Answer:
B. the force of friction of the road on the tires
Explanation:
Unless the car engine is like jet engine, the main force that accelerates the car forward is the force of friction of the road on the tires, which is ultimately driven by the force of engine on the tires shaft. As the engine, and the shaft are part of the system, their interaction is internal. According to Newton laws of motion, the acceleration needs external force, in this case it's the friction of the road on the tires.
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.
Answer:
my opi
Explanation:
Mass is like the inside of the weight. weight is just home much it weighs
Lear vv
F
B
C
D
A
E
G
greatest pressure ^^
I’m really sorry if I’m wrong
Explanation:
since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.
We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.
range can be calculated by the formula :-

u is the velocity during its take off and
is the angle at which its thrown
Given that
- u = 8m/ s
= 40°
calculating range using the above formula


value of sin 80 = 0. 985



Hence,
