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777dan777 [17]
3 years ago
15

Example 1:

Physics
1 answer:
strojnjashka [21]3 years ago
8 0

Answer:

50 Ohms

Explanation:

Given that

the original resistance of the wire is 2 ohms, then

it is known that the equivalent resistance in a parallel circuit is

1/R(eq) = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5.......+ 1/Rn

Giving that there are 5 pieces, then we stop at number 5.

Since R(eq) is 2, we substitute for that, and then we have

1/2 = 1/R + 1/R + 1/R + 1/R + 1/R

1/2 = 5/R

R = 5 * 2

R = 10 ohms

Since each R is 10 ohms, and the wire was cut into 5 pieces, we can then say that the original resistance of the wire is

R + R + R + R + R

10 + 10 + 10 + 10 + 10

total original resistance is 50 Ohms

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Melting, freezing, and boiling are _____________ changes. Question 3 options: A. compound
vfiekz [6]
Melting freezing and boiling are molecular changes


3 0
3 years ago
Read 2 more answers
If the velocity of an object is 9 m/s and its momentum is 72 kgm/s, what is its mass
MAVERICK [17]

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

<h3>What is momentum?</h3>

In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.

It is a vector quantity, possessing a magnitude and a direction.

The mathematical expression for momentum is:

p = m . v

where,

  • p is the linear momentum of the object.
  • m is the mass of the object.
  • v is the velocity of the object.

An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.

p = m . v

m = p / v

m = (72 kg.m/s) / (9 m/s) = 8 kg

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

5 0
2 years ago
Tarzan wants to swing on a vine across a river. He is standing on a ledge above the water's edge, and the river is 5.00 m wide.
Mandarinka [93]

Answer:

E D G E N U I T Y

Explanation:

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7 0
3 years ago
A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops
Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).  

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg andv_{2} =0.4v_{1} from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the T_{sat}=179.88^{0} C \leq T_{1} from the table 1 we get:v_{1} =0.30661(m^{3}/Kg). Now we have second conditions as: P_{2}=1(MPa), T_{2}=250^{0}C so from the same table we can see the state still superheated and we getv_{2}=0.23275(m^{3}/Kg), knowing that it's a isobaric process we can find the compression's work as:W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ) so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and:W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ) and the second process is isochoric:W_{2}=zero,nowW_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ). Finally to get the temperarure at the final state in part (b) we get:v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa) from table 2 (see attached) we comparev_{f} andv_{g} at the saturated water table and find the following:v_{f}=0.001093(m^{3}/Kg), so we know that the final state phase is a satured mixture and we get the temperature at the final state as:T_{2} =T_{sat} =151.83^{0}C.

3 0
4 years ago
If a CFOC was launched and travels 65 meters and is in the air for 3 seconds, what is the launch velocity and angle?
labwork [276]

Answer:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

Explanation:

From the question given above, the following data were obtained:

Range (R) = 65 m

Time of flight (T) = 3 s

Acceleration due to gravity (g) = 10 m/s²

Lauch velocity (u) =?

Lauch Angle (θ) =?

R = u²Sin2θ /g

65 = u² × Sin2θ /10

Recall:

Sin2θ = 2SinθCosθ

65 = u² × 2SinθCosθ / 10

65 = u² × SinθCosθ / 5

Cross multiply

65 × 5 = u² × SinθCosθ

325 = u² × SinθCosθ .....(1)

T = 2uSinθ / g

3 = 2uSinθ / 10

3 = uSinθ / 5

Cross multiply

3 × 5 = uSinθ

15 = u × Sinθ

Divide both side by Sinθ

u = 15 / Sinθ....... (2)

Substitute the value of u in equation (2) into equation (1)

325 = u² × SinθCosθ

u = 15 / Sinθ

325 = (15 / Sinθ)² × SinθCosθ

325 = 225 / Sin²θ × SinθCosθ

325 = 225 × SinθCosθ / Sin²θ

325 = 225 × Cosθ / Sinθ

Cross multiply

325 × Sineθ = 225 × Cosθ

Divide both side by Cosθ

325 × Sineθ / Cosθ = 225

Divide both side by 325

Sineθ / Cosθ = 225 / 325

Sineθ / Cosθ = 0.6923

Recall:

Sineθ / Cosθ = Tanθ

Tanθ = 0.6923

Take the inverse of Tan

θ = Tan¯¹ 0.6923

θ = 35°

Substitute the value of θ into equation (2) to obtain the value of u.

u = 15 / Sinθ

θ = 35°

u = 15 / Sin 35

u = 15 / 0.5736

u = 26.15 m/s

Summary:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

8 0
3 years ago
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