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3 years ago

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Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of t

his circle with the symbol for “initial position”. Move the object 5.0cm to the right and stop. Label this circle with the correct symbol for “final position.” (A) What was the initial position of the object? (B) What is the final position of the object? (C) What is the distance traveled by the object? (D) What is the displacement of the object? (E) Of the three underlined quantities, which are numerically equal?

1 answer:

Viefleur [7K]3 years ago

6 0

It’s is D bc it is and idk you should do it

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Air is heated in a glass bottle. The heat energy added to the air is 2.0 × 104 joules. What is the change in internal energy of

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The change in internal energy of the gas is $\Delta U = 2.0 \cdot 10^4 J$ .

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3 years ago

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The on

astraxan [27]

Answer: (a) $F=7(10)^{-7}N$

(b) $F=1.344(10)^{-6}N$

(c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father $m_{1}=100kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between them $r=0.2m$ , the force $F_{F}$ exerted by the father is:

$F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}$ (2)

$F_{F}=0.0000007N=7(10)^{-7}N$ (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter $m_{J}=1.898(10)^{27}kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between Jupiter and Earth (where the baby is) $r_{E}=6.29(10)^{11}m$ , the force $F_{J}$ exerted by the Jupiter is:

$F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}$ (4)

$F_{J}=0.000001344N=1.344(10)^{-6}N$ (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

$F_{J}=0.000001344N=1.344(10)^{-6}N$ and $F_{F}=0.0000007N=7(10)^{-7}N$ we can see $F_{J}$ is greater than $F_{F}$ . However, the difference is quite small as well as the force exerted on the baby.

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3 years ago

a runner moves 2.88 m/s north. she accelerates at 0.350 m/s2 at -52.0 angle. at the point where she is running directly east, wh

Naily [24]

Answer:

11.7 m

Explanation:

I assume north is the y direction and x is the east direction, so Δx refers to the displacement in the east direction.

First, find the time it takes for the velocity to change from directly north to directly east.

Given (in the y direction):

v₀ = 2.88 m/s

v = 0 m/s

a = 0.350 m/s² sin(-52.0°) = -0.276 m/s²

Find: t

v = at + v₀

(0 m/s) = (-0.276 m/s²) t + (2.88 m/s)

t = 10.4 s

Given (in the x direction):

v₀ = 0 m/s

a = 0.350 m/s² cos(-52.0°) = 0.215 m/s²

t = 10.4 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10.4 s) + ½ (0.215 m/s²) (10.4 s)²

Δx = 11.7 m

7 0

3 years ago