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Degger [83]
3 years ago
13

Determine the density of an object that has a mass of 149.8 g and displaces 12.1 mL of water when placed in a graduated cylinder

.
Physics
1 answer:
Crazy boy [7]3 years ago
6 0

Answer:

"12.4 g/mL" would be the right solution.

Explanation:

The given values are:

Mass of the object,

= 149.8 g

Volume of the object,

= 12.1 mL

Now,

The Density will be:

⇒ Density=\frac{Mass}{Volume}

On substituting the given values in the above formula, we get

⇒               =\frac{149.8}{12.1}

⇒               =12.4 \ g/mL

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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find
sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge q_{1}=9\times10^{-6}\ C at origin

Second charge q_{2}=6\times10^{-6}\ C

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

E=\dfrac{kq}{r^2}

0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}

\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}

\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1

\dfrac{1}{d}=1.82

d=\dfrac{1}{1.82}

d=0.55\ m

Hence, The coordinates of the point is (0,0.55).

3 0
3 years ago
An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
Scorpion4ik [409]

Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

8 0
3 years ago
Time Intervals in Ice Ages
stepladder [879]
It started off with 68% less than it did at the peak, and later created a void and melted the remainder of the ice at about 92%
8 0
3 years ago
The power of the kettle was 2.6 kW
faltersainse [42]

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f  = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

7 0
2 years ago
Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.2cm and a standard deviation o
Svet_ta [14]

Answer:

97.5%

Explanation:

By the empirical rule (68-95-99.7),

  1. 68% of data are within <em>μ </em>- <em>σ</em> and <em>μ </em>+ <em>σ</em>
  2. 95% of data are within <em>μ </em>- 2<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
  3. 99.7% of data are within <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 2<em>σ</em>

<em>σ </em> and <em>μ</em> are the standard deviation and the mean respectively.

From the question,

<em>μ</em> = 7.2 cm

<em>σ</em> = 0.38 cm

7.96 = 7.2 + (<em>n</em> × 0.38)

<em>n</em> = 2

Hence, 7.96 represents <em>μ </em>+ 2<em>σ</em>.

P(X < <em>μ </em>+ 2<em>σ</em>) = P(X < <em>μ</em>) + P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>)

P(X < <em>μ</em>) is the percentage less than the mean = 50%.

P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>) is half of P(<em>μ </em>- 2<em>σ</em> < X < <em>μ </em>+ 2<em>σ</em>) = 95% ÷ 2 = 47.5%.

Considering this, for apples that are no more than 7.96 cm,

P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%

<em />

5 0
2 years ago
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