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Masteriza [31]
4 years ago
14

In a fission experiment observed by Hahn and Strassman, uranium-235 was bombarded by neutrons. The products of this ission react

ion included two light nuclei, barium-137
and krypton-84. What was another product of this nuclear reaction?
water
energy
hydrogen-1
helium-4
Physics
1 answer:
wlad13 [49]4 years ago
7 0

Answer:

helium-4 (90%) or tritium (7%).

Explanation:

hope it helped u buddy

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many musical instruments use "boxes" as a part of an instrument. Some examples are acoustic guitars and pianos. From your experi
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Answer :

Essay on Sound instruments/sound boxes:

Musical instruments also some times called <em>"sound box"</em>,  because the box modifies the sound of an instrument. It also helps in transfer of sound to the surrounding air.

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For example <em>xylophone</em>, which is a musical instrument with wooden bars. These wooden bars are arranged similar to the piano. under each bar of the xylophone there is a resonator tube, which amplifies the sound.


8 0
3 years ago
Parallax of the stars is created by
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Parallax of the stars is created by :
B. Different locations of Earth in its orbit
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hope this helps
4 0
4 years ago
Read 2 more answers
Two violinists are trying to tune their instruments in an orchestra. One is producing the desired frequency of 440.0 hz. The oth
Katyanochek1 [597]

Answer:

Percentage change in tension is 3.8%

Explanation:

We have given initially frequency f_1 = 440 Hz

Let tension in the string at this frequency is T_1

Now second frequency is f_2=448.4Hz

Frequency in string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}

From the relation we can say that

\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}

\frac{440}{448.4}=\sqrt{\frac{T_1}{T_2}}

{\frac{T_2}{T_1}}=1.038

Percentage change in tension is equal to

=\frac{T_2-T_1}{T_1}=\frac{T_2}{T_1}-1=(1.038-1)\times 100=3.8 %

So percentage change in tension is 3.8%

4 0
3 years ago
A 0.50-kg croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it i
NeTakaya

The impulse given to the ball is equal to the change in its momentum:

J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s

This is also equal to the product of the average force and the time interval ∆t :

J = F(ave) ∆t

so that if F(ave) = 200 N, then

∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s

7 0
2 years ago
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