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3 years ago

Explain how your weight can change even though your mass remains the same.

Physics

2 answers:

lara31 [8.8K]3 years ago

6 0

Answer:

See explanation below

Explanation:

Recall that Weight is a Force that results from the action of a gravitational field on our body mass. Therefore if one changes locations to places where the acceleration of gravity is different, our weight can change although the mass of our body stays the same.

For example, going to a high mountain, where the acceleration of gravity is a little smaller than at sea level, will produce such change. Also, going to another planet with different gravitational field, or going to the Moon (where the acceleration of gravity is about 1/6 of that on Earth which therefore will reduce our weight without reducing our mass)

matrenka [14]3 years ago

5 0

Since weight is calculated with the equation:

Q=mg

the weight can also change when the gravitational acceleration (g) changes.

g can change at example when you travel to a different planet or (slightly) when gaining attitude on Earth (e.g. when climbing a mountain or flying on a plane you get lighter).

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Which statement about a right triangle is NOT true?

Alex787 [66]

Answer:

The square of the hypotenuse is equal to the sum of the squares of the other two lengths.

Explanation:

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3 years ago

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The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge

nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to $1.129m^2$

(b) Charge on the capacitor is equal to $1.5\times 10^{-6}C$

Explanation:

We have given spacing between the plates d = 0.05 mm = $5\times 10^{-5}m$

Value of capacitance $C=1\mu F=10^{-6}F$

(A) Capacitance of a parallel plate capacitor is equal to $C=\frac{\epsilon _0A}{d}$

So $10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}$

$A=1.129m^2$

So surface area of the plate will be equal to $1.129m^2$

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3 years ago

Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the

marysya [2.9K]

$Volume of balloon = \frac{4}{3} \pi r^{3}$

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V = 68+ 0.178*V

0.934 V = 68

V = 72.81 $m^3$

\frac{4}{3} \pi r^{3}[/tex]= 72.81

r = 2.59 m

So radius of helium balloon = 2.59 m

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3 years ago

Calculate the heat, in kilocalories, that is absorbed if 183 g of ice at 0.0 ∘C is placed in an ice bag, melts, and warms to bod

boyakko [2]

Answer:

The total amount of heat needed will be $Q_T=21.411kcal$ .

Explanation:

We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

$m=183g$

$l_f=80\frac{cal}{g} =334\frac{J}{g}$

$l_w=1\frac{cal}{g} =4.184\frac{J}{g}$

<em>i) </em>The fusion heat will be:

$Q_f=l_fm=14640cal=14.640kcal$

<em>ii)</em> The heat needed to warm the water from $T_i=0^{\circ}C$ to $T_i=37^{\circ}C$ will be:

$Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal$

So, the total amount needed will be the sum of these two results:

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