During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.
Answer:
M V = m v conservation of momentum (Caps-cannon Small-projectile)
V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s recoil velocity of cannon
KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2 = 40 kg m^2/s^2 = 40 J
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.
Answer:
D
Explanation:
The negative feedback is responsible for maintaining equilibrium (stability) in a system as it lessens effects, which is contrary to positive feedback which increases reaction and moves a system further away from equilibrium (stability), It does so by amplifying the effects of a product or event and occurs when something needs to happen quickly. e.g
- Insulin lowers down blood sugar levels, so when the body detects that it has too much sugar, the pancreas is prompted to release insulin and only stops when balance is achieved; hence, negative feedback.
- When there is a tear on the skin, a chemical is released. This chemical causes platelets in the blood to activate, hence they release a chemical which signals more platelets to activate, until the wound is clotted, positive feedback.
Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
Now, we have to find Jada's speed with respect to the water.
First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
To find the speed, we take the magnitude of this velocity vector we have
|v|= 
|v|= 
= 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.
