3 years ago