All categories

3 years ago

What is the identity of a sample that has a volume of 92.5 cm3 and a mass of 249.8 g?

Physics

2 answers:

schepotkina [342]3 years ago

7 0

Its aluminum its correct. :)

denpristay [2]3 years ago

3 0

Here, We know, Density = Mass / Volume
Here, mass = 249.8 g
volume = 92.5 cm³

Substitute their values,
d = 249.8 / 92.5
d = 2.7 g/cm³

In short, Your Answer would be "Aluminum"

Hope this helps!

You might be interested in

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

6 0

3 years ago

Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (

lesya [120]

Answer:

a) When moving towards a high pressure center the pressure values increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.

When moving towards a high pressure center the pressure values increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

6 0

3 years ago

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o

LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is: