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eduard
3 years ago
9

A coyote is looking for some food for its young. It spots a pocket mouse near a cactus and starts to track and hunt the mouse fo

r food.
In this scenario, the coyote is considered:
a. predator
b. prey
c. neither
Physics
2 answers:
MariettaO [177]3 years ago
5 0

Answer:

A. Predator

Explanation:

The coyote's prey is the mouse in this situation.

elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

b.

Explanation:

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SSSSS [86.1K]
Number 20 is B and 21 is C
5 0
3 years ago
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The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods
max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a                                  ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B                               ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

 

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

 

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0
3 years ago
Read 2 more answers
crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied​
Klio2033 [76]

Answer:

 F = 5226.6 N

Explanation:

To solve a lever, the rotational equilibrium relation must be used.

We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise

         F d₁ = W d₂

where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.

In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is

d₂ = 200 cm = 2 m

therefore the distance to the applied force is

          d₁ = L -d₂

         d₁ = 5 -2

         d₁= 3m

we clear from the equation

          F = W d₂ / d₁

          W = m g

          F = m g d₂ / d₁

we calculate

      F = 800 9.8 2/3

      F = 5226.6 N

4 0
3 years ago
Can someone help? I’ll give brainlest .
MrRissso [65]

Answer:

I think it’s 6.8 m/s2

Explanation:

Please give me brainlist if the answer is right.

5 0
3 years ago
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You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
Anarel [89]

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

R = 4.48 \times 10^6 m

h = 630 km

now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

M = 2.34 \times 10^24 kg

now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

a = 7.7 m/s^2

now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

R = 24.3 m

3 0
4 years ago
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