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2 years ago

HELP PLS HELP. WILL GIVE BRAINLIEST.

Physics

2 answers:

statuscvo [17]2 years ago

6 0

Let's see

According to law of conservation of linear momentum

PE=KE
KE=mgh
KE=0.5(9.8)(0.07)
KE=4.9(0.07)
KE=0.343J
KE=343mJ

Licemer1 [7]2 years ago

6 0

Answer:

Explanation:

This problem requires the Law of Conservation of Energy.

⇒ Potential energy = Kinetic Energy

⇒ mgh = 1/2mv²

⇒ 0.5 x 10 x 0.07 = KE

⇒ KE = 5 x 0.07

⇒ KE = <u>0.35 J</u>

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The circuit you should use to find the open-circuit voltage is

fiasKO [112]

Answer:

Incomplete questions check attachment for circuit diagram.

Explanation:

We are going to use superposition

So, we will first open circuit the current source and find the voltage Voc.

So, check attachment for open circuit diagram.

From the diagram

We notice that R3 is in series with R4, so its equivalent is given below

Req(3-4) = R3 + R4

R(34) = 20+40 = 60 kΩ

Notice that R2 is parallel to the equivalent of R3 and R4, then, the equivalent of all this three resistor is

Req(2-3-4) = R2•R(34)/(R2+R(34))

R(234) = (100×60)/(100+60)

R(234) = 37.5 kΩ

We notice that R1 and R(234) are in series, then, we can apply voltage divider rule to find voltage in R(234)

Therefore

V(234) = R(234) / [R1 + R(234)] × V

V(234) = 37.5/(25+37.5) × 100

V(234) = 37.5/62.5 × 100

V(234) = 60V.

Note, this is the voltage in resistor R2, R3 and R4.

Note that, R2 is parallel to R3 and R4. Parallel resistor have the same voltage, then voltage across R2 equals voltage across R34

V(34) = 60V.

Now, we also know that R3 and R4 are in series,

So we can know the voltage across R4 which is the Voc we are looking for.

Using voltage divider

V4 = Voc = R4/(R4 + R(34)) × V(34)

Voc = 40/(40+60) × 60

Voc = 24V

This is the open circuit Voltage

Now, finding the short circuit voltage when we short circuit the voltage source

Check attachment for circuit diagram.

From the circuit we notice that R1 and R2 are in parallel, so it's equivalent becomes

Req(1-2) = R1•R2/(R1+R2)

R(12) = 25×100/(25+100)

R(12) = 20 kΩ

We also notice that the equivalent of Resistor R1 and R2 is in series to R3. Then, the equivalent resistance of the three resistor is

Req(1-2-3) = R(12) + R(3)

R(123) = 20 + 20

R(123) = 40 kΩ

We notice that, the equivalent resistance of the resistor R1, R2, and R3 is in series to resistor R4.

So using current divider rule to find the current in resistor R4.

I(4) = R(123) / [R4+R(123)] × I

I(4) = 40/(40+40) × 8

I(4) = 4mA

Then, using ohms law, we can find the voltage across the resistor 4 and the voltage is the required Voc

V = IR

V4 = Voc = I4 × R4

Voc = 4×10^-3 × 40×10^3

Voc = 160V

Then, the sum of the short circuit voltage and the open circuit voltage will give the required Voc

Voc = Voc(open circuit) + Voc(short circuit)

Voc = 24 + 160

Voc = 184V.

3 0

3 years ago

An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum

Tju [1.3M]

Answer:

The pressure is $P_2 = 4.25 \ a.t.m$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 11.2\ a.t.m$

The temperature is $T_1 = 299 \ K$

Let the first volume be $V_1$ Then the final volume will be $2 V_1$

Generally for a diatomic gas

$P_1 V_1 ^r = P_2 V_2 ^r$

Here r is the radius of the molecules which is mathematically represented as

$r = \frac{C_p}{C_v}$

Where $C_p \ and\ C_v$ are the molar specific heat of a gas at constant pressure and the molar specific heat of a gas at constant volume with values

$C_p=7 \ and\ C_v=5$

=> $r = \frac{7}{5}$

=> $11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )$

=> $P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2$

=> $P_2 = 4.25 \ a.t.m$

8 0

3 years ago

An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi

sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

now we know that

object distance from mirror is

$d_o = -50 cm$

Focal length of the mirror is given as

$f = 40 cm$

now we have

$\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}$

$\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}$

$d_i = 22.22 cm$

6 0

3 years ago

You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle

nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A