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Tom [10]
3 years ago
15

A 60kg student is 5m from a 120kg door, how much gravitational forces act between the two?

Physics
1 answer:
slavikrds [6]3 years ago
7 0

Gravitational force = G · m₁ · m₂ / r²

' G ' is the 'universal gravitational constant, 6.673 x 10⁻¹¹ m³/kg-s²

m₁ = 60 kg

m₂ = 120 kg

r = 5 m

F = (6.673 x 10⁻¹¹ m³/kg-sec²) · (60 kg) · (120 kg) / (5m)²

F = (6.673 x 10⁻¹¹ · 60 · 120 / 25) · (m³ · kg · kg / kg · s² · m²)

F = 1.922 x 10⁻⁸ (kg · m / s²)

<em>F = 1.922 x 10⁻⁸ Newton</em>  (about 0.0000000692 ounce of force)

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By what percent is the torque of a motor decreased if its permanent magnets lose 6.5 % of their strength
djyliett [7]
6 is the answer I remember the answer from when I took this and it was easy
4 0
2 years ago
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

5 0
3 years ago
A stone is thrown straight upward and reaches a maximum height of 31.8 m above its
atroni [7]

Answer:

aral ka muna ng mabuti para maintindihan mo

7 0
3 years ago
Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees​,
Iteru [2.4K]

Answer:

x=65.55m

Explanation:

Let x be the distance to the shore

From trigonometry properties:

tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m

3 0
3 years ago
A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for
Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0
3 years ago
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