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3 years ago

If the initial velocity of an object was -2 meters per second

1 answer:

3 0

A :-) for this question , we should apply
a = v - u by t
Given - u = -2 m/s
v = -10 m/s
t = 16 sec
Solution -
a = v - u by t
a = -10 - -2 by 16
a = -12 by 16
( cut 12 and 16 because 2 x 6 = 12 and
2 x 8 = 16 )
( cut 6 and 8 because 2 x 3 = 6 and
2 x 4 = 8 )
a = 3 by 4
a = 0.75 m/s^2

.:. The acceleration is 0.75 m/s^2.

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A thin electrical heating element provides a uniform heat flux qo" to the outer surface of a duct through which air flows. The d

Ivanshal [37]

Answer:

a)q= 2800 W/m²

b)To=59.4°C

Explanation:

Given that

L = 10 mm

K= 20 W/m·K

T=30°C

h= 100 W/m²K

Ti=58°C

Heat flux q

q= h ΔT

q= 100 x (58 - 30 )

q= 2800 W/m²

As we know that heat transfer by Fourier law given as

Q= K A ΔT/L

Lets take outer temperature is To

So by Fourier law

To= Ti + qL/K

Now by putting the values

To= Ti + qL/K

$To= 58 + 2800 \times \dfrac{ 0.01}{20}$

To=59.4°C

5 0

3 years ago

D. 1<br> 1.212.52<br> -2<br> 12<br> 12

harkovskaia [24]

Answer:

what?

Explanation:

7 0

3 years ago

a flashlight bulb with a potential difference of 6.0 v across it has a resistance of 8.0 how much current is the bulb filament

Andreas93 [3]

Answer:

Explanation:

Ohm's law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points. The constant of proportionality is called the Resistance (R) of the conductor.

The formula can be expressed as:

V = I.R

The intensity of the current through a resistor can be calculated by solving the equation for I:

$\displaystyle I=\frac{V}{R}$

A flashlight bulb is an example of a resistor with the specific task to emit light when it's hot enough.

The potential difference (or voltage) is V=6 V and the resistance is R=8Ω, thus the current is:

$\displaystyle I=\frac{6~V}{8~\Omega}$

I = 0.75 Amp

3 0

3 years ago

Identify the solute and solvent in a solution made of 15g of oxygen and 5g<br> of helium

Marrrta [24]

Taking into account the definition of solvent and solute, helium is the solute and oxygen is the solvent.

<h3>Definition of solute and solvent</h3>

The solution is a homogeneous mixture of two or more components that have the same chemical and physical properties in a single phase. A solution is made up of two main components: the solute and the solvent.

That is, the solute and the solvent are the components of a chemical solution, that is, of a homogeneous mixture that occurs when one or more substances are dissolved in another substance.

The solute is the substance that dissolves in a solution. In solution, the solute is usually found in a lower proportion than the solvent.

Solvent is the substance in which a solute dissolves, resulting in a chemical solution. Generally, the solvent is the component that is found in the highest proportion in the solution. It will always be the solvent who will determine the state of the matter that will have the solution.

<h3>Solute ans solvent in this case</h3>

In this case, you consider that the solute is the component in less quantity and the solvent is the major component.

You know that a solution made of 15 g of oxygen and 5 g of helium. Then, helium is the solute and oxygen is the solvent.

Learn more about solvent and solute:

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8 0

2 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

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