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rewona [7]
3 years ago
9

The force of Earth's gravity on a capsule in space will lessen as it moves farther away. If the capsule moves to twice its dista

nce, the force toward Earth becomes?
Physics
1 answer:
Bess [88]3 years ago
3 0

Answer: One quarter of the force

Explanation:

According to Newton's law of Gravitation, the force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the product of their masses and inversely proportional to the square of the distance:

F=G\frac{(m1)(m2)}{r^2}    (1)

Where Gis the gravitational constant

This means that the gravity force decreases when the distance between these two bodies increases.

In this context, if the distance between the capsule and the Earth increases twice, the new distance will be 2r.

Substituting this distance in (1):

F=G\frac{(m1)(m2)}{(2r)^2}    (2)

F=G\frac{(m1)(m2)}{4r^2}    

<u>Finally:</u>

F=\frac{1}{4}G\frac{(m1)(m2)}{r^2} >>>This means the force toward Earth becomes one quarter "weaker"

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A student starts at position (4,4) m and undergoes three displacements:
quester [9]

Answer:

The final position is  (2, 7)m

Explanation:

When we work with coordinate pairs, the addition works as:

(a, b) + (c, d) = (a + c, b + d)

So, for example, if we start at (a, b), and we have a displacement d = (1, 1)

we just need to solve:

(a, b) + (1, 1) = (a + 1, b + 1)

Now, in this case, we start at (4, 4)m

first, we have d1 = (2, -3) m

After this displacement, the position is:

(4, 4)m + (2, -3)m = (4 + 2, 4 - 3)m = (6, 1)m

Now we have a displacement d2 = (-5, 0) m

After this, the position is:

(6, 1)m + (-5, 0)m = (6 -5, 1 + 0)m = (1, 1)m

After this, we have the final displacement d3 = (1, 6) m, so the final position will be:

(1, 1)m + (1, 6)m = (1 + 1, 1 + 6)m = (2, 7)m

Below you can see a rough sketch of the path that the student take, where he/she starts at point A.

6 0
3 years ago
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are
igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

4 0
4 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
mojhsa [17]

Answer:

The magnitude of the electric field between the membranes is  1.13 x 10⁶ N/C

Explanation:

Given;

the distance of separation of the parallel plate capacitor, d =  10nm

the charge density, σ = 10⁻⁵C/m²

the magnitude of the electric field between the membranes, is calculated using the formula below;

E = σ / ε₀

Where;

ε₀ is permittivity of free space, = 8.85 x 10⁻¹² C²/Nm²

E is magnitude of the electric field between the membranes

σ is surface charge density

E = (10⁻⁵) / (8.85 x 10⁻¹²)

E = 1.13 x 10⁶ N/C

Therefore, the magnitude of the electric field between the membranes is  1.13 x 10⁶ N/C

6 0
3 years ago
The radius of he Earth orbit around the sun (assumed circular) is 1.50 X 10^8km, with T=365d. What is the radial acceleration of
Karolina [17]

Answer:

ar = 5.86*10^-3 m/s^2

Explanation:

In order to calculate the radial acceleration of the Earth, you first take into account the linear speed of the Earth in its orbit.

You use the following formula:

v=\sqrt{\frac{GM_s}{r}}         (1)

G: Cavendish's constant = 6.67*10^-11 m^3 kg^-1 s^-2

Ms: Sun's mass = 1.98*10^30 kg

r: distance between Sun ad Earth = 1.50*10^8 km = 1.50*10^11 m

Furthermore, you take into account that the radial acceleration is given by:

a_r=\frac{v^2}{r}             (2)

You replace the equation (1) into the equation (2) and replace the values of all parameters:

a_r=\frac{1}{r}\frac{GM_s}{r}=\frac{GM_s}{r^2}\\\\a_r=\frac{(6.67*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}{(1.50*10^{11}m)^2}\\\\a_r=5.86*10^{-3}\frac{m}{s^2}

The radial acceleration of the Earth, towards the sun is 5.86*10^-3 m/s^2

7 0
3 years ago
A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft
pentagon [3]

Answer:

2.877 m/s

Explanation:

According to the laws of conservation of linear momentum,

the momentum of the moving objects before impact is equal to the momentum of the objects after impact (Assuming no external forces were applied)

Let both players are tackled and moving in V velocity

  • M and m - masses of the players
  • U and  u -  velocities of them respectively (both velocities are towards east direction )

momentum before impact = momentum after impact

                          →MU + →mu  = →(M+ m )v

 91.5  * 2.73 + 63.5 * 3.09 =  (91.5 + 63.5) * V

                                       →V = 2.877 m/s (To East)

3 0
3 years ago
Read 2 more answers
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