An object has a kinetic energy of 793 J and a mass of 9 kg, how fast is the object<br> moving?

The characteristic that gives an element its distinctive properties is its number of:_________

oee [108]

The characteristic that gives an element its distinctive properties is its number of protons because the number of protons of any element represents its atomic number.

<h3>What is the atomic number?</h3>

The total number of protons present in an atom is known as the atomic number of that atom. The atomic number has no correlation either with the number of neutrons or the number of electrons present inside an atom.

Since the number of protons in any element corresponds to its atomic number, this property provides an element with its particular features.

Learn more about the atomic number from here,

brainly.com/question/14190064

#SPJ4

5 0

1 year ago

How do the gravitational and electrical forces between a proton and neutron compare

uysha [10]

Assume distance of seperarion is 1m

F.elec = kQq/r^2

charge of a proton: 1.6×10^-19 C
charge of a neutron: 0 C

F.elec = 0 N

F.grav = GMm/r^2

mass of a proton: 1.672621898×10^-27 kg
mass of a neutron: 1.674927471×10^-27 kg

F.grav = (6.67408×10^-11)×(1.674927471×10^-27)×(1.672621898×10^-27)÷(1^2)

F.grav = 1.8699588×10^-64 N

4 0

4 years ago

Read 2 more answers

The maximum displacement in an oscillatory motion is A = 0.49 m. Determine the position x at which the kinetic energy of the par

amid [387]

Answer:

x = 0.40 m

Explanation:

When the displacement is maximum, the particle is momentarily at rest, which means that at this point (assuming no friction present) all the mechanical energy is elastic potential, which can be written as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} (1)$

Since in absence of friction, total mechanical energy must keep constant, this means that at any time, the sum of the kinetic and potential energy, must be equal to (1), as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (KE)_{f} + U_{f} (2)$

If KEf = U/2f, replacing in (2), we get:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (U/2)_{f} + U_{f} = \frac{3}{2} *U_{f} (3)$

At any point, the elastic potential energy is given by the following expression:

$U_{f} = \frac{1}{2} *k*x^{2} (4)$

where k= spring constant (N/m) and x is the displacement from the

equilibrium position.

Replacing (4) in (3), simplifying and rearranging, we get:

$E_{tot} = U_{o} = \frac{1}{2} *A^{2} = \frac{3}{4} *x^{2} (5)$

Finally, solving for x, we get:

$x = \sqrt{\frac{2}{3} } * A = \sqrt{\frac{2}{3} } * 0.49m = 0.40 m (6)$

8 0

3 years ago

You have an unpolarized light source and you wish to send a beam of this light with intensity Io through a number of sheets of p

tekilochka [14]

Answer:

Explanation:

a ) The angle between the polarization axis of two adjecent sheet

= 90 / 3 = 30 degree.

The formula for intensity of polarised light from unpolarised light ( first transmission

I₁ = I₀ /2

I₀ is intensity of unpolarised light and I₁ is intensity of light after first time polarization .

The relation of I₁ and I₂ is as follows

I₂ = I₁ cos²30

= I₀/2 x3/4

=3 I₀/8

Relation between I₃ and I₂ is as follows

I₃ = I₂ cos²30

= 3I₀ / 8 x 3/4

= 9 I₀ / 32

= 0 .28 I₀

In case of stack of 4 plates

angle between two plates = 90/4 = 22.5 degree

I₁ = I₀ /2

I₂ = I₁ cos²22.5

= I₀ /2 x .85

I₃ = I₂ cos²22.5

= I₀ /2 x .85 x .85

= .36 I₀

7 0

3 years ago

When antimatter interacts with an equal mass of ordinary matter, both matter and antimatter are converted completely into energy

Ilya [14]

Answer:

v = 5.88 10⁷ m / s

Explanation:

For this exercise we use the relation

E = m c²

also indicate that all energy is converted into kinetic energy

E = K = ½ (M-2m) v²

where m is the mass of antimatter and M is the mass of the ship's mass. Factor two is due to the fact that equal amounts of matter and antimatter must be combined

we substitute

m c² = ½ (M-2m) v²

v² = $2 \frac{m}{M+2m} \ c^2$

let's calculate

v = $\sqrt{2 \ \frac{4 \ 10^4 }{2 \ 10^6 + 2 \ 4 \ 10^4} \ (3 \ 10^8)^2}$

v = $\sqrt{ 34.615 \ 10^{14}}$

v = 5.88 10⁷ m / s

8 0

3 years ago

An object has a kinetic energy of 793 J and a mass of 9 kg, how fast is the object moving?