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sveticcg [70]
3 years ago
13

Why can’t you use a model to learn everything about an object?

Physics
2 answers:
kogti [31]3 years ago
4 0

Answer:

Models have always been important in science and continue to be used to test hypotheses and predict information. Often they are not accurate because the scientists may not have all the data. It is important that scientists test their models and be willing to improve them as new data comes to light.

Explanation:

goldenfox [79]3 years ago
3 0

Answer:Models have always been important in science and continue to be used to test hypotheses and predict information. Often they are not accurate because the scientists may not have all the data. It is important that scientists test their models and be willing to improve them as new data comes to light.

Explanation:

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A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag
Readme [11.4K]

Answer:

(a)0.0675  J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675  J

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0
3 years ago
A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular
ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0

Where,

\theta = Angular Displacement

\alpha =Angular Acceleration

\omega_0 = Angular velocity

\theta_0 =Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

\theta = \frac{1}{2}\alpha t^2

Re-arrange for \alpha,

\alpha = \frac{2\theta}{t^2}

Replacing our values,

\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}

\alpha = 0.139rad/s

Therefore the ANgular acceleration of the mass is 0.139rad/s^2

4 0
2 years ago
Holes drilled several kilometers into Earth’s crust provide direct evidence about Earth’s interior in the form of
ivanzaharov [21]
C I hope this helps cccccccccccccccccccc
6 0
3 years ago
Read 2 more answers
Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that
lawyer [7]

As we know by the first law of thermodynamics

Q = \Delta U + W

here we know that

Q = heat given to the system

\Delta U = U_f - U_i

W = work done by the system

now here we can say

\Delta U = 180 - 78 = 102 J

W = 64 J

now we can say that heat will be given as

Q = 64 + 102 = 166 J

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0
3 years ago
Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places
Olenka [21]

Answer:

The magnitude of V_{2} is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage V_{out} = 17.33

Resistance R_{1} = 3 kΩ

Voltage gain A_{v} = 4.33

For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

     A_{v} = 1+\frac{R_{f} }{R_{1} }

   4.33 = 1+ \frac{R_{f} }{3 k }

     R_{f} ≅ 10 kΩ

For finding input voltage we use,

   A_{v} = \frac{V_{out} }{V_{2} }

    V_{2} = \frac{17.33}{4.33}

    V_{2} = 4 V

The Phase of V_{2} is zero because output voltage phase is 360°

Therefore, the magnitude of V_{2} is 4 V and phase of input voltage is zero

7 0
3 years ago
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