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Vaselesa [24]
1 year ago
15

Two spherical objects have masses of 200 kg and 500 kg. Their centers are separated by a distance of 25 m. Find the gravitationa

l attraction between them.
Physics
1 answer:
ElenaW [278]1 year ago
8 0

Answer:

1.07 x 10⁻⁸N

Explanation:

Given parameters:

Mass 1 = 200kg

Mass 2  = 500kg

Distance of separation  = 25m

Unknown:

Gravitational attraction between the two bodies  = ?

Solution:

To solve this problem, we use the equation of the universal gravitation;

                 F  = \frac{G mass 1  x mass 2}{r^{2} }  

G is the universal gravitation constant  = 6.67 x 10⁻¹¹Nm²kg⁻²

r is the distance

 Now insert the parameters and solve;

    F  = \frac{6.67 x 10^{-11} x 200 x 500 }{25^{2} }   = 1.07 x 10⁻⁸N

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irga5000 [103]
Sometimes arithmetic problems can be solved much more easily using the dimensional analysis approach. You focus on the units of the given information. Then, you manipulate them applying the laws of algebra where like units cancel, in order to end up with the unit of the unknown.

Given:
-50 nc/step
31 steps
Unknown: charge

Thus,
Charge = -50 nc/step * 31 steps =<em> -1550 nc</em>
7 0
2 years ago
Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mu
MArishka [77]

Answer:

Horizontal component: F_x = 58\ N

Vertical component: F_y = 33.5\ N

Explanation:

To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.

Therefore, for the horizontal component, we have:

F_x = F * cos(angle)

F_x = 67 * cos(30)

F_x = 58\ N

For the vertical component, we have:

F_y = F * sin(angle)

F_y = 67 * sin(30)

F_y = 33.5\ N

So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.

4 0
2 years ago
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
The amount of pressure required to move a 6800 lb force with a 6" d piston is ___ psi.
Katena32 [7]
The pressure needed in PSI = Pounds of force needed divided by the cylinder Area
The Cylinder rod Area is 21.19  sq inches
Thus, the pressure= 6800/21.19
                              = 320.91 PSI

7 0
2 years ago
If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he
shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

F = \frac{GMm}{R^2} = mg

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

(mg)'= \frac{GMm}{(R+h)^2}

we need to find "h"

taking the ratio of two:

\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

5 0
2 years ago
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