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Kamila [148]
3 years ago
8

A bird carrying a fish (5kg) drops it from 107 meters in the air how fast does the fish hit the ground

Physics
1 answer:
notka56 [123]3 years ago
4 0

Answer:

The velocity of the fish hitting the ground is , v = 45.795 m/s        

Explanation:

Given data,

The mass of the fish, m = 5 kg

The height of the bird from the surface, h = 107 m

Using the III equation of motion,

                          v² = u² + 2gs

                          <em> v = √(u² + 2gs)</em>

Substituting the values,

                           v = √(0² + 2 x 9.8 x 107)  

                              = 45.795 m/s

Hence, the velocity of the fish hitting the ground is, v = 45.795 m/s        

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an object moving at 10. km/hr has a kinetic energy of 10. J. what is the kinetic energy of the same object if it is moving at 20
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Kinetic energy is related to velocity by:
KE = (1/2)mv^2

solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
10= 50m
10/50 = m
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at 20 km/hr

KE = (1/2)(1/5)(20)^2
KE = (1/10)(400)
KE = 40 J
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3 years ago
A block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block
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No friction present means: Ek = Ep

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

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