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KatRina [158]
2 years ago
15

0.6% of 36 = ? Please answer today This is due tomorrow!,​

Physics
1 answer:
ludmilkaskok [199]2 years ago
8 0

Answer:

21.6

Explanation:

0.6% * 36 = 21.6

are you looking for percentage or just answer??

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Are passing through optical Centre deviate why ​
nikklg [1K]

Answer:

Actually the same happens when the ray passes through optical centre. This can be observed in a thick lens. In thin lenses the perpendicular distance between extended incident ray and extended emergent ray is negligible. So we can say that light ray passes through optical centre without deviation.

Explanation:

7 0
2 years ago
A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav
oksano4ka [1.4K]

Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

t = 0.99 s

Now, we can calculate the initial speed:

8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

<u>v0 = 10 m/s</u>

The takeoff speed is 10 m/s

b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

<u>y = 1.2 m</u>

The maximum height above the ground is 1.2 m.

4 0
3 years ago
what is the mechanical advantage of a crowbar when a worker uses 10N of force to pry open a window that has a resistance of 500N
Oksana_A [137]

Answer:

50

Explanation:

The mechanical advantage of a machine is given by

MA=\frac{F_{out}}{F_{in}}

where

F_{out} is the output force

F_{in} is the input force

For the crowbar in this problem,

F_{in}=10 N is the force in input applied by the worker

F_{out}=500 N is the force that the machine must apply in output to overcome the resistance of the window and to open it

Substituting into the equation, we find

MA=\frac{500}{10}=50

3 0
3 years ago
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If the magnitude of F1 is greater than the magnitude of F2, then the box is
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Assuming that the vectors are acting along the same axis, we could just simply add or subtract the vectors. Since the F1 is greater than F2, there would be motion, there would be acceleration, and that the direction of motion is along the F1.

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2 years ago
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The relationship between distance from the sun and orbital period is that as the distance from the sun increases, the orbital pe
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Second law is it:))))))))
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