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Arada [10]
3 years ago
7

gravity is a force between any two objects with mass. why doesn't a person feel a gravitational force between him/herself and an

other person?a)a person doesn't exert a gravitational force.b)the two gravitational forces cancel each other out.c)the gravitational forces of people is so small it is overshadowed by that of earth.d)there are so many people we are actually balanced by all the different gravitational forces.
Physics
2 answers:
Doss [256]3 years ago
7 0
If you and your brother both have 70 kg of mass, then . . .

-- you both weigh about 154 pounds on Earth.

-- You both weigh about 25 pounds on the Moon.

-- If you stand about 2 meters (6-1/2 feet) apart, then the force
of gravity pulling you together is about  8.2 x 10⁻⁸ newton.

That's about  0.00000029 ounce.

THAT's why.

The correct choice is 'c'.
antiseptic1488 [7]3 years ago
4 0

Answer:

c)the gravitational forces of people is so small it is overshadowed by that of earth.

Explanation:

The gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the masses of the two objects

r is the distance between the two objects

From the formula, we see that the gravitational force depends on the masses of the objects: since the mass of the Earth (5.97\cdot 10^{24} kg is much much larger than the average mass of one person (80-100 kg), then the gravitational force exerted by the Earth on a person is also much much larger than the gravitational force between two people.

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A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in
sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]            ----------------------    (1)

in which R₁ is the resistance at temperature T₁

R_{T(2) is the resistance at temperature T₂

Given that V= IR

R = \frac{V}{I}

Therefore, the resistance at temperature 28°C is;

R_{28}= \frac{120V}{1.36A}

= 88.24Ω

R_{T(2) = \frac{120V}{1.23A}

= 97.56Ω

From (1) above;

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]      

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

\frac{0.1056}{4.5*10^{-4}}= T-28^0C

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

P= I^2_2R_{T^0C

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0
3 years ago
Can someone help me with this? (The answer marked in red text is a incorrect answer)
Tju [1.3M]

Answer:

the answer is A

Explanation:

because I just know

3 0
2 years ago
Three diffrent examples of accelerated motion
LekaFEV [45]

Answer:

The three different examples of the accelerated motion are Falling/dropping of ball, Standing in circular rotating space, moving around the circle.

Explanation:

Acceleration is the change in velocity, which is related to the speed and direction in which the object is travelling. Hence, speeding up, slowing down and turning are few types . A simple example would be dropping a ball: as it falls its speed increases, which is a type of acceleration. A more complicated example would be standing in a circular, rotating space station. A point on the station moves in a circle, meaning that as it travels it must be turning (to remain in circular motion) making this another example of acceleration

3 0
2 years ago
As temperature increase what energy increases
garik1379 [7]
Kinetic energy would increase sir.
8 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
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