Answer:
I'M SORRY I CAN'T SEE THE PICTURE BUT IT WILL MOST LIKELY BE THE DIFFERENCE BETWEEN THE NUMBERS
Explanation:
for instance if there was someone pushing a desk with a net force of 9 towards the right, the net force would come to a total of 9 since there is no one on the other side pushing the desk. Meaning the desk would go right. I'm sorry if this isn't what you needed D:
Answer:
1.54 m/s²
Explanation:
The free-fall acceleration is calculated as
g = w²r
Where w is the angular velocity of the satellite and r is the radius of the moon.
The angular velocity can be calculated as
Where T is the period, so
T = 110 min = 110 x 60 s = 6600 s
Then,
Finally, the radius of the moon is r = 1.7 x 10⁶ m, so the free-fall acceleration is
Therefore, the answer is 1.54 m/s²
Vertically, the object is in equilibrium, so that the net force in this direction is
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
where <em>n</em> is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is <em>mg</em> = 550 N, so <em>n</em> = 550 N as well.
Horizontally, the net force would be
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where <em>p</em> is the magnitude of the applied force and <em>f</em> is the magnitude of (kinetic) friction opposing <em>p</em>. Now,
<em>f</em> = 0.012<em>n</em> = 0.012 (550 N) = 6.6 N
so that you need to apply a force of <em>p</em> = 6.6 N to keep the object sliding at a steady pace.
Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
Answer:
تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار
Explanation:
