Ask question

Ask question

All categories

2 years ago

13

the volume of ice block is 2400cm^3 and its density is 0.9 g/cm^3. how much part of it remains above the surface of water when i

t is kept in water?

1 answer:

stiks02 [169]2 years ago

4 0

Answer:

240 cm³

Explanation:

Weight = Buoyancy

mg = ρVg

m = ρV

(0.9 g/cm³ × 2400 cm³) = (1 g/cm³) V

V = 2160 cm³

The submerged volume is 2160 cm³, so the volume above the surface is 240 cm³.

You might be interested in

You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

3 0

2 years ago

what will be the moment of inertia of a body if it rotates at a uniform rate of 10rad/sec^2 by a torque of 120Nm?

Naya [18.7K]

Answer:

12 kgm²

Explanation:

here angular acceleration = 10rad/sec²

torque= 120Nm

moment of inertia=?

we know,

torque= angular acceleration× moment of Inertia

or, moment of inertia = torque/angular acceleration

= 120/10

= 12kgm²

7 0

2 years ago

A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul

jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, $m_a$ = 126 kg

Speed he acquires, $v_{a}$ = 2.70 m/s

Mass of the space capsule, $m_{c}$ = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

$P_f = m_av_a + m_cv_c$

$P_I$ = 0

$m_av_a + m_cv_c = 0$

$v_c =\frac{- m_a v_a}{m_c}}\\\\$

$= \frac{126* 2.70}{1800}$

$= - 0.189$ m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

= 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

= 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = $\frac{1}{2} m v^2$

$= \frac{1}{2}$ × 126 × $(2.70) ^2$

= 459.27 J

The Kinetic Energy of the capsule;

K.E = $\frac{1}{2} m v^2$

= $\frac{1}{2}$ ×1800× $(0.189) ^2$

= 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0

1 year ago

Why is the picture above not an accurate representation of charges and their field lines?

pochemuha

Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.

7 0

2 years ago

Read 2 more answers

Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro

blagie [28]

Answer:

$21.2\times 10^{-6}$ T

Explanation:

$i$ = magnitude of current in each wire = 2.0 A

$a$ = length of the side of the square = 4 cm = 0.04 m

$r$ = length of the diagonal of the square = $\sqrt{2}$ a = $\sqrt{2}$ (0.04) = 0.057 m

$B$ = magnitude of magnetic field by wires at A and C

$B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )$

$B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )$

$B = 10\times 10^{-6}$ T

$B'$ = magnitude of magnetic field by wire at B

$B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )$

$B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )$

$B' = 7.02\times 10^{-6}$ T

Net magnitude of the magnetic field at D is given as

$B_{net} = \sqrt{B^{2}+B^{2}} + B'$

$B_{net} = \sqrt{2} B + B'$

$B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})$

$B_{net} = 21.2\times 10^{-6}$ T

8 0

3 years ago