All categories

3 years ago

Which statement describes the effects of forces on an object

Physics

1 answer:

lukranit [14]3 years ago

8 0

Answer:

Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.

Explanation:

You might be interested in

Elaborate on the suitability of "cola" type drinks to polish chrome surfaces.

nexus9112 [7]

The answer is t<span>he phosphoric acid in cola easily removes dirt and grime. The </span>"cola" type drinks is suitable to polish chrome surfaces. Cola contains a very diluted solution of phosphoric acid, which can remove rust, dirt, and grime from chrome surface.

3 0

4 years ago

Please help with Physics Circuits!

Zigmanuir [339]

1) Let's start by calculating the equivalent resistance of the three resistors in parallel, $R_2, R_3, R_4$ :
$\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}$
From which we find
$R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
$R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega$
So, the correct answer is D)

2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$

And these are connected in series with a resistor of $10 \Omega$ , so the equivalent resistance of the circuit is
$R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega$

And by using Ohm's law we find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A$
So, the correct answer is C).

3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
$R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega$

And by using Ohm's law we find the current flowing in the circuit:
$I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A$

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
$V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V$
So, the correct answer is D).

4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}$
From which we find
$R_{23} = 7.55 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is:
$R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega$

The current in the circuit is given by Ohm's law
$I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A$

Now we can compare the voltage drops across the resistors. Resistor 1:
$V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V$
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
$V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V$
Resistor 4:
$V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V$

So, the greatest voltage drop is on resistor 1, so the correct answer is D).

5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
$P_3 = I_3^2 R_3 = \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W$
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.

5 0

3 years ago

Read 2 more answers

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu

Rus_ich [418]

Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is $I = 3.535 *10^{-6} \ W/m^2$

Explanation:

From the question we are told that

The temperature is $T = 2.81 \ K$

Now According to Stefan's law

$Power(P) = \sigma * A * T^4$

Where $\sigma$ is the Stefan Boltzmann constant with value $\sigma = 5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}$

Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

$I = \frac{P}{A}$

=> $I = \frac{\sigma * A * T^4}{A}$

=> $I = \sigma * T^4$

substituting values

$I = 5.67 *10^{-8} * (2.81)^4$

$I = 3.535 *10^{-6} \ W/m^2$

4 0

3 years ago

A 4.2 kg sled is being pulled along a snow-covered road with a rope that exerts a horizontal force of 6.0 n and, at that moment,

Stella [2.4K]

<span>the body is moving horizontally, it doesnt matter watever kind of horizontal forces are acting.
Therefore the normal force is equal to the weight
N=mg=4.2*9.8=41N
Note: the other data in the problem have no relevance
answer
</span> the normal force on the sled is 41N

6 0

4 years ago

Read 2 more answers

A transformer has 100 turns in its primary coil and 75 turns in its secondary coil. If the input voltage is 12.0 V, what is the

Sindrei [870]

Is 54 por que es igual

7 0

3 years ago