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svp [43]
2 years ago
13

A supplier wants to make a profit by buying metal by weight at one altitude and selling it at the same price per pound weight at

another altitude. The supplier should
a. buy at high altitude and sell at low altitude
b. buy at low altitude and sell at high altitude
c. disregard altitude because it makes no difference
Physics
1 answer:
Veronika [31]2 years ago
3 0

The correct option is C.

In physics, the gravitational law makes us to understand that gravity change with altitude. The higher the altitude, the more the gravity one will experiences. Change in altitude also produce change in weight but all these changes are not really significant. Therefore, there will be no significant difference between weighing metals at different altitude, this is because the effect of gravity on them will be the same.

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Using the superposition method, calculate the current through R5 in Figure 8-71
Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

r_1 = \frac{2.2 * 1}{2.2 + 1}

r_1 = 0.6875 k ohm

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}

r_2 = 0.95 k ohm

now the current flowing through the battery is

i = \frac{2}{1 + 0.95} = 1.02 mA

now this will divide into R3 and R2 so current flowing in R3 will be

i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA

now this will again divide in R4 and R5

so current in R5 will be

i_5 = \frac{R_4}{R_4 + R_5}* i_1

i_5 = 0.18 mA

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm

also after its series with R3

r_2 = 1 + 0.6875 = 1.6875 k ohm

now it is in parallel with R5 on other side

r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm

now current through the battery will be given as

i = \frac{3}{1 + 0.95} = 1.53 mA

now it is divide in r2 and R5

so current in R5 is given as

i_5 = \frac{r_2}{r_2 + R_5}*i

i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53

i_5 = 0.67 mA

now the total current in R5 will be given by super position which is

i = 0.67 + 0.18 = 0.85 mA

so there is 0.85 mA current through R5 resistance

5 0
2 years ago
The acceleration of an object as a function of time is given by a(t) = (1.00 m/s2)t2. If displacement of the object between time
jolli1 [7]

not enough information is given to determine the velocity of the object at time to=0.00s

3 0
2 years ago
A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle
denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/s^{2} = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = \frac{8.04N}{70.2Kg} = 0.1130801680m/s^2.

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0
2 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
QveST [7]

Answer:

Please see attachment

Explanation:

Please see attachment

8 0
2 years ago
A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington
Roman55 [17]

Let \theta be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

The current has velocity vector (relative to the Earth)

\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath

The swimmer's resultant velocity (her velocity relative to the Earth) is then

\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}

\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ

which is approximately 41º west of north.

6 0
2 years ago
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