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boyakko [2]
3 years ago
11

Elastic collision formula find each velocity

Physics
1 answer:
pshichka [43]3 years ago
7 0
We know the initial velocity of the golf ball and its mass, but we don't know the final velocitiesof either ball, and the trick to make these calculations go faster for an elastic collision is to use this equation, which says the initial velocity of one of the objects before the collision

Hoped this helped!
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A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 proto
svetoff [14.1K]

Answer:

Positively charged particle

Explanation:

¹⁰₆C → ¹⁰₅B + e⁺ + ve

Since there is a decrease in the atomic number, there has been nuclear transmutation and a proton has be broken into a neutron and positron with the release of neutrino which has zero charge and very small mass. This is a beta plus decay with the release of Positively charged particle

5 0
4 years ago
Read 2 more answers
What is vector quantities?
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Answer: Vector quantity is a measurement that has both magnitude and direction.

Explanation:

Vector quantity is a measurement that has both magnitude and direction. Examples are force, acceleration, displacement and velocity.

5 0
3 years ago
If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i
kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from 15 cm to 21 cm will be , E_{1}=27J

The energy used for stretch the spring from 21 cm to 27 cm will be , E_{2}=45J

using the energy of spring formula ,we find that

27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))

45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))

Dividing both the equation will get,

\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm

Therefore, the natural length of the spring is, 9 cm.

4 0
3 years ago
A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by
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Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

f_o = Focal length of objective = 0.9 cm

f_e = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52

The angular magnification of the compound microscope is -252.52

6 0
3 years ago
The diagram below shows different weights on the see-saw. Will the see-saw move?
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Answer:

yes it will

Explanation:

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3 years ago
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