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2 years ago

A fish cannot be killed by throwing an arrow into the water why?

Physics

2 answers:

juin [17]2 years ago

8 0

Answer:

A big-sized fish wouldn't be killed even if one arrow is shot at it. it takes 5-10 arrows to kill a big-sized fish. A fish can swim faster in water than an arrow when it is shot at water.

Explanation:

olga55 [171]2 years ago

3 0

Answer:

a fish cannot be killed by throwing an arrow into the water because you have to throw the arrow at the fish not just in the water….

Explanation:

common sense TwT

duhhh

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Why does the sun appear to rise in the east and set in the west every day?

anygoal [31]

Answer: even heating

Explanation:

the earth rotates

4 0

3 years ago

Read 2 more answers

At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel

sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

7 0

3 years ago

A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

$l_o =52 \ m$

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

3 years ago

Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

The range is 35.35 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is: