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Lelu [443]
3 years ago
15

A researcher is analyzing data from a high-energy particle collision. the result of the analysis gives a value of 8.8 x 10-19 c

± 0.1 x 10-19 c for the charge of one of the emitted particles. should the researcher accept the value?
Physics
1 answer:
Harrizon [31]3 years ago
5 0
Elementary charge used to determine charges of other objects is equal to a charge of electron or proton. It's value is roughly 1.6* 10^{-19} C. All other charges are whole-number multipliers of this elementary charge, meaning that we multiply elementary charge by {...,-2,-1,0,1,2,...}.

To find out if the measured charge can be accepted we need to divide it with elementary charge to see if we get whole number as result.
There are three possible values of measured charge: 8.7* 10^{-19} C\\ 8.8* 10^{-19} C \\ 8.9* 10^{-19} C

N_{1} = \frac{8.7* 10^{-19} C }{1.6* 10^{-19} C } =5.44 \\ N_{2} = \frac{8.8* 10^{-19} C }{1.6* 10^{-19} C } =5.5 \\ N_{3} = \frac{8.9* 10^{-19} C }{1.6* 10^{-19} C } =5.56

As we can see none of the possible values of a measured charge is whole-number multiplier of elementary charge so the researcher should not accept the value.
This charge can be achieved by using quarks which have value of 1/3 of elementary charge but they do not remain stable for long enough.
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Technically, this delivers a lot of energy into the Earth, but it’s spread out over a large enough area that it doesn’t do much more than leave footprints in a lot of gardens. A slight pulse of pressure spreads through the North American continental crust and dissipates with little effect. The sound of all those feet hitting the ground creates a loud, drawn-out roar which lasts many seconds.
6 0
4 years ago
A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli
trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

  • mass of the toy, m = 0.1 kg
  • the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}

Substitute the given values and solve the speed;

v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

7 0
3 years ago
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

5 0
3 years ago
A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. When the temperature of th
Elena L [17]
P1v1/t1 = p2v2/t2

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be aware, temperature must be in Kelvin.
7 0
3 years ago
Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi
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Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

\Delta X=X_{k}-X_{C}=0.021miles

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