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3 years ago

It would be really helpful if u help me solving this question. PLEASE!!!

Physics

1 answer:

sweet [91]3 years ago

7 0

Answer: The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

The room temperature will be the point at which the themometric substance remains constant when rising from ice point.

Explanation:

Apparatus available:

Unmarked thermometer

250 cm3 glass beaker

crushed ice

water

heatproof mat

clamp, boss and stand

meter rule

Added apparatus

Bunsen burner

Stirrer

Method

The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

Procedures

Set up the apparatus of illustrated in the attached figure.

Immerse the unmarked thermometer into the ice in the beaker.

When the level indicated by the thermometric substance remains steady after some time, a mark will be made at that point. This mark will corresponds to the ice point (lower fixed point) and is assigned the value of 0 °C.

You may add little water and continue to stir gently.

The themometric substance will start to rise and stop when it reaches room temperature. Mark the point but do not assign any value

Place the beaker on bunsen burner and boil the water. The themometric substance will continue to rise and remain constant at upper fixed point

This mark will corresponds to the steam point (upper fixed point) and is assigned the value of 100 °C.

Divide between the lower fixed point and upper fixed point into equal intervals. Then you can see the value of room temperature.

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A -3.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

Korvikt [17]

Answer:

a) -537 N/C , b) - 327.8 N/C and c) 723.7 N/C

Explanation:

The electric field is a vector magnitude, so we must add them as vectors. The electric field equation is

E = k q / r²

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m²/C², that the magnitude of the load and r the distance between the load and the test load

Let's find the field created by each charge at the point x = 0.200 m

charge 1.

This charge is at the origin, the distance is

x₁ = 0.200 m

q₁ = 3 nC = 3 10-9 C

E1 = k q₁ / x₁²

E1 = 8.99 10⁹ 3 10⁻⁹ / 0.2²

E1 = 674.25 N / C

Charge 2

This load is the point a = 0.800 m, so the distance to the test charge at 0.200 m

x₂ = 0.800 - 0.200

x₂ = 0.600 m

q₂ = -5.5 nC = -5.5 10-9 C

E2 = 8.99 10⁹ 5.5 10⁻⁹ / 0.600²

E2 = 137.35 N / C

We already have the value of each field, add them vectorially, remember that if the charges are of the same sign they repel and if they are of the opposite sign they attract, the field E1 is directed to the left and the field E2 is directed to the right

Et = -E1 + E2

Et = -674.25 + 137.25

Et = -537 N / C

The field is headed to the left

b) we perform the same procedure for another distance value

Charge 1

x = 1.20

E1 = 8.99 10⁹ 3 10⁻⁹ / 1.2²

E1 = 18.73 N / C

Chage 2

x = 0.8 - 1.2 = -0.4 m

E2 = 8.99 10⁹ 5.5 10⁻⁹ / 0.4²

E2 = 309.03 N / C

Total field, E1 is on the left and E2 goes on the left

Et = -E1 -E2

Et = -18.73 - 309.03

Et = - 327.8 N / C

The field is headed to the left

c) point at x = -0.200 m

charge 1

x = -0.200 m

E1 = 8.99 10⁹ 3 -10⁻⁹ / (-0.2)²

E1 = 674.25 N / C

charge 2

x = 0.800 - (-0.2) = 1,000 m

E2 = 8.99 10⁹ 5.5 10⁻⁹ / 1²

E2 = 49.45 N / C

Ei goes to the right and E2 goes to the right

Et = E1 + E2

Et = 674.25 +49.45

Et = 723.7 N / C

The field is headed to the right

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3 years ago

A 45.0-kg woman stands up in a 69.0-kg canoe that is 5.00 m long. She walks from a point 1.00m from one end to a point 1.00 m fr

Sonja [21]

Answer:

The center of the canoe is displaced 1.18m in the opposite direction to the movement of the woman.

Explanation:

Let's assume the woman is on the left of the canoe and that the origin of our system is in the middle of the canoe.

The position of the center of gravity of the whole system remains in the same position, so:

$m_{c}*r_{oc}+m_{w}*r_{ow}=m_{c}*r_{fc}+m_{w}*r_{fw}$

Since our origin is in the middle of the canoe, all of the positions become:

$r_{oc}=0$

$r_{ow}=-(2.5-1)=-1.5$

$r_{fc}=-X$ (It is <em>negative</em>, assuming that the canoe will move to the <em>left</em>).

$r_{fw}=-X+(2.5-1)=-X+1.5$

Being X the displacement we want to find.

Replacing all of these values, we get:

$69*0+45*(-1.5)=69*(-X)+45*(-X+1.5)$

Solving for X:

X=1.18m Since the result is positive, we conclude that our assumption was correct.

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I think your answer would Be A. Because historians mostly study written documents, while archaeologists uncover fossils and buildings.

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Answer:

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Neko [114]

Answer:

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