Fibrous joint functions as a suture to tightly bind bones together so they do not move.
Just add all of them up and there is your answer I just added it but I want u to work it out to..
Answer:
the <em>ratio F1/F2 = 1/2</em>
the <em>ratio a1/a2 = 1</em>
Explanation:
The force that both satellites experience is:
F1 = G M_e m1 / r² and
F2 = G M_e m2 / r²
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- r is the orbital radius
- M_e is the mass of Earth
Therefore,
F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]
F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]
F1/F2 = m1/m2
F1/F2 = 1000/2000
<em>F1/F2 = 1/2</em>
The other force that the two satellites experience is the centripetal force. Therefore,
F1c = m1 v² / r and
F2c = m2 v² / r
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- v is the orbital velocity
- r is the orbital velocity
Thus,
a1 = v² / r ⇒ v² = r a1 and
a2 = v² / r ⇒ v² = r a2
Therefore,
F1c = m1 a1 r / r = m1 a1
F2c = m2 a2 r / r = m2 a2
In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,
F1 = F1c
G M_e m1 / r² = m1 a1
a1 = G M_e / r²
also
a2 = G M_e / r²
Thus,
a1/a2 = [G M_e / r²] / [G M_e / r²]
<em>a1/a2 = 1</em>
Answer:
<em>Magnitude of the Frictional force is 200 N</em>
Explanation:
The frictional force is the force that tries to oppose relative motion between two surfaces that are contacting. The coefficient of static friction is the coefficient of friction of a body that is not moving.
Newton's third law of motion states that action and reaction forces are equal and opposite. So the frictional force felt on the filing cabinet will be equal to the applied force pulling the cabinet.
Frictional force = Force applied
Force applied = 200 N
Therefore, the magnitude of the friction force on the filing cabinet is 200 N
In the experiment of free fall bob released a bag of mass 1 lb
so here we can say that initial speed of the bag is Zero
time taken by the bag to free fall is given as
t = 1.5 s
also the acceleration of free fall is given as
a = 9.8 m/s^2
now we will use kinematics equation here for finding the distance of free fall
so the bag will fall down by total distance of 11.025 m from its initial released position.
