Answer:
6.6 N
Explanation:
Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:


So the resultant has components

So the magnitude of the resultant is

And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.
Answer:
An estimate for the time it will take for a spacecraft to travel from Earth to Mars is approximately 138.8 days
Explanation:
The distance between Earth and the Moon = 684,400 km
The distance between Earth and Mars = 220.58 × 10⁶ km
The distance between Earth and Pluto = 5.2241 × 10⁹ km
The ratio of the distance between Earth and Pluto and the distance between Earth and Mars = (5.2241 × 10⁹ km)/(220.58 × 10⁶ km) ≈ 23.683
It took 2006 to 2015 (9 years) to travel from Earth to Pluto, therefore, it can take approximately (9 years)/(23.683) ≈ 0.38 of a year which is ((9 years)/(23.683)) × 365.2422 ≈ 138.8 days for a spacecraft to travel from Earth to Mars
Answer:
basically they have too much mass in them
Explanation:
They are held tightly together by strong forces of attraction. They are held in fixed positions but they do vibrate. Because the particles don't move, solids have a definite shape and volume, and can't flow. Because the particles are already packed closely together, solids can't easily be compressed.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.