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OleMash [197]
3 years ago
9

Name some applications of white surfaces.

Physics
2 answers:
timofeeve [1]3 years ago
7 0
White marble is a white surface
yarga [219]3 years ago
3 0
White and black surface?
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Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

\frac{m_A g}{A_A}=\frac{m_B g}{A_B}

Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0
3 years ago
How can we relate density and pressure in liquids ??
Masteriza [31]
Pressure increases with increasing depth. h2=2hh
6 0
3 years ago
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Which statement below best reflects the energy of the rock shown in the diagram?
Natali [406]

Answer:

32

Explanation:

32

5 0
2 years ago
A star's parallax angle is 0.8. How far away is the star in parsecs? Astronomy
kramer

Distance= 1/arc seconds

1/.8= 1.25 parsecs away

3 0
3 years ago
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When a hot object is placed in a water bath whose temperature is 25◦C, it cools from 100 to 50◦C in 150 s. In another bath, the
mixas84 [53]
Using Newton's Law of Cooling, the given are:
Temperature of the first bath (Ta1): 25C
a(final temperature)= 50C, b (initial temperature)=100C
x-y is the difference in time. In the first bath, 150s

dT/dt=-k(T-Ta)
\int\limits^a_ b{(dT/(T-Ta1))} \, dt = \int\limits^x_y{-k} \, dt
ln (T-25) [from a=100 to b=50]= -k*150
ln ((50-25)/(100-25))= -k*150

Solving for k=7.324081X10^-3

For the second bath: find Ta2 with dt=120s

dT/dt=-(7.324081x10^-3)*(T-Ta2)
\int\limits^a_ b{(dT/(T-Ta2))} \, dt = \int\limits^x_y{-7.324081x10^-3} \, dt
ln (T-Ta2) [from a=100 to b=50]= -(7.324081x10^-3)*120
ln ((50-Ta2)/(100-Ta2))= -(<span>7.324081x10^-3)</span>*120

Solving for Ta (second bath temperature)= 14.49 degrees Celsius

6 0
3 years ago
Read 2 more answers
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