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andre [41]
3 years ago
12

Pressure is always measured in comparison to a standard pressure. True False

Physics
1 answer:
irakobra [83]3 years ago
8 0

Answer:

true

Explanation:

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What minimum volume must the slab have for a 75.0 kgkg woman to be able to stand on it without getting her feet wet
Alisiya [41]

V = 0.9375 m3 is minimum volume .

<h3>What is density ?</h3>
  • We use the word "density" to indicate how much space (or "volume") an object or substance occupies in relation to the amount of matter contained therein (its mass).
  • Density can also be defined as the quantity of mass per unit of volume. A dense object is one that is both hefty and small.

Given,

            1000 kg/m3 = density of water

             920 kg/m3 = density of ice

BF = Woman Weight + Slab Weight

Pvg = 75 + Pvg

V = 75 kg / (1000 - 920)

V = 75 / 80

V = 0.9375 m3

Therefore, V = 0.9375 m3 is minimum volume .

Learn more about density

brainly.com/question/952755

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6 0
2 years ago
Which of these could NOT be used to find the mechanical advantage of an inclined plane?
34kurt
The "c) percent efficiency" could not be used to find the mechanical advantage of an inclined plane. There are two formulae that could be used to determine the mechanical advantage of an inclined plane which stated as MA = Length/rise and Wout=Win. MA is the mechanical advantage, Wout is the output force, Win is the input force, and "rise" is the height of the inclined plane<span>.</span>
8 0
3 years ago
An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce
ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}

From them we can get v^2=v_0^2+2ad.

We have:

v-v_0=at\\t=\frac{v-v_0}{a}

And substitute:

x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}

We multiply both sides by 2a, and continue:

2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2

Being d the displacement x-x_0, we have v^2=v_0^2+2ad

For our exercise, we will write this as:

d=\frac{v^2-v_0^2}{2a}

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m

For the time we use:

t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s

6 0
3 years ago
A 15.0 volt storage battery is connected to three resistors, 7.75?, 15.5?, and 21.7?, respectively. the resistors are joined in
Rainbow [258]
Total resistance= 7.75+15.5+21.7=44.95
Current = 15V/44.95=0.334A
4 0
3 years ago
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of
enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

\frac{n_1}{n_2}=\frac{V_1}{V_2}

In the old transformer the ratio of voltages was:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1

In the new transformer the ratio of voltages is:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1

If the old current output was 425 kV, the ratio of current was:

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1

Then, the ratio of the new output over the old output is:

\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

P_L=I^2\cdot R

Then, the ratio of losses is (R is constant for both power losses):

\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74

The losses in the new line are 74% the losses in the old line.

7 0
3 years ago
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