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3 years ago

Physics question about density

Physics

2 answers:

scZoUnD [109]3 years ago

7 0

So the right answer is 4kg/m^3.

Look at the attached picture

Hope it will help you

Good luck on your assignment

ohaa [14]3 years ago

6 0

Answer:

The answer is option 4.

Explanation:

Firstly, you have to convert the mass of tea bagto kilogram as the question wants the unit in kg/m³.

$1000g = 1kg$

$100g = 1kg \div 10$

$100g = 0.1kg$

So the mass of tea bag is 0.1 kg. Next you have to use density formula ρ = mass/volume. Then you have to substitute the values into the formula :

$ρ = \frac{mass}{volume}$

Let mass = 0.1kg,

Let volume = 0.025m³,

$ρ = \frac{0.1}{0.025}$

$ρ = 4kg \: per \: {m}^{3}$

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The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s

JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

We will use second equation of motion to find the height:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

$h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\$

h = 593.50 m

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

$v_{f} = v_{i} + gt$

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

$v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s$

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height covered during last second = h₁₁ = ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

$h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\$

h₁₁ = 103 m

Now, we use first equation of motion for complete motion:

$v_{f} = v_{i} + gt$

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

$v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)$

vf = 107.91 m/s

8 0

3 years ago

When electric and _______ fields both exist, it is called an electromagentic

Sergeeva-Olga [200]

Answer:

the answer is magnetic

3 0

3 years ago

1. Una carga Q1 = + 12 μC se coloca a una distancia r = 0.024 m desde una carga Q2 = + 16 μC. a) Determina la magnitud de la fue

lyudmila [28]

Answer:

1. a. 3,000 N

b. Repulsión

2. 46.875 × 10⁶ N/C

3. a. 81,000 J

b. 6.75 × 10⁹ V

Explanation:

1. Los parámetros dados son;

Q₁ = +12 μC, Q₂ = +16 μC

La distancia entre las cargas, r = 0.024

La magnitud de la fuerza electrostática, F, entre cargas se da como sigue;

$F = k \times \dfrac{Q_1 \cdot Q_2}{r^2}$

Donde, k = constante de Coulomb = 9.0 × 10⁹ N · m² / C²

Por lo tanto, obtenemos;

F = 9.0 × 10⁹ × 12 × 10⁻⁶ × 16 × 10⁻⁶ / 0.024² = 3.000

La magnitud de la fuerza electrostática, entre las cargas, F = 3000 N

(b) Dado que tanto Q₁ como Q₂ son cargas positivas, y las cargas iguales se repelen entre sí, la fuerza es la repulsión.

2) La intensidad de un campo eléctrico, E, se da como sigue;

$E = \dfrac{k \cdot Q}{r^2}$

La magnitud de la carga, Q = 24 μC

La distancia donde se mide el campo, r = 48 mm = 0.048 m

Por lo tanto, E = 9.0 × 10⁹ × 12 × 10⁻⁶ / 0.048² = 46,875,000 N / C

La intensidad de un campo eléctrico, E = 46,875,000 N / C = 46.875 × 10⁶ N / C

3. La magnitud de las cargas son;

Q₁ = 24 mC

Q₂ = -12 μC

La distancia entre las cargas, r = 0.032 m

un. El potencial eléctrico de una carga, $U_E$ , se da de la siguiente manera;

$U_E = k \times \dfrac{Q_1 \cdot Q_2}{r}$

Por lo tanto;

$U_E$ = 9.0×10⁹ × 24 × 10⁻³ × (-12) × 10⁻⁶ /0.032 = -81,000

La energía potencial eléctrica entre la carga, Q₁ y Q₂= -81,000 J

b. El potencial eléctrico de Q₁ en Q₂, V₁ = $k \times \dfrac{Q_1 }{r}$

Por lo tanto, V₁ = 9.0×10⁹ × 24 × 10⁻³/0.032 = 6.75 × 10⁹

El potencial eléctrico de Q₁ en Q₂, V₁ = 6.75 × 10⁹ V

3 0

3 years ago

Why are different constellations of stars seen during different seasons?

slamgirl [31]

Actually, they're not. There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around. And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night.

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ? Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ? Now they're straight in the
direction of the sun. So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.

5 0

3 years ago

Read 2 more answers

Wavelength is measured in hertz true or false?

g100num [7]

Wavelength is measured in various units of distance (mm, cm, m, etc.). Frequency uses hertz, but that is not what the question is asking. Therefore, the answer is false.

3 0

3 years ago