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DIA [1.3K]
3 years ago
9

xperiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are gi

ven. No other forces are acting on the object. In which cases may the object possibly remain at rest?
Physics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201)\neq0

d) 2+2+(-4)=0

e) 2+2+(-2)\neq0

f) 2+2+(-3) \neq0; 2+(-2)+3\neq0

g) 2+2+(-5)\neq0; 2+(-2)+5\neq0

h)200 + 200 +(-5)\neq0; 200+(-200)+5\neq0

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In our problem, I=4.00 A and R=330 \omega, so the potential difference is
\Delta V = IR=(4.00 A)(330 \omega)=1320 V
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Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and
Liono4ka [1.6K]

a) 5 \Omega, 1.6 A

b) 6 \Omega, 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

R=R_1+R_2

where

R_1=4\Omega

R_2=1 \Omega

Therefore, the equivalent resistance is

R=4+1=5 \Omega

Now we can use Ohm's Law to find the current flowing through the circuit:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=5\Omega is the equivalent resistance of the circuit

Substituting,

I=\frac{8}{5}=1.6 A

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

R=R_1+R_2+R_3

where:

R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega

Substituting, we find the equivalent resistance:

R=4+1+1=6 \Omega

Now we can find the current through the circuit by using again Ohm's Law:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=6\Omega is the equivalent resistance

Substituting,

I=\frac{8}{6}=1.33 A

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

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3 years ago
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