A 1500 kg boat rests on a slip-rail which is inclined at 20° to the horizontal ground. The

Wave crests in a pond are separated by 0.2m . The period of these waves is 0.2 s . What is the speed at which these waves travel

Ne4ueva [31]

1m per minute

or .5 m per 30 seconds

5 0

2 years ago

A campground official wants to measure the distance across an irregularly-shaped lake, but can't do so directly. He picks a poin

sleet_krkn [62]

Answer:

149 m

Explanation:

The distances across the lake is forming a triangle.

let the distance between the point and the left side be 'x'

and the distance between the point and the right be 'y'

and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:

∠Z = 83°

x = 105 m

y = 119 m

Now, applying the Law of Cosines, we get

z² = x² + y² - 2xycos(Z)

Substituting the values in the above equation, we get

z² = 105² + 119² - 2×105×119×cos(83°)

or

z = √22140.48

or

z = 148.796 m ≈ 149 m

The point is 149 m across the lake

8 0

2 years ago

HELP!!! 30 POINTS+BRAINLIEST!!!!!QUICK!!

tia_tia [17]

Answer:

A:7.2 $ms^{-1}$

B:14.25 $ms^{-1}$

C:1.45 $sec$

D:10.3 $m$

E:2.9 $sec$

F:20.88 $m$

Explanation:

Let $v$ be the velocity and $\alpha$ be the angle between the velocity and ground.

Question A:

Horizontal component of velocity is given by $vcos(\alpha )$ .

So,horizontal component is $16\times cos(63)=16\times 0.45=7.2ms^{-1}$

Question B:

Vertical component of velocity is given by $vsin(\alpha )$ .

So,vertical component is $16\times sin(63)=16\times 0.89=14.25ms^{-1}$

Question C:

Time required is given by $\frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds$

Question D:

Maximum height is given by $\frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m$

Question E:

Time of flight is twice the time required to reach maximum height= $2\times 1.45=2.9 seconds$ .

Question F:

The distance between the player and ball after landing is called range and is given by $\text{horizontal component of velocity}\times \text{time of flight}=$ $7.2\times 2.9=20.88m$

8 0

2 years ago

Which part of the EM spectrum is used in night vision goggles?

umka2103 [35]

Answer:

Night vision goggles capture infrared light given off by heat

6 0

2 years ago

When burl and paul stand on opposite ends of a sign-painting scaffold, the tensions in the supporting ropes combine to equal zer

Savatey [412]

The tensions in the supporting ropes combine to equal zero, if the tensions are equal.

<h3>Tension exerted by Burl and Paul </h3>

The sum of the tension exerted by Burl and Paul is determined as follows;

∑T = Ta + Tb

The net tension will be zero, when Ta is equal to Tb

0 = Ta - Tb

Ta = Tb

Thus, the tensions in the supporting ropes combine to equal zero, if the tensions are equal.

Learn more about tension here: brainly.com/question/918617

#SPJ1

6 0

1 year ago