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Helga [31]
3 years ago
9

Which best explains Susan B Anthony purpose in her speech “Women’s rights to the suffrage”

Engineering
2 answers:
Furkat [3]3 years ago
6 0

Answer:

She wrote and delivered a speech in 1873, which came to be known as the “Women's Rights to the Suffrage” speech. In her address, she lets the audience know of her “crime” of voting. She reminds the listener that the Constitution of the United States says “we the people” and does not exclude women as people

Arte-miy333 [17]3 years ago
3 0

Answer:

D. To show why she was justified in voting.

Explanation:

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A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac
antoniya [11.8K]

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a )  ------ ( 1 )

From the above data given will  be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1  becomes

Py =  -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

    = 3270 N/m^2

6 0
3 years ago
An LED camping headlamp can run for 18 hours, powered by three AAA batteries. The batteries each have a capacity of 1000 mAh, an
KIM [24]

Answer:

a) the power consumption of the LEDs is 0.25 watt

b) the LEDs drew 0.0555 Amp current

Explanation:

Given the data in the question;

Three AAA Batteries;

<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------

so V_total = 3 × 1.5 = 4.5V

a) the power consumption of the LEDs

I_battery = 1000 mAh / 18hrs    { for 18 hrs}

I_battery = 1/18 Amp    { delivery by battery}

so consumption by led = I × V_total

we substitute

⇒ 1/18 × 4.5

P = 0.25 watt

Therefore the power consumption of the LEDs is 0.25 watt

b) How much current do the LEDs draw

I_Draw = I_battery = 1/18 Amp = 0.0555 Amp

Therefore the LEDs drew 0.0555 Amp current

5 0
2 years ago
Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr
baherus [9]

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

5 0
3 years ago
Technologies that allow for instant worldwide communication include
hram777 [196]

Answer:

mobile phones and internet access

Explanation:

I got it right on my quiz

7 0
3 years ago
Read 2 more answers
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc
a_sh-v [17]

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = C_p<em>d</em>T

q = \int\limits^{T_2}_{T_1} {C_p} \, dT   = C_p (T₂ - T₁)

From the above equations, the underlying assumption is that  C_p remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let C_{cal} be heat constant of calorimeter

Q₂ = C_{cal} ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m C_p ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m C_p' ΔT

C_p = (16.73 - 6.14) / (15.085 x 3.10)

C_p = 0.22646 KJ mol⁻¹ k⁻¹

6 0
3 years ago
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