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zavuch27 [327]
3 years ago
11

If a kestrel eats a mouse that eats grass the kestrel is a

Physics
2 answers:
vovangra [49]3 years ago
5 0
<span> Second-level consumer </span>
Vinil7 [7]3 years ago
4 0

Second Level Consumer

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The stars in the sky are organized into groups of stars called constellations which appear near each other in the sky but are no
bezimeni [28]

Answer:

The International Astronomical Union (IAU)  has accepted 88 constellations in the sky.

Explanation:

Constellations has been used since the beginnings of civilizations and each one of them named them as they considered appropiate. It means Greeks' constellations were different than the ones described by Chinese, so it was necessary to gather all these constellations and make a great record with all of them, but there was a problem: Some constellations from different civilizations overlaped because they shared the same stars. There was necessary to put some order on this and that is when in 1922 the International Astronomical Union (IAU) defned a set of 88 moderm constellations  that would become the international standard to look at the night sky. Each one of them is unique and does not share stars with the other constellations.  

6 0
2 years ago
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0
3 years ago
Read 2 more answers
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
2 years ago
Convert 12cm to picometers​
Salsk061 [2.6K]

Answer:

1.2e+11

Explanation:

8 0
3 years ago
Read 2 more answers
What term refers to the part of a spacecraft that is occupied by the crew for takeoff and landing?
Semenov [28]
Command module ✅

service module

lunar module

annum module
7 0
2 years ago
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