Please help ASAPa substance of mass 10kg with the dimension as 5cm by 4cm by 10cm. What is the density of true substance?​

Answers

Answer 1
Answer:

Answer:

5cm by 4cm by 10cm = 200

200 / 10 = 20

20 :>


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A car tire is fully inflated to a pressure of 32 psi at 25 °C. After driving for 2 hours, the volume of the tire is unchanged, but pressure of the tire is now 36 psi. What is the new temperature of the gas inside the tire?What caused this change?

Answers

Answer:

50.5°C

Explanation:

Start with the ideal gas law:

PV = nRT

Rearranging:

P/T = nR/V

Since the right side is constant (volume and mass don't change):

P₁ / T₁ = P₂ / T₂

This equation is also known as Gay-Lussac's law. Plugging in the values (make sure to use absolute pressure and temperature):

(32 psi + 14.7 psi) / (25 + 273.15) K = (36 psi + 14.7 psi) / T

Solving for T:

T = 323.69 K

T = 50.5°C

What happens to an earthquake's S-waves when they strike the core?A) They bounce off
B) They are absorbed
C) The change into P-waves
D) They change into L-waves

Answers

Answer:

B) They are absorbed

Explanation:

B is correct because S waves can only travel through solid materials.  When they hit liquid materials they are absorbed in and undetectable.

Answer:

B) They are absorbed

Explanation:

When they hit the liquid core, S-waves are no longer able to be detected. While P-waves bounce off the liquid core, S-waves are absorbed at the core.

When you apply the ipde process, you may decide to?

Answers

When you apply the IPDE Process, you may decide to change speed, change direction, or communicate with others. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.

Answers

Answer:

508Hz

Explanation:

A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - this phenomenon is beat production

frequency is the number of oscillation a wave makes in one seconds.

f1-f2=beats

therefore f1=512Hz

f2=?

beats=4Hz

512Hz-f2=4Hz

f2=512-4

f2=508Hz

the original frequency of the violin is 508Hz

Final answer:

The original frequency of the violin was 508 Hz. This is based on the principle of beats, where the beat frequency is the absolute difference in frequency between the two sources - in this case, the tuning fork and the violin string.

Explanation:

The original frequency of the violin string can be found using the principle of beats. The frequency of the beats is equal to the absolute difference in frequency between the two sources - in this case, the tuning fork and the violin string.

Initially, the beat frequency was heard as 4 Hz. This indicates that the original frequency of the violin was either 512 Hz + 4 Hz = 516 Hz, or 512 Hz - 4 Hz = 508 Hz. However, when the violin string was tightened, the beat frequency decreased to 2 Hz, which means the frequency of the note it was producing increased.

Therefore, the violin must have initially been producing a note with lower frequency (508 Hz), and even after tightening the string, the note it now produces (510 Hz) remains lower than that of the tuning fork.

Learn more about Beat Frequency here:

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2 What does how much, how
long and how
often have to do with exposure to chemicals?

Answers

Well how much, how long, and how often  

For example how much is like how much a object is left in a small room with chemicals

How long is the time

how often is how much the same cycle is done repeatedly

Two cars leave towns 680 kilometers apart at the same time and travel toward each other. one car's rate is 10 kilometers per hour less than the other's. if they meet in 4 hours, what is the rate of the slower car?

Answers

D = distance between the cars at the start of time = 680 km

v₁ = speed of one car

v₂ = speed of other car = v₁ - 10

t = time taken to meet = 4 h

distance traveled by one car in time "t" + distance traveled by other car in time "t" = D

v₁ t + v₂ t = D

(v₁ + v₂) t = D

inserting the values

(v₁ + v₁ - 10) (4) = 680

v₁ = 90 km/h

rate of slower car is given as

v₂ = v₁ - 10

v₂ = 90 - 10 = 80 km/h

Final answer:

The slower car travels at 75 km/hr while the faster car travels at 85 km/hr. They meet up after both traveling for 4 hours, thereby covering the 680 kilometers between them.

Explanation:

The subject of this question is algebra - specifically involving rates of speed and time. Here's how you would find the answer:

  1. Begin by understanding that it is not two separate trips but merely one trip where they are traveling towards each other. This means collectively they've traveled 680 kilometers.
  2. Denote the slower car's rate as 'r'. The faster car would then have a rate of 'r + 10' km/hr.
  3. Since they are traveling towards each other for the same duration, you can set up the formula 'distance = rate * time' for BOTH cars, i.e.:
    4r (slower car's distance covered) + 4(r + 10) (faster car's distance covered) = 680km.
  4. Solve the equation for 'r'.

The result would be 75 km/hr for the slower car and 85 km/hr for the faster car. They meet up after both traveling for 4 hours, thereby covering the 680 kilometers between them.

Learn more about Algebra here:

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