I am Pushing down at a 20

A particle of charge q moving through a magnetic field B experiences the force where v is the velocity of the particle. Show tha

MrRa [10]

Answer:

dx/Dt x B . x =0

Explanation:

Let's calculate the work and the magnetic force, the expression for magnetic force is

F = qv x B

Bold indicate vector quantities, the expression for the job is

W = F. X

Let's replace in this equation

W = q v x B . X

The definition of speed is

v = dX / dt

With what work is left

W = q dX / dt x B . X

As we can see the vector product gives us a vector perpendicular to dX and its scalar product by X of zero

Second part

The speed a vector and although the magnitude is constant the change of direction implies a change in the speed.

Let's calculate the magnitudes of speed (speed)

F = qv B sin θ

F = ma

q v B sin θ = ma

a = qvB / m senT

This acceleration is perpendicular to the magnetic field and the velocity, so it does not change if magnitude but its direction, it is directed to the center of the circle.

| v | = q vB/m sin θ

3 0

3 years ago

According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something

Tasya [4]

Answer:

The pressure is $p_1 = 4051.4 \ Pa$

Explanation:

From the question we are told that

The gauge pressure at the mouth is $p_1$

The radius of the column is $r_2 = 4 \ mm = 0.004 \ m$

The speed of the liquid outside the body is $v_2 = 3.1 \ m/s$

The area of the column is $A_2$

The area inside the mouth $A_1 = 10 A_2$

Generally according to continuity equation

$v_1 A_1 = v_2 A_2$

=> $v_ 1 = v_2 * \frac{A_2}{A_1}$

=> $v_ 1 = 3.1 * \frac{1}{10}$

=> $v_ 1 = 0.31 \ m/s$

So

$A_1 = 10A_2$

=> $\pi * r_1^2 = 10(\pi * r_2^2)$

=> $r_1 = 10 * r_2$

substituting values

$r_1 = 10 * 0.004$

$r_1 =0.04 \ m$

Now the height of inside the mouth is $h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m$

Now the height of the column is $h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m$

Generally according to Bernoulli's equation

$p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]$

Now $\rho = 1000 \ kg m^{-3}$ which is the density of water

$p_2$ is the gauge pressure of the atmosphere which is zero

So

$p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-$

$[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]$

$p_1 = 4051.4 \ Pa$

8 0

3 years ago

You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th

vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

5 0

3 years ago

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An object is thrown straight up with an initial velocity of 10 m/s, and there is an air resistance force causing an acceleration

lana [24]

Answer:

Vf= 7.29 m/s

Explanation:

Two force act on the object:

1) Gravity

2) Air resistance

Upward motion:

Initial velocity = Vi= 10 m/s

Final velocity = Vf= 0 m/s

Gravity acting downward = g = -9.8 m/s²

Air resistance acting downward = a₁ = - 3 m/s²

Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²

( Acceleration is consider negative if it is in opposite direction of velocity )

Now

2as = Vf² - Vi²

⇒ 2 * (-12.8) *s = 0 - 10²

⇒-25.6 *s = -100

⇒ s = 100/ 25.6

⇒ s = 3.9 m

Downward motion:

Vi= 0 m/s

s = 3.9 m

Gravity acting downward = g = 9.8 m/s²

Air resistance acting upward = a₁ = - 3 m/s²

Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²

Now

2as = Vf² - Vi²

⇒ 2 * 6.8 * 3.9 = Vf² - 0

⇒ Vf² = 53. 125

⇒ Vf= 7.29 m/s

8 0

3 years ago

During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t

tia_tia [17]

1.A) 4.9 m

AL2006 Ace

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.

ANYTHING you drop does that, if air resistance doesn't hold it back.

Read more on Brainly.com - brainly.com/question/11776597#readmore

2 idk sorry

5 0

3 years ago

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