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stellarik [79]
3 years ago
10

If the final position vector of a moving object has a smaller magnitude than the initial position vector, then the change in the

object's position vector has a positive magnitude.
Physics
1 answer:
BARSIC [14]3 years ago
5 0

Explanation:

The given statement is absolutely true.  this is because magnitude of a vector is always non negative, it can not be zero unless its a zero vector. So, in the given question, final position vector of a moving object has a smaller magnitude than the initial position vector, so, magnitude is neither zero nor negative. Hence, it has a positive magnitude.

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Answer:

A. 1.6 N/cm

Explanation:

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3 years ago
What is the guage pressure of an object with a mass of 30 kg and 5ft height?
wariber [46]

Answer: D=pg

Explanation:

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3 years ago
Hello!
Mnenie [13.5K]

Answer:

T_0=80695.17162...

Explanation:

Given equation:

\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124

To solve the given equation:

\textsf{Apply log rules}: \quad e^{\ln (x)}=x

\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}

Multiply both sides by T₀:

\implies T_0-100=T_0e^{-0.00124}

Add 100 to both sides:

\implies T_0=T_0e^{-0.00124}+100

Subtract T_0e^{-0.00124} from both sides:

\implies T_0-T_0e^{-0.00124}=100

Factor out the common term T₀:

\implies T_0(1-e^{-0.00124})=100

Divide both sides by (1-e^{-0.00124})

\implies T_0=\dfrac{100}{1-e^{-0.00124}}

Carry out the calculation:

\implies T_0=\dfrac{100}{1-0.99876...}

\implies T_0=\dfrac{100}{0.001239231...}

\implies T_0=80695.17162...

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3 years ago
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3 years ago
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Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0
3 years ago
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