All categories

4 years ago

What kind of deformation does a cube of Jell-O exhibit when it jiggles?

Physics

1 answer:

nadya68 [22]4 years ago

3 0

It depends on chemistry... A physical deformation to the Jell-O.

You might be interested in

A poker is a long thin tool used to move pieces of coal or logs burning in a fire. To be as safe as possible, the poker should b

nata0808 [166]

Answer:

See the answer below

Explanation:

A poker that will effectively and safely function to move pieces of coal or logs in a burning fire must be fireproof itself. Hence, to be as safe as possible, such <u>poker should be made from a material that is fireproof</u> and that does not conduct a lot of heat. Otherwise, the poker will catch fire/becomes too hot during the course of usage.

3 0

2 years ago

What causes interstellar dust and clouds to condense in the planets and stars

lorasvet [3.4K]

That would be a nebula, which is an interstellar cloud of hydrogen gas, dust, and plasma. It is the first stage of a star's cycle.

3 0

4 years ago

the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr

Dmitry [639]

Answer:

The approximate number of decays this represent is $N= 23*10^{10}$

Explanation:

From the question we are told that

The amount of Radiation received by an average american is $I_a = 2.28 \ mSv$

The source of the radiation is $S = 5.49 MeV \ alpha \ particle$

Generally

$1 \ J/kg = 1000 mSv$

Therefore $2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg$

Also $1eV = 1.602 *10^{-19}J$

Therefore $2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg$

An Average american weighs 88.7 kg

The total energy received is mathematically evaluated as

$1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x$

Cross-multiplying and making x the subject

$x = 88.7 * 1.423*10^{16} eV$

$x = 126.2*10^{16}eV$

Therefore the total energy deposited is $x = 126.2*10^{16}eV$

The approximate number of decays this represent is mathematically evaluated as

N = $\frac{x}{S}$

Where n is the approximate number of decay

Substituting values

$N = \frac{126 .2*10^{16}}{5.49*10^6}$

$N= 23*10^{10}$

7 0

4 years ago

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

Keith_Richards [23]

Answer:

1. Recollapsing universe

2. Critical universe

3. Coasting universe

Explanation:

Recollapsing universe has dark matter density greater than critical density. While critical universe has its matter density equal to the critical sensity. Coasting universe on the other hand has much smaller matter density compared to critical density.

Note that the critical density is approximately 10^-20 grams/cm3

3 0

3 years ago

Plz and ty ... free points, be nice

harkovskaia [24]

Answer:

Thank you!

Explanation:

5 0

3 years ago

Read 2 more answers