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3 years ago

What are the properties of bungee gum,hmmm?

2 answers:

8 0

It’s both a solid and a liquid. It can thicken and soften depending on how it’s handled. It can be used to cover wounds to stop bleed, and used to drown enemies. Bungee Gum has the properties of both rubber and gum.

sergejj [24]3 years ago

5 0

Answer:

It's both a solid and a liquid.

It can thicken and soften depending on how it's handled.

It can be used to cover wounds to stop bleeding.

It can be used to drown enemies.

It might be edible like normal gum. ...

It's elasticated.

Explanation:

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What is the relationship between the electric force and each one of the charges?

Bess [88]

Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.

3 0

3 years ago

Assignment S of write the Symbol Told, mercury and Cooper, Iron, Lead

m_a_m_a [10]

Answer:

Check explanation

Explanation:

Gold - Au (Aurum)

Mercury - Hg (Hydrargyrum)

Copper - Cu (Cuprum)

Iron - Fe (Ferrum)

Lead - Pb (Plumbum)

These elements in the periodic table are some of the elements represented by letters not in line with their names.

This is because, these elements were known in ancient times and therefore, they are represented by letters from their ancient names.

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3 years ago

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve

Vlad1618 [11]

Answer:

$v=0.9833\ c$

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side, b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 $kg/m^3$

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184= \sqrt{1-\frac{v^2}{c^2}}$

$0.033= 1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

$v=0.9833\ c$

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

3 0

3 years ago

Introduction to Symbolic Answers, Part B, Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

lesantik [10]

2cos2(o)-1 is the answer

4 0

4 years ago

What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh

Dafna1 [17]

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled s = 9.5 m.

Substituting these values in the equation

$s= u \timest+0.5at^{2}$

$t= \sqrt{\frac{2s}{g} }$

$t=\sqrt{\frac{2\times9.5}{9.8} }$

⇒ t= 1.392 sec

Velocity needed

We know the time taken (1.392 s) to travel 1.5 m,

So velocity = 1.5 m / 1.392 s = 1.08 m/s

hence velocity of the diver must be at least 1.08 m/s

4 0

3 years ago