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Pavlova-9 [17]
3 years ago
5

A 235 kg object and a 1.37×1012 kg are located 2.59×104 m away from each other. What is the force due to gravity between the two

objects?
Physics
2 answers:
V125BC [204]3 years ago
8 0

Answer:

the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Explanation:

Given;

mass of the first object, m₁ = 235 kg

mass of the second object, m₂ = 1.37 x 10¹² kg

distance between the two object, r = 2.59 x 10⁴ m

The gravitational force between the two object is calculated as;

F= \frac{Gm_1m_2}{r^2}

where;

G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²

F= \frac{(6.67\times 10^{-11})(235)(1.37\times 10^{12})}{(2.59\times 10^4)^2} \\\\F = 3.2 \times 10^{-5} \ N

Therefore, the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Alexus [3.1K]3 years ago
4 0

Answer:

F = 3.2 x 10⁻⁵ N

Explanation:

The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:

F = \frac{Gm_{1}m_{2}}{r^{2}}

where,

F = gravitational force = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m₁ = mass of object 1 = 235 kg

m₂ = mass of object 2 = 1.37 x 10¹² kg

r = distance between objects = 2.59 x 10⁴ m

Therefore,

F = \frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(235\ kg)(1.37\ x\ 10^{12}\ kg)}{(2.59\ x\ 10^{4}\ m)^{2}}

<u>F = 3.2 x 10⁻⁵ N</u>

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meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
Compute the torque about the origin of the gravitational force F--mgj acting on a particle of mass m located at 7-xî+ yj and sho
Andrews [41]

Answer:

Explanation:

Force, F = - mg j

r = - 7x i + y j

Torque is defined as the product f force and the perpendicular distance.

It is also defined as the cross product of force vector and the displacement vector.

\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}

\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)

[tex]\overrightarrow{\tau  }= 7 m g x k

Here, we observe that the torque is independent of y coordinate.

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3 years ago
Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi
77julia77 [94]

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

3 0
3 years ago
Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, an
suter [353]

Answer:

J = 357.5 kg*m/s

Explanation:

  • The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.
  • Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:

       p_{o} = m*v_{o} = 55 kg * (-3.0 m/s)  (1)

  • By the same token, the final momentum is as follows:

       p_{f} = m*v_{f} = 55 kg * (3.5 m/s)  (2)

  • As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):

      J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)

5 0
3 years ago
The period of a satellite circling planet Nutron is observed to be 84 s when it is in a circular orbit with a radius of 8.0 x 10
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Answer:

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Explanation:

Given:

Period, T = 84s

Radius of satellite orbit, r = 8*10^6

Using the relation :

M = 4π²r³ / GT²

Where G = Gravitational constant, 6.67 * 10^-11

M = 4*π^2*(8*10^6)^3 / 6.67 * 10^-11 * 84^2

M = (20218.191872 * 10^18) / 47063.52 * 10^-11

M = 0.4295937 * 10^18 - (-11)

M = 0.4295937 * 10^29

M = 4.295937 * 10^28 kg

M = 4.3 * 10^28 kg

Mass of planet Nutron = 4.3 * 10^28 kg

8 0
3 years ago
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