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Alina [70]
3 years ago
13

Which has the same relationship as circle:cylinder? Triangle:sphere, square:window, rectangle:box, pentagon:counter

Physics
1 answer:
ddd [48]3 years ago
5 0
I think it is square:window
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Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
4 years ago
Why must mine tailings be stored and disposed of carefully?
gavmur [86]
After thorough researching, the mine tailings must be stored and disposed of carefully because they have lots of chemical and various toxic materials. They can also leach to the aquifers. The correct answer to the following given statement above is they have chemicals which are dangerous.
7 0
3 years ago
The work done to compress a spring with a force constant of 290.0 N/m a total of 12.3 mm is: a) 3.57 J b) 1.78 J c) 0.0219 J d)
iren2701 [21]

Answer:

Work done, W = 0.0219 J

Explanation:

Given that,

Force constant of the spring, k = 290 N/m

Compression in the spring, x = 12.3 mm = 0.0123 m

We need to find the work done to compress a spring. The work done in this way is given by :

W=\dfrac{1}{2}kx^2

W=\dfrac{1}{2}\times 290\times (0.0123)^2

W = 0.0219 J

So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.

7 0
3 years ago
you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?
laiz [17]
At the same speed because it will slow down as it approaches the peak then speed up as it goes down again


it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something
8 0
3 years ago
An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
mash [69]

Answer:

1.73 m/s²

Explanation:

Given:

Δx = 250 m

v₀ = 0 m/s

t = 17 s

Find: a

Δx = v₀ t + ½ at²

250 m = (0 m/s) (17 s) + ½ a (17 s)²

a = 1.73 m/s²

5 0
3 years ago
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