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slavikrds [6]
2 years ago
11

What is the gravitational force between two students, John and Mike, if John has a mass of 81.0 kg, Mike has a mass of 93.0 kg,

and their centers are separated by a distance of .620 m?
Physics
1 answer:
Marianna [84]2 years ago
8 0

Answer:

1.31×10¯⁶ N

Explanation:

From the question given above, the following data were obtained:

Mass of John (M₁) = 81 Kg

Mass of Mike (M₂) = 93 Kg

Distance apart (r) = 0.620 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force (F) =?

The gravitational force between the two students, John and Mike, can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 81 × 93 / 0.62²

F = 6.67×10¯¹¹ × 7533 / 0.3844

F = 1.31×10¯⁶ N

Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N

You might be interested in
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

5 0
3 years ago
Question 3 of 15
omeli [17]

Answer:

B) Degrees

Explanation:

The directions of the vectors are often defined in terms of due East, due North, due West and due South. A direction exactly in between of North and East can be described as Northeast, similarly we can describe directions in terms of Northwest, Southeast and South west.

From these, the direction of a vector can be easily expressed in degrees, which is measured counter clockwise about its tail from due East. Considering that we can say that East is at 0° , North is at 90° , West is at 180 and South is at 270° counter clockwise rotation from due East.

So, we know that the direction of a vector lying somewhere between due East i.e 0° and due North i.e 90°, will be measured in degrees, which will have a value between 0°-90°

4 0
2 years ago
Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it conta
shtirl [24]

Answer:

1020 km

Explanation:

A complete rotation of the wheel equals a distance of 1 circumference.

The circumference is

C = \pi d

where <em>d</em> is the diameter of the wheel.

300,000 rotations = 300000\pi d = 300000\times\pi\times1.08\text{ m} = 1017876.0\ldots\text{ m}

In kilometers, this is = 1017876/1000 km = 1020 km

6 0
3 years ago
The equations are not balanced. Which equation would have the same coefficients in the same order as 2CO2 + 3H20 → C2H6O + 3O2?
KiRa [710]
<span>A: Al + FeO → Al2O3 + Fe

Hope it helps!
</span>
5 0
3 years ago
Read 2 more answers
Will give brainliest!
tamaranim1 [39]
Howdy!!

your answer is ----


total resistance 1/R = 1/9+1/9+1/9
= 3/9

=> R = 9/3 = 3 ohm
according to ohms law

voltage = current × resistance

= 4 × 3 = 12 volt


hope it help you
8 0
3 years ago
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