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2 years ago

What type of wave has more energy, an ultraviolet wave or an x-ray?

Physics

1 answer:

muminat2 years ago

3 0

As the wave length increases the energy of the wave decreases as the equation that relates the c=λυ λ is the wave length and υ is the frequency (which is directly proportional to the energy).
In the wave length spectrum, x-ray has a shorter wave length, meaning that x-ray has a higher energy than ultraviolet waves.

Hope this helps.

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PLS ANSWER WILL MARK BRANLIEST!!!!!!!!!!!!

Angelina_Jolie [31]

1) describe the life cycle of a star before it collapses into a black hole.

1) describe the life cycle of a star before it collapses into a black hole.ans: A star's life cycle is determined by its mass. The larger its mass, the shorter its life cycle. A star's mass is determined by the amount of matter that is available in its nebula, the giant cloud of gas and dust from which it was born. Over time, the hydrogen gas in the nebula is pulled together by gravity and it begins to spin. As the gas spins faster, it heats up and becomes as a protostar. Eventually the temperature reaches 15,000,000 degrees and nuclear fusion occurs in the cloud's core. The cloud begins to glow brightly, contracts a little, and becomes stable. It is now a main sequence star and will remain in this stage, shining for millions to billions of years to come. This is the stage our Sun is at right now.

2) describe the life cycle of a star before it becomes a dwarf.

ans: The life cycle of a low mass star (left oval) and a high mass star (right oval). ... As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf.

3) what is the likely outcome of our sun?

ans: All stars die, and eventually — in about 5 billion years — our sun will, too. Once its supply of hydrogen is exhausted, the final, dramatic stages of its life will unfold, as our host star expands to become a red giant and then tears its body to pieces to condense into a white dwarf.

5 0

2 years ago

Alright eksqijakojqnlqozjzbw.wlisj

saw5 [17]

Answer:

eksqijakojqnozjzbw.wlisjaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Explanation:

6 0

2 years ago

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

$= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

3 0

3 years ago

A 5.00 g bullet traveling 355 m/s is stopped by lodging in the side of a building. The heat produced is shared between the build

steposvetlana [31]

Answer:

Explanation:

Given

mass of bullet $m=5\ gm$

speed of bullet $v=355\ m/s$

bullet is stopped by building and heat produced is shared between building and bullet

Kinetic Energy of bullet is converted into Thermal energy

Kinetic Energy of bullet $K=\frac{1}{2}mv^2$

$K=0.5\times 5\times 10^{-3}\times (355)^2$

$K=315.06\ J$

So 315.06 J of Energy is converted in to thermal energy

7 0

3 years ago

A gas contained within a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where

Sidana [21]

Answer:

a) 90 kJ

b) 230.26 kJ

Explanation:

The pressure at the first point $P_{1}$ = 10 bar —> 10 x 102 = 1020 kPa

The volume at the first point $V_{1}$ = 0.1 m^3

The pressure at the second point $P_{2}$ = 1 bar —> 1 x 102 = 102 kPa

The volume at the second point $V_{2}$ = 1 m^3

Process A.

constant volume V = C from point (1) to P = 10 bar.

Constant pressure P = C to the point (2).

Process B.

The relation of the process is PV = C

Required

For process A & B

(a) Sketch the process on P-V coordinates

(b) Evaluate the work W in kJ.

Assumption

Quasi-equilibrium process

Kinetic and potential effect can be ignored.

Solution

For process A.

V=C

There is no change in volume then

$W_{a(1)}= 0\\P=10^{2}$

The work is defined by

$W_{a(2)}=\int\limits^V_V {P} \, dV$

$W_{A(2)} =$ ║ $10^{2}$ V║limit 1--0.1

$W_{A(2)} =$ 90 kJ

Process B

PV=C

By substituting with point (1) C = 10^2 x 1= 10^2

The work is defined by

$W_{b}=\int\limits^V_V {P} \, dV\\P=10^{2} V^{-1}\\ W_{b}=\int\limits^V_V {10^{2} V^{-1}} \, dV\\\\$

$W_{A(2)} =$ ║ $10^{2}$ ln(V)║limit 1--0.1

=230.26 kJ

3 0

3 years ago