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Nataliya [291]
3 years ago
6

Add these measurements, using significant digit rules: 1.0090 cm + 0.02 cm = cm

Physics
1 answer:
marin [14]3 years ago
7 0

Answer:

1.029

Explanation:

1.0090 can also be looked at as "1.009"

0.02 can also be looked at as "0.020"

I think of it as 20+9 which is 29. There for your answer should be 1.029. There are no measurement rules applying to this equation since they are both in centimeters. So you don't have to convert anything.

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An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
ira [324]

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The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,

where h_1 and h_2 are the height of the column of oil and the unkown liquid, respectively. Writing for \gamma_{unk}, we have

\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.

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The magnetic lines of force always travel from north to south true or false
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2 years ago
A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an
anyanavicka [17]

Answer:

The value of tangential acceleration \alpha_{t} =  40 \frac{m}{s^{2} }

The value of radial acceleration \alpha_{r} = 80 \frac{m}{s^{2} }

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We know that tangential acceleration is given by the formula \alpha_{t} = r \alpha

Where r =  radius of the disk

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⇒ \alpha_{t} = 0.8 × 50

⇒ \alpha_{t} = 40 \frac{m}{s^{2} }

This is the value of tangential acceleration.

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\alpha_{r} = \frac{V^{2} }{r}

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\alpha_{r} = 80 \frac{m}{s^{2} }

This is the value of radial acceleration.

7 0
3 years ago
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