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joja [24]
3 years ago
13

Name at least two limitations of using models in science.

Physics
2 answers:
saw5 [17]3 years ago
7 0

Explanation:

Missing Details. Most models can't incorporate all the details of complex natural phenomena. 

Most Are Approximations. Most models include some approximations as a convenient way to describe something

olganol [36]3 years ago
3 0

Answer:

Models may be oversimplified, leading to misunderstanding about the subject. Sometimes models cannot represent all aspects of an object or system. Finally, models are only as accurate as current scientific knowledge.

Explanation:

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Ozone, O3 methane, CH4 oxygen gas, O2 Are classified as
erma4kov [3.2K]
I would think compounds.
5 0
3 years ago
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A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
Crazy boy [7]
A) Work energy relation;
 Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
 F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
 F = 900/0.04
    = 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
 F = Ma
Thus; a = F/m
             = 22500/(5×10^-3)
             = 4,500,000 m/s²
But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
6 0
3 years ago
A man ties one end of a strong rope 8.17 m long to the bumper of his truck, 0.524 m from the ground, and the other end to a vert
Kamila [148]

Answer:

2442.5 Nm

Explanation:

Tension, T = 8.57 x 10^2 N

length of rope, l = 8.17 m

y = 0.524 m

h = 2.99 m

According to diagram

Sin θ = (2.99 - 0.524) / 8.17

Sin θ = 0.3018

θ = 17.6°

So, torque about the base of the tree is

Torque = T x Cos θ x 2.99

Torque = 8.57 x 100 x Cos 17.6° x 2.99

Torque = 2442.5 Nm

thus, the torque is 2442.5 Nm.

8 0
3 years ago
4. According to Newton third law what is the action force? The swimmer pushes against the water the water pushes back on the swi
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A.The swimmer pushes the water
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3 0
3 years ago
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A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s,    a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

           v₀ₓ =  10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

           v_y = v_{oy} - g t

           v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)

at the highest point of the trajectory the vertical speed must be zero

           v_y = 0

therefore the velocity of the body is

          v = (10 i ^ + 0j ^) m / s

the acceleration is

          a = (0 i ^ - g j⁾

          a = (0i ^ - 9.8 j ^) m / s²

5 0
2 years ago
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