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3 years ago

CISTU U

Physics

1 answer:

marusya05 [52]3 years ago

5 0

solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

density of ball= mass of ball/Volume of ball

or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

viscous force(F)=6 × pi × eta × r × v

=6×3.14×eta×0.05×25

=23.55×eta

To attain the terminal velocity,

Fiscous force=Weight

or, 23.55× eta = 41.6

or, eta = 1.76

whete eta is the coefficient of viscosity.

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A marble resting near the edge of .90 m high table is given an initial horizontal speed of 1.24 m/s. What will be it’s horizonta

Maru [420]

Answer:

0.53 m

Explanation:

First of all, we have to consider the vertical motion of the ball, in order to find the time it takes for the marble to reach the ground. The initial height is $h=0.90 m$ , the initial vertical velocity is zero, while the acceleration is $g=-9.81 m/s^2$ , so the vertical position at time t is given by

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By demanding y(t)=0, we find the time t at which the ball reaches the ground:

$0=h-\frac{1}{2}gt^2$

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(0.9 m)}{9.81 m/s^2}}=0.43 s$

Now we can find the horizontal range of the marble: we know the initial horizontal speed (v=1.24 m/s), we know the total time of the motion (t=0.43 s), and since the horizontal speed is constant, the total distance traveled on the horizontal direction is

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3 years ago

What is it called cavity left behind in the rock after an organism's hard part has dissolved.

Lapatulllka [165]

Mold

Explanation:

A mold is a cavity that is left behind in the rock after an organism hard part has been dissolved. These are important fossils that useful in relative dating.

Some hard parts of organism are preserved in form of molds in soft sediments.
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3 years ago

The greenhouse effect presentation summarized?

emmainna [20.7K]

Answer:

What is the greenhouse effect?

The greenhouse effect is the way in which heat is trapped close to Earth's surface by “greenhouse gases.”

Explanation:

These heat-trapping gases can be thought of as a blanket wrapped around Earth, keeping the planet toastier than it would be without them. Greenhouse gases include carbon dioxide, methane, nitrous oxides, and water vapor. (Water vapor, which responds physically or chemically to changes in temperature, is called a "feedback.") Scientists have determined that carbon dioxide's warming effect helps stabilize Earth's atmosphere. Remove carbon dioxide, and the terrestrial greenhouse effect would collapse. Without carbon dioxide, Earth's surface would be some 33°C (59°F) cooler.

Credit: NASA

I hope this helps, if it doesn't then just message me and ill be more than happy to help :)

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2 years ago

A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

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3 years ago

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mixas84 [53]

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3 years ago