Ask question

Ask question

All categories

3 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]3 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

1. A car is stopped at a red light. When the light turns green, the car starts

Lubov Fominskaja [6]

Answer:

Because of inertia (Newton's First Law of motion)

Explanation:

According to Newton's First Law of motion:

"An object at rest (or in motion at constant velocity) will tend to stay at rest (or tend to keep moving with same velocity) unless acted upon an unbalanced force"

In this problem, the object we are analyzing is the coffee cup.

At the beginning, the cup is at rest, together with the car.

Later, the car starts moving when the light turns green.

If we apply Newton's First Law of motion to the cup, we see that the coffee cup tends to keep its state of rest: for this reason, as the car moves forward, the coffee in the cup will spill backward, into the rear seat. This property of an object to mantain its state of motion is also called as inertia.

8 0

3 years ago

In the picture at left. The launch height is 10m ball B mass 5kg has a launch velocity of 12 m/s and ball A mass 8 kg has a laun

Hitman42 [59]

Answer:

24

Explanation:

8 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

Distance is the length of a path followed by a particle. The displacement of a particle is defined as its change in position in

worty [1.4K]

Answer:

a. True

Explanation:

Distance is described with only magnitude. It is defined as the total path covered by an object, in other words it is the length of a path followed by a particle.

Displacement is described with both magnitude and direction. It is distance traveled in a specified direction or change in position in some time interval.

Therefore, the correct option is " a. True"

7 0

3 years ago

A boy standing on the bridge kicks a stone into the water below. He kicks the stone with a horizontal velocity of 8.06 m/s. It l

Ronch [10]

Time taken to reach water :

$t = \dfrac{D}{v}\\\\t=\dfrac{6.78}{8.06}\ s\\\\t=0.84\ s$

Now, initial vertical speed , u = 0 m/s.

By equation of motion :

$h = ut +\dfrac{at^2}{2}$

Here, a = g = acceleration due to gravity = 9.8 m/s².

So,

$h = 0(t) +\dfrac{9.8\times 0.84^2}{2}\\\\h=3.46\ m$

Therefore, the height of the bridge is 3.46 m.

Hence, this is the required solution.

6 0

3 years ago