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Aleks04 [339]
3 years ago
10

Which of the following BEST describes a deer?

Physics
2 answers:
Verizon [17]3 years ago
7 0

Answer: herbivore, primary consumer

Explanation:

so the answer is D

ad-work [718]3 years ago
6 0
If we go down the list,

It wouldn’t be Primary Producer because Deer don’t really produce anything

Can’t be Carnivore because deer don’t eat meat

Same with tertiary

They do eat plants. So it would have to be the last one
You might be interested in
1 point<br> You throw a ball up in the air with a velocity of 30 m/s. How high does it<br> go?
choli [55]

Answer:

Explanation:

2as=vf^2-Vi^2

vf=30 m/s

vi= 0 m/s

a=g=9.8 m/s^2

s=vf^2-Vi^2/2a

s=(30)²-(0)²/2*9.8

s=900-0/19.6

s=45.9=46 m

4 0
3 years ago
The position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗ =[2.90m+(0.0900m/s2)t2]i^
Arlecino [84]

The solution for this problem is:

 
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s² 
this is for t in seconds and r in meters 

v = dr/dt = [0.180t i - 0.0450t² j] m/s² 

tan(-36.0º) = -0.0450t² / 0.180t 
0.7265 = 0.25t 
t = 2.91 s is the velocity vector of the insect

7 0
3 years ago
Read 2 more answers
Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1
miv72 [106K]

Answer:

w = 4,786 rad / s ,  f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

       L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

      L₀ = 0 + I₂ w₂

Final after coupling

      L_{f} = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

      I = ½ M R²

How the moment is preserved

      L₀ = L_{f}

      I₂ w₂ = (I₁ + I₂) w

      w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

      d₁ = 60 cm = 0.60 m

      d₂ = 40 cm = 0.40 m

      f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

      w₂ = 2 π f₂

      w₂ = 2π 3.33

      w₂ = 20.94 rad / s

Let's replace

       w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

       w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

      w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

      w = 20.94 1.28 / 5.6

      w = 4,786 rad / s

Angular velocity and frequency are related.

      w = 2π f

      f = w / 2π

      f = 4.786 / 2π

      f = 0.76176 Hz

7 0
3 years ago
A box is hanging from two strings. String #1 pulls up and left, making an angle of 50° with the horizontal on the left, and stri
creativ13 [48]

Answer:

mb = 3.75 kg

Explanation:

System of forces in balance

ΣFx =0  

ΣFy = 0

Forces acting on the box

T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left

T₂  = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.

Wb :Weightt of the box (vertical downward)

x-y T₁ and T₂ components

T₁x= T₁cos50°

T₁y= T₁sin50°

T₂x= 30*cos75° = 7.76 N

T₂y= 30*sin75° = 28.98 N

Calculation of the Wb

ΣFx = 0  

T₂x-T₁x = 0

T₂x=T₁x

7.76 = T₁cos50°

T₁ = 7.76 /cos50° = 12.07 N

ΣFy = 0  

T₂y+T₁y-Wb = 0

28.98 + 12.07(cos50°) = Wb

Wb = 36.74 N

Calculation of the mb ( mass of the box)

Wb = mb* g

g: acceleration due to gravity = 9.8 m/s²

mb = Wb/g

mb = 36.74 /9.8

mb = 3.75 kg

8 0
3 years ago
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

4 0
3 years ago
Read 2 more answers
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