Can we use position as another word for phase?

The number of magnetic field force lines passing through the given surface determines: 1. magnetic flux 2. magnetic induction 3.

Assoli18 [71]

A magnetic flux would be the correct answer

8 0

3 years ago

"Videos of hoverboard riders who were injured when they fell while operating their hoverboards at a low speeds "went viral" over

bagirrra123 [75]

Answer:

The answer is "No, Hoverboards are risky, and riders are in danger of falling".

Explanation:

It's also known as a self-balanced scooter, it handheld electrical devices traveling on two wheels are hoverboards. It dominated the industry around 2015 and since then has become more and more successful. A rider is balanced on a frame between these wheels, driven by battery-powered lithium-ion batteries.

8 0

2 years ago

A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball

lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

$M_{1} u_{1} +M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

$M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

Since initial velocity for M1 ball is zero, then

$M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

and

$M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

So, on solving all the above equation, we get an equation for velocity and that is

$\frac{2M_{2}u_{2} }{(M_{1}+M_{2} }$ =final velocity of ball with mass 2.5 kg

$v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s$

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is $\frac{1}{2}*2.5*(4.67)^{2} =27.09 J$

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0

3 years ago

An electric motor rotating a workshop grinding wheel at 1.00 × 10² rev/min is switched off. Assume the wheel has a constant nega

ololo11 [35]

An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.

<h3>What is negative angular acceleration?</h3>

A particle that has a negative angular velocity rotates counterclockwise.
Negative angular acceleration () is a "push" that is hence counterclockwise.
The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
It is known that a change in the linear acceleration correlates to a change in the linear velocity.

Let t be the time taken to stop.

ω = 0 rad/s

Use the first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

To learn more about angular acceleration, refer to:

brainly.com/question/21278452

#SPJ4

7 0

1 year ago

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f

hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass $m_1$ = 30 kg

mass $m_2$ = 50 kg

$\mu$ = 0.1

From the question; we can arrive at two cases;

That :

$m_{2} a _ \ {net} }= m_2g - T$ ----- equation (1)

$m_{1} a _ \ {net} }= T - mg sin \theta - F$ ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a = $[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]$ g

80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0

2 years ago