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GenaCL600 [577]
3 years ago
13

Which of the following is NOT a skill scientists use to learn about the world?

Physics
1 answer:
sdas [7]3 years ago
8 0
There are many skills that scientists don't use to learn about the world. 
These skills include playing the harmonica, riding a unicycle, and asking
others to choose from a list and then not telling them what's on the list.
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Select the statement that correctly describes how light travels? light can travel in a vacuum, and its speed is constant even if
Irina18 [472]
<span>Light can travel in a vacuum, and ... strange as it may seem ...
its speed is always the same, even if the light source is moving. </span>
3 0
3 years ago
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You are helping two friends from our class with a physics problem where a cart is pushed up a ramp. In examining the motion of t
stiks02 [169]

Answer: Acceleration will have 2 components, vertical and horizontal.

Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.

Net-horizontal acceleration will  be equal to the horizontal component of the applied acceleration.

Explanation:

Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.

<u>For vertical acceleration:</u>

If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.

In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.

<u>For horizontal acceleration:</u>

This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.

But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.

8 0
3 years ago
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Normal body temperature. the average normal body temperature measured in the mouth is 310 k . what would celsius and fahrenheit
faltersainse [42]
The question is simply asking us to convert one unit, in this case temperature, to another unit. To do this, we need conversion factors to multiply, divide or relate to the original measurement. We do as follows:

Celsius = Kelvin - 273.15
310 - 273.15 = 36.85 degrees celsius

Fahrenheit = <span> (°</span>C<span> × </span>9<span>/5) + 32
</span> (36.85<span> × </span>9<span>/5) + 32 = 98.33 degrees fahrenheit
</span>
Hope this helps.

7 0
3 years ago
Humans cannot convert the Sun's energy into glucose for themselves because humans lack _____.
rosijanka [135]
Human lack chlorophyll, which is the function that collects energy from the Sun to conduct photosynthesis.
3 0
3 years ago
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You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown
Kipish [7]

Answer:

a

The mass is  m_2 =21.75*10^{-27} \ kg

b

The velocity is  v_2 = 3.0*10^{6} m/s

Explanation:

From the question we are told that

     The speed of the protons is  u_1 =  2.10*10^{7} m/s

     The mass of the protons is  m

     The speed of the rebounding protons are v_1 =  -1.80 * 10^{7} \ m/s

The negative sign shows that it is moving in the opposite direction

     

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

        m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1

Where m_1 is the mass of a single proton

          So substituting values

       m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1

        m_2 =13 m_1

The mass of on proton is  m_1 = 1.673 * 10^{-27} \ kg

So     m_2 =13 ( 1.673 * 10^{-27} )

        m_2 =21.75*10^{-27} \ kg

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

      m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2

Now  u_2 because before collision the the nucleus was at rest

So

        m_1 u_1 =  m_1 v_1 + m_2v_2

=>    v_2 =  \frac{m_1(u_1 -v_1)}{m_2}

Recall that m_2 =13 m_1

So

       v_2 =  \frac{m_1(u_1 -v_1)}{13m_1}

=>         v_2 =  \frac{(u_1 -v_1)}{13}

substituting values

              v_2 =  \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}

              v_2 = 3.0*10^{6} m/s

   

7 0
3 years ago
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