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3 years ago

9

Two runners ran side by side, each holding one end of a horizontal pole. How would this affect the direction of the runners? Exp

lain.

1 answer:

zhuklara [117]3 years ago

6 0

Answer:

They will run parallel to each other as the none of a straight pole cannot be bent in such a way where one side can turn without the other turning.

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Which pair of countries are both part of the crib Caribbean cultural regions in North America

Anni [7]

I believe Cuba and the Bahamas.

6 0

3 years ago

Which direction will the box move?

Yuki888 [10]

Net force

F1-F2-F3
5-10-20
5-30
-25N

As it's negative the box will move left

3 0

2 years ago

(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma

andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

a)

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

c)

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

4 0

3 years ago

An airplane takes off from Dallas Texas to fly to new york city traveling ne for 2,760 km the Same plane returns that day to Dal

Kipish [7]

<h3>Answer</h3>

1104 km/hour

<h3>Explanation</h3>

Distance between Dallas Texas to New York = 2760 km

Time the plane took from Dallas to New York = 2 hours

Time the plane took from New York back to Dallas = 2.5 hours

Formula to use

<h3>distance = speed x time </h3>

Speed the plane took from Dallas to New York

2760 = 2 x speed

speed = 2760 / 2

= 1380 km/hour

Speed the plane took from New York to Dallas (ROUND TRIP)

2760 = 2.5 x speed

speed = 2760 / 2.5

= 1104 km/hour

3 0

3 years ago

Read 2 more answers

In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

3 years ago