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2 years ago

7

If the social distancing length between two students is doubled from two metered to four meters, does the gravitational force be

tween the two students increase or decrease?
Explain your reasoning!

1 answer:

jonny [76]2 years ago

6 0

Answer:

the gravitational force decreases

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A 8kg cat us running 4 m/s. How much kenetic energy is that

Flauer [41]

Using the Definition of Kinetic Energy, we have:

$E_{c}= \frac{mv^2}{2} \\ E_{c}= \frac{8*4^2}{2} \\ \boxed {E_{c}=64J}$

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2 years ago

When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le

77julia77 [94]

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

$x+2.93$

Both cover a magnitude of 8.32 ft, therefore

$x +(x+2.97) = 8.32$

Now solving for x we have,

$x + (x + 2.93) = 8.32$

$2x + 2.93 = 8.32$

$2x = 8.32 - 2.93$

$x = \frac{ 8.32 - 2.93}{2}$

$x = 2.695 ft$

Therefore the shorter stick is 2.695ft long.

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3 years ago

Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

elena-14-01-66 [18.8K]

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, $1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}$

$1\ kcal/h=1.16\ J/sec$

J/sec is nothing but watts.

So, $1\ kcal/h=1.16\ watts$

and $88\ kcal/h=88\times 1.16\ watts = 102.08\ watts$

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

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2 years ago

Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

storchak [24]

Answer:

$\large \boxed{42\, \mu \text{C}}$$

Explanation:

The formula for the force exerted between two charges is

$F=k \dfrac{ q_1q_2}{r^2}$

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

$F=k\dfrac{q^{2}}{r^2}$

$\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}$

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2 years ago

So long as there are(is) ______ there is no movement.

patriot [66]

Answer:

answer is friction. MCQ A is answer

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2 years ago