Look at the Grand Canyon cross section above. What was happening to the Grand Canyon area during the Permian Age?

point y is in the interior of xwz. given that and are opposite rays and mxwy4(mywz), what is mywz?

Xelga [282]

point y is in the interior of xwz. given that and are opposite rays and mxwy4(mywz), mywz = 36°

Opposite rays:- Opposite rays are the rays, which shares a common initial point, but points towards opposite directions. Angle between two opposite rays is 180°.

According to the question,

m∠XWY = 4(m∠YWZ) (Given)

As the two rays WX and WZ are opposite rays, the initial points of those are same. Here in the question, it is W.

∴∠XWZ = 180°

∴m∠XWY + m∠YWZ = 180°

⇒ 4(m∠YWZ) + m∠YWZ = 180°

⇒ 5(m∠YWZ) = 180°

⇒ m∠YWZ = 180°/5

⇒ m∠YWZ = 36°

Thus we can conclude that, m∠YWZ = 36°.

To know more about opposite rays refer below link:

brainly.com/question/28216147

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8 0

2 years ago

Your favorite team has made the playoffs, and you decide that you should host a party. Before kickoff, you turn on your wide-scr

zhuklara [117]

Answer:

5 hr

Explanation:

I just did it In my head

4 0

3 years ago

If you sit outside and get a sunburn, what type of electromagnetic waves are to blame? x–rays microwaves infrared light ultravio

N76 [4]

Then, it would be INFRARED.........

4 0

3 years ago

Read 2 more answers

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago

A mountain climber of mass 60.0 kg slips and falls a distance of 4.00 m, at which time he reaches the end of his elastic safety

Sever21 [200]

The energy that the rope absorbs from the climber is Ep=m*g*h where m is mass of the climber, g=9.81m/s² and h is the height the climber fell. h=4 m+2 m because he was falling for 4 meters and the rope stretched for 2 aditional meters. The potential energy stored in the rope is Er=(1/2)*k*x², where k is the spring constant of the rope and x is the distance the rope stretched and it is
x=2 m. So the equation from the law of conservation of energy is:

Ep=Er

m*g*h=(1/2)*k*x²

k=(2*m*g*h)/x² = (2*60*9.81*6)/2² = 7063.2/4 =1765.8 N/m

So the spring constant of the rope is k=1765.8 N/m.

7 0

3 years ago