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3 years ago

A 0.8 kg object displaces 500 mL of water. What is its specific gravity?

1 answer:

8 0

The specific gravity is how the density of the object compares to the density of water. Water's density is 1gram per milliliter. We just need to figure out the density of the object.

The object is .8 kg and it displaces 500mL of water, so the density is the mass divided by the volume. Since the density of water is given in grams, we have to convert the objects mass from kg to g and then we can get the density.

.8kg * 1000g/kg = 800 grams

So

800g/500ml = 1.6grams/mL this is the density.

So divide the density of your object by the density of water, which is 1g/mL, you get 1.6 as the specific gravity. This means the object is 1.6 times more dense than water.

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Anettt [7]

Answer:

270 mi/h

Explanation:

Given that,

To the south,

v₁ = 300 mi/h, t₁ = 2 h

We can find distance, d₁

$d_1=v_1\times t_1\\\\d_1=300\times 2\\\\d_1=600\ \text{miles}$

To the north,

v₂ = 250 mi/h, d₂ = 750 miles

We can find time, t₂

$t_2=\dfrac{d_2}{v_2}\\\\t_2=\dfrac{750\ \text{miles}}{250\ \text{mi/h}}\\\\t_2=3\ h$

Now,

Average speed = total distance/total time

$V=\dfrac{d_1+d_2}{t_1+t_2}\\\\V=\dfrac{600+750}{2+3}\\\\V=270\ \text{mi/h}$

Hence, the average speed for the trip is 270 mi/h.

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2 years ago

A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$

Charge of second Particle $q_2=-8\ \mu C$

distance between them $d=4\ cm$

$k=9\times 10^{9}$

magnetic field due to first charge at mid-way between two charged particles is

$E_1=\frac{kq_1}{r^2}$

$r=\frac{d}{2}=\frac{4}{2}=2\ cm$

$E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_1=27\times 10^7\ N/C$ (away from it)

Electric field due to $q_2=-8\ \mu C$

$E_2=\frac{kq_2}{r^2}$

$E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_2=-18\times 10^7\ N/C$ (towards it)

$E_{net}=E_1+E_2$

$E_{net}=9\times 10^7\ N/C$ (away from first charge)

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