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3 years ago

Which of the following is NOT a characteristic of scientific knowledge? please help me

2 answers:

8 0

Answer: "Scientific knowledge can answer all of society's questions" this is not a characteristic of scientific knowledge.

Explanation: Scientific knowledge is durable in most cases, this is true because the things demonstrated with science usually can be repeated by another scientist, so this knowledge is actually durable. But there are cases where the knowledge has a little mistake, ar a new discovery refutes one part of an older theory, so the knowledge can be changed or discarded in some cases.

For the other one, scientific knowledge can not answer all of the society's questions, where some of those, like "the meaning of life" or things like that need a more philosophical approach, so the second sentence is false.

kakasveta [241]3 years ago

4 0

I think the correct answer from the choices listed above is the second option. It would be that scientific knowledge can answer all of society's questions that is not a <span>characteristic of scientific knowledge. Hope this answers the question. Have a nice day.</span>

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DochEvi [55]

Answer:

Explanation:

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3 years ago

A football player runs down a field with a speed of 8 m/s. How long will it take him to run 20m?

Shtirlitz [24]

To find the answer for this question, you simply need to divide 20 by 8, which is the speed he is traveling.

20 / 8 = 2.5

The football player will run 20 yards in 2 1/2 seconds.
Hope that helped! =)

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3 years ago

Read 2 more answers

What is a an reason why a object gets extinct

adoni [48]

Answer:

An object may become extinct due to the lack of quantity of the object and the object no longer exists so it is now extinct.

7 0

3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

$\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

7 0

3 years ago

A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.

IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

$(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}$

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

$(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]$

7 0

3 years ago