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egoroff_w [7]
2 years ago
13

A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a

later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Physics
1 answer:
olga nikolaevna [1]2 years ago
4 0

Answer:

A

Explanation:

Ke = 1/2 MV^2

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Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing of N 42º E from the western sta
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Answer:

Explanation:

AB = 110 miles

Let the distance of the western station from fire is d.

As according to the diagram, use Sine law

\frac{d}{Sin 15}=\frac{110}{Sin 133}

d = 110 x 0.2588 / 0.73

d = 39 miles

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In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi
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components of the speed of the coin is given as

v_x = v cos60

v_x = 6.4 cos60 = 3.2 m/s

v_y = vsin60

v_y = 6.4 sin60 = 5.54 m/s

now the time taken by the coin to reach the plate is given by

t = \frac{\delta x}{v_x}

t = \frac{2.1}{3.2}

t = 0.656 s

now in order to find the height

h = vy * t + \frac{1}{2} at^2

h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2

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3 0
3 years ago
A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

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