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hichkok12 [17]
3 years ago
9

You are given two rectangular blocks of shiny metal, Block A and Block B, and are asked to determine which one will float in a b

ath of mercury (a metal that is liquid at room temperature). Block A is 10 cm long, 6 cm wide and 1 cm high and has a mass of 630 g. Block B is 5 cm long, 5 cm wide, 3 cm high, and has a mass of 604 g. If mercury has a density of 13.6 g/cm3, which of the two blocks will float when placed in mercury?
Physics
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

Explanation:

Volume of block A = 10 x 6 x 1 = 60 cm³

Mass of block A = 630 g

density of mass A = mass / density

= 630 / 60 = 10.5g / cm³

Volume of block B = 5 x 5 x 3 = 75 cm³

Mass of block A = 604 g

density of mass A = mass / density

= 604 / 75 = 8.05 g / cm³

Since density of both A and B are less than that of mercury , both will float in mercury.

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Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it conta
shtirl [24]

Answer:

1020 km

Explanation:

A complete rotation of the wheel equals a distance of 1 circumference.

The circumference is

C = \pi d

where <em>d</em> is the diameter of the wheel.

300,000 rotations = 300000\pi d = 300000\times\pi\times1.08\text{ m} = 1017876.0\ldots\text{ m}

In kilometers, this is = 1017876/1000 km = 1020 km

6 0
3 years ago
Which combination of these devices monitors waves that have a compressional component?
kompoz [17]

Answer:

The answer is A

Explanation:

4 0
3 years ago
Read 2 more answers
How would you find the total energy stored in the
likoan [24]

Answer:

The energy of the capacitors connected in parallel is 0.27 J

Given:

C = 2.0\micro F = 2.0\times 10^{- 6} F

C' = 4.0\micro F = 4.0\times 10^{- 6} F

Potential difference, V = 300 V

Solution:

Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:

C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F

The energy of the capacitor, E is given by;

E = \frac{1}{2}C_{eq}V^{2}

E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J

6 0
3 years ago
As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce
WINSTONCH [101]

Answer:

a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

The force of friction has the expression

                  fr = μ N

We replace

                 μ mg = ma

                 a = μ g

                 g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0
3 years ago
Part b only please if you have time
andrew-mc [135]
To convert parametric to Cartesian systems, you need to find a way to get rid of the t's.

In this case, the t's are inside trigonometric functions, so we're going to use a very famous trig identity you should memorize:

{sin(t)}^{2} + {cos(t)}^{2} = 1

If we plug sin(t) and cos(t) into that equation only x and y variables will be left!

BUT there's one thing. The given cos(t + pi/6) has nasty extra stuff in it. However, part a gives you a tip on how to relate x and y to a nice clean cos(t)

So if we do a little rearranging:

\sin(t) = \frac{y}{2} \\ \cos(t) = \frac{x + y}{2 \sqrt{3} }

Now we can plug these into the famous trig identity!

{( \frac{y}{2}) }^{2} + {( \frac{x + y}{2 \sqrt{3} } )}^{2} = 1

Do a little bit of adjustments to get that final form asked for, and you'll be able to find those integers of a and b. ;)
7 0
3 years ago
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