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Nata [24]
2 years ago
11

Students had two batteries and two different resistors. During four trials, they build four different circuits and measure the c

ircuit’s current in Amps according to the following table.
Trial Number

Voltage (V)

Resistance (Ω)

Current (A)

1

1.5

200


2

1.5

100


3

3.0

200


4

3.0

100




For which trial would the students measure the smallest current in the circuit? (AKS 10a)



A.
Trial 1

B.
Trial 2

C.
Trial 3

D.
Trial 4
Physics
1 answer:
Darina [25.2K]2 years ago
5 0

Answer:

bhi jo bhi of gp oh oh gi IG 7u to uff do if goo td to yd do FP ae rt 7g hi pic vo icon

Explanation:

bh hi h bhi vc di oh x At jb jo iv hp of di of dr hi o hc x gh ki vc hi jo

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Define the term “force”.
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Energy that is applied to an object.

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Which is true regarding the penetrating power of radiation? (2 points) Select one:
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Answer:d

Explanation:

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Gamma Particles are lightest among three so they can Penetrate most .

The order of Penetration is given by

Alpha< Beta < Gamma              

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3 years ago
The coal with the highest energy available per unit burned is:________
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A Ignite. Because. When you put coal in a gas pit, it ignites and burns. However, there is a chance that all of us fed
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1 year ago
A 4113 N piano is to be pushed up a(n) 3.99 mfrictionless plank that makes an angle of 25.5â—¦with the horizontal.Calculate the
Serhud [2]

Answer:

W = 14.8 kJ

Explanation:

W = F S cos ∅

W = 4113 x 3.99 x cos 25.5

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3 0
3 years ago
A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
3 years ago
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