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2 years ago

What caused the blackout in Canada?

Physics

1 answer:

Maru [420]2 years ago

7 0

Answer:

Hello, the tripping of a 230-kilovolt transmission line.

Explanation:

the tripping of a 230-kilovolt transmission line near Ontario, Canada, at 5:16 p.m., which caused several other heavily loaded lines also to fail. Hopefully this helps you find what your looking for!.

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A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees. What is the vertical component of t

Svetllana [295]

If a football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees, then the vertical component of the initial velocity would be 12.65 m/s

<h3>What is Velocity?</h3>

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.

As given in the problem A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees

The horizontal component of the velocity is given by

Vx = Vcosθ

The vertical component of the velocity is given by

Vy = Vsinθ

As we have to find the vertical component of the velocity given that speed of 16.71 m/s at an angle of 49.21 degrees from the ground

Vy = 16.71 × sin49.21°

Vy = 12.65 m/s

Thus, the vertical component of the velocity would be 12.65 m/s

Learn more about Velocity from here

brainly.com/question/18084516

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4 0

1 year ago

Determina la velocidad de un avión que recorre 459 km en 67 min

satela [25.4K]

Answer:

122397.7

Explanation:

4 0

2 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

2 years ago

A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at

Naddika [18.5K]

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

4 0

2 years ago

Time Intervals in Ice Ages

stepladder [879]

It started off with 68% less than it did at the peak, and later created a void and melted the remainder of the ice at about 92%

8 0

2 years ago