An engine uses 5280 j of input heat to do 702 j of work. what is its actual efficiency'

What is the direction of the magnetic field if an electron moving in the positive x direction experiences a magnetic force in th

Alex787 [66]

Given :

An electron moving in the positive x direction experiences a magnetic force in the positive z direction.

To Find :

The direction of the magnetic field.

Solution :

We know, force is given by :

$\vec{F}=q(\vec{v}\times \vec{B)}$

Here, q = -e.

$\vec{F}=(-e)(\vec{v}\times \vec{B)}\\\\\hat{k}=(-e)(\hat{i}\times \vec{B})$

Now, for above condition to satisfy :

$\hat{i}\times \vec{B}=-\hat{k}$

So, $\vec{B}=-\hat{j}$

Therefore, direction of magnetic field is negative y direction.

Hence, this is the required solution.

7 0

3 years ago

Janet jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.8 s later. How high is th

Archy [21]

Initial height of platform is 15.87m or 16m.

8 0

2 years ago

Aristotle said that a moving earthly or `mundane' object with nothing pushing or pulling on it will always A slow down and stop.

andrew11 [14]

Answer:

A slow down and stop

Explanation:

When there is no force acting on something it automatically begins to slow down and then stops.Essentially, Aristotle's perspective of motion is that "it requires a force to move an object in an unnatural" way— or, plainly, that "movement involves strength." Indeed, if you propel a book, it keeps moving. Once you stop trying to push, it comes to a stop.

5 0

2 years ago

The pulley system has a mechanical advantage of 5. Determine how much input force is required to achieve 1000 N of lifting force

olga55 [171]

Answer:

Input force of pulley system = 200 N

Explanation:

Given:

Mechanical advantage of pulley system = 5

Output force from pulley system = 1,000 N

Find;

Input force of pulley system

Computation:

Mechanical advantage = Output force / Input force

Mechanical advantage of pulley system = Output force from pulley system / Input force of pulley system

5 = 1,000 / Input force of pulley system

Input force of pulley system = 1,000 / 5

Input force of pulley system = 200 N

3 0

2 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

2 years ago