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3 years ago

PLEASE HELP WITH THIS QUESTION.

Physics

1 answer:

Sophie [7]3 years ago

6 0

Answer:

#2 and #3 respectively

Explanation:

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I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0

3 years ago

A car travels 200km in 3.0 hours. Determine the average velocity of the car

Masja [62]

Answer:

Explanation:

The average velocity of the car can be found by using the formula

$a = \frac{d}{t } \\$

d is the distance

t is the time taken

From the question we have

$a = \frac{200}{3} \\ = 66.66666...$

We have the final answer as

Hope this helps you

4 0

2 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

3 0

2 years ago

How many significant figures does 0.00340 have?

True [87]

Greetings!

"<span>How many significant figures does 0.00340 have?"...

A significant figure is:
-A non zero number
-A zero in between non zero numbers
-Trailing zeros to the right of the decimal point

Therefore, this number has three significant figures:
3,4,0

Hope this helps.
-Benjamin</span>

4 0

3 years ago

Read 2 more answers

A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?

klasskru [66]

The final velocity of the train at the end of the given distance is 7.81 m/s.

The given parameters;

initial velocity of the train, u = 6.4 m/s
acceleration of the train, a = 0.1 m/s²
distance traveled, s = 100 m

The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;

v² = u² + 2as

v² = (6.4)² + (2 x 0.1 x 100)

v² = 60.96

v = √60.96

v = 7.81 m/s

Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.

Learn more here:brainly.com/question/21180604

4 0

2 years ago

PLEASE HELP WITH THIS QUESTION. ​

PLEASE HELP WITH THIS QUESTION.