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Ne4ueva [31]
3 years ago
12

Convert the following measurements. Only type numbers and decimals into the answer boxes:

Physics
2 answers:
Semmy [17]3 years ago
7 0

1. .134

2. 22.34

3. 330

4. 4350

ivolga24 [154]3 years ago
6 0

Answer:

2m 4m2Explanation:rteji3ujnrej

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A
asambeis [7]

Answer:

22Volts

Explanation:

The pd at the terminal is known as the emf

Since there are Ten 2.2V cells

Terminal voltage = number of cells * pd of one cell

Terminal voltage = 10 * 2.2

Terminal voltage = 22V

Hence the pd at the battery terminals​ is 22Volts

4 0
3 years ago
One of the way atoms bond with each other would be through:
Galina-37 [17]

Answer:

ehdgywi

Explanation:

djhcuowhciwurvgwyirgvy mm. ncmsmsmx. n. mssmsmiwvfiywrvvkjbwviverbladcnviwrgqecocqeboodqeugচঠচবি

8 0
3 years ago
Which statements are true about galaxies, stars, and the universe? (more than one true answer by the way)
Elina [12.6K]

Answer:

statements <em><u>2, 3, 4, and 7</u></em> are true

Explanation:

8 0
3 years ago
The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known
Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

m_1 = Mass of electron = 9.11\times 10^{−31}\ kg

v_1 = Speed of electron = 5.7\times 10^7\ m/s

p_2 = Neutrino has a momentum = 7.3\times 10^{-24}\ kg m/s

M = total mass = 2.34\times 10^{-26}\ kg

In the x axis as the momentum is conserved

Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s

In the y axis

Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s

The resultant velocity is

R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s

The recoil speed of the nucleus is 2240.92365 m/s

3 0
3 years ago
Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp
jekas [21]

Answer:

a=\frac{mBg-mAgSin\theta}{mA+mB}

Explanation:

Given two mass on an incline code mA and mB and an angle of inclination \theta. g. Assume that mA is the weight being pulled up and mB the hanging weight.

-The equations of motion from Newton's Second Law are:

mBg-T=mBa where a is the acceleration.

#Substituting for T (tension) gives:

mBg-mAsin\theta-mAa=mBa

#and solving for a:

a=\frac{mBg-mAgSin\theta}{mA+mB} which is the system's acceleration.

8 0
4 years ago
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