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Kaylis [27]
3 years ago
11

Lindsay and coworkers "slime in the first-grade teacher's desk" experiment showed that presenting

Physics
1 answer:
tia_tia [17]3 years ago
4 0
The answer that completes the statement is this: A PHOTOGRAPH OF THE PARTICIPANT'S FIRST-GRADE CLASS INCREASED THE LIKELIHOOD OF FALSE MEMORIES. This experiment is similar to Maryanne Gary's experiment which also states that viewing memory as a tape or video recording can be mistaken and may create false memories.
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A 20-kg barrel is rolled up a 20-m ramp to the back of a truck whose floor is 5.0 m above the ground. What work is done in loadi
oee [108]

The angle of inclination is calculated using sin function,

sin θ = 5 m / 20 m = 0.25

θ = 14.4775° 

 

<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>

F net = 49N 

<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>

<span>Work = 980 J </span>

4 0
3 years ago
In general, which statement does your data support ?
hjlf

Answer:

A

Explanation:

As distance increases, velocity increases.

5 0
3 years ago
A rock weighing 98 newtons is pushed off the edge of a bridge 50 meters above the ground. What was the potential energy of the r
lidiya [134]
Ep = 4900 because Ep = wh
4 0
3 years ago
Read 2 more answers
A bungee jumper jumps off a bridge and bounces up and down several times.
Stella [2.4K]

The energy that was lost due to air resistance while she was bouncing is determined as 3,360 J.

<h3>Conservation of energy</h3>

The amount of energy lost due to  air resistance while she was bouncing is determined from the principle of conservation of energy.

ΔE = P.E - Ux

ΔE = mgh - ¹/₂kx²

ΔE = (50)(9.8)(16) - ¹/₂(35)(16)²

ΔE = 3,360 J

Thus, the energy that was lost due to air resistance while she was bouncing is determined as 3,360 J.

Learn more about energy here: brainly.com/question/13881533

#SPJ1

3 0
2 years ago
wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area o
tresset_1 [31]

Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

 (b)  What is the total charge stored in the parallel combination?

Answer:

a

   i    Q_1 =  2.124 *10^{-11} \  C

   ii    Q_2 =  4.4604 *10^{-11} \ C

b

  Q_{eq} = 6.5844 *10^{-11} \ C

Explanation:

From the question we are told that

   The  voltage of the battery is  V  = 12.0  \ V

    The  plate area of each capacitor is  A  =  5.30 \ cm^2  =  5.30 *10^{-4} \ m^2

    The  separation between the plates is  d =  2.65 \ mm =  2.65 *10^{-3} \ m

     The permittivity of free space  has a value  \epsilon_o  =  8.85 *10^{-12} \  F/m

     The  dielectric constant of the other material is  z =  2.10

The  capacitance of the  first capacitor is mathematically represented as

       C_1  =  \frac{\epsilon  *  A }{d }

substituting values

        C_1  =  \frac{8.85 *10^{-12 } *   5.30 *10^{-4} }{2.65 *10^{-3} }

       C_1  =  1.77 *10^{-12} \  F

The  charge stored in the first capacitor is  

       Q_1 =  C_1 *  V

substituting values

        Q_1 =  1.77 *10^{-12} * 12

       Q_1 =  2.124 *10^{-11} \  C

The capacitance of the second  capacitor is mathematically represented as

       C_2  =  \frac{ z * \epsilon  *  A }{d }

substituting values

       C_1  =  \frac{  2.10 *8.85 *10^{-12 } *   5.30 *10^{-4} }{2.65 *10^{-3} }

       C_1  =  3.717 *10^{-12}  \ F

The  charge stored in the second capacitor is  

      Q_2 =  C_2 *  V

substituting values

     Q_2 = 3.717*10^{-12} *  12

     Q_2 =  4.4604 *10^{-11} \ C

Now  the total charge stored in the parallel combination is mathematically represented as

     Q_{eq} =  Q_1 + Q_2

substituting values

    Q_{eq} =  4.4604 *10^{-11} + 2.124*10^{-11}

     Q_{eq} = 6.5844 *10^{-11} \ C

7 0
3 years ago
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