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jok3333 [9.3K]
3 years ago
12

If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f

ast must the fish spit the water to hit its target? The insect is at the highest point of the trajectory of the spit water. Use g = 10 m/s2.
Physics
1 answer:
Burka [1]3 years ago
3 0

Answer:

The speed is  v =  9.8 \ m/s

Explanation:

From the question w are told that

    The angle  made is \theta  =  30^o

     The distance  above the surface of the water is  h_{max} = 1.2 \ m

     The  value of  g = 10 \  m/s^2

   

The maximum height attained by the fish is mathematically evaluate as

       h_{max} =  \frac{v^2 sin ^2 \theta }{2g }

Making v which is the speed of the fish the subject of the formula

      v =  \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }

  substituting values

     v =  \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2  } }

     v =  9.8 \ m/s

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How high can a body vertically thrown with a speed of 40m/s raise after 3 sec (neglecting air
Tcecarenko [31]

y = 75.9 m

Explanation:

y = -(1/2)gt^2 + v0yt + y0

If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.

y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)

= -44.1 m + 120 m

= 75.9

5 0
2 years ago
Which is the correct answer?
Nezavi [6.7K]

Answer:

Point A

Explanation:

The work done by stretching or compressing a spring is given by E=1/2kx²

The potential energy is numerically equal to the work done.

This means that the higher the bigger the value of the extension, x, the higher the energy contained.

In this scenario the modulus of x is considered.

Among the given values of x the modulus of -5 is the largest.

thus it gives the highest value of energy.

7 0
3 years ago
A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect
AleksAgata [21]

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

8 0
2 years ago
In a game of egg-toss, you and a partner are throwing an egg back and forth trying not to break it. Given your knowledge of mome
lutik1710 [3]
F=dP/dt.  So you want the momentum to change as slowly as possible in time to minimize the force.  So as you catch the egg, let your hand move backward with it for awhile, slowly bringing it to a stop.  If you hold your hand steady when you catch it the force due to the impact could break it.
3 0
2 years ago
A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0
2 years ago
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