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Sidana [21]
1 year ago
5

The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt

ed. If the wavelength of the emitted gamma ray is 3.44×10−12 m , what is the decrease in mass, ΔmXe, of the xenon nucleus as a result of this decay?
Physics
1 answer:
nevsk [136]1 year ago
6 0

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

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A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
1 year ago
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic
andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0
2 years ago
Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp
MatroZZZ [7]

Answer:

E) True.   Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

          Em₀ = K

          Em₀ = ½ m v²

Final

         Emf = U = mg h

         Em₀ = emf

        ½ m v² = mgh

        h = v² / 2g

For ball A

         h_A = v² / 2g

For ball B

        h_B = (2v)² / 2g

        h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False  has 4 times the gravitational potential energy than ball A

D) False.  It goes 4 times higher

E) True.

6 0
1 year ago
Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th
Inessa [10]

Answer:

F = 2.01*10^-16N -^k

Explanation:

In order to calculate the magnetic force perceived by the bee, you use the following formula:

F=qv\ X\ B            (1)

q: charge of the bee = 1pC = 1*10^-12 C

The average speed of a bee and the magnetic field of the earth are:

v = 6.70m/s

B = 30*10^-6 T

The bee is flying to the west (-^i). You consider that the magnetic field direction is to the north (^j). Then, the direction of the magnetic force is:

-^i X ^j = -^k

You replace the values of the parameters in the equation (1), in order to calculate the magnitude of the force:

|F|=qvB=(1*10^{-12}C)(6.70m/s)(30*10^{-6}T)=2.01*10^{-16}N

The magnetic force perceived by the bee is 2.01*10^-16N in the -^k direction, that is, toward the ground

5 0
2 years ago
A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude
Pani-rosa [81]
The amplitude of a wave corresponds to its maximum oscillation of the wave itself. 
In our problem, the equation of the wave is
y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
y(x,t)=0.750 cm
and therefore this value corresponds to the amplitude of the wave.
4 0
2 years ago
Read 2 more answers
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