Answer:
Answer:196 Joules
Explanation:
Hello
Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem
the work is the product of a force applied to a body and the displacement of the body in the direction of this force
assuming that the force goes in the same direction of the displacement, that is upwards
W=F*D (work, force,displacement)
the force necessary to move the object will be
![F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules](https://tex.z-dn.net/?f=F%3Dmg%28mass%20%2Agravity%29%5C%5CF%3D4kgm%2A9.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%5C%5C%20F%3D39.2%20Newtons%5C%5Creplace%5C%5C%5C%5CW%3D39.2%20N%2A5m%5C%5CW%3D196%5C%20Joules)
Answer:196 Joules
I hope it helps
Explanation:
When a truck travels in equal distances in equal intervals of time then we say that the body has got a uniform velocity. In the above example a truck is traveling at 5 miles in all the positions at A, B, and C and all in the intervals of 5 minutes each.
<span>Answer: Burrhus Frederic Skinner's Operant Conditioning.
</span><span>B.F. Skinner believed that to understand behavior, in the best way, is to look at the root causes or reasons of an action and its outcomes.
</span>
Skinner proposes the Law of Effect-Reinforcement. Here,he differentiated the positively reinforced behavior or the strengthened behavior, the negatively reinforced behavior (removal of the unpleasant experience), and weakened behavior because of punishment.
<span>
In positive reinforcement, behavior is strengthened through providing an outcome, an effect that an individual finds rewarding. Negative reinforcement also strengthens behavior because the unpleasant experience was removed. Punishment on the other hand is an opposite to reinforcement. Instead of increasing the response, it eliminates it or weakens it.
</span>
Answer:
The speed of the large cart after collision is 0.301 m/s.
Explanation:
Given that,
Mass of the cart, ![m_1 = 300\ g = 0.3\ kg](https://tex.z-dn.net/?f=m_1%20%3D%20300%5C%20g%20%3D%200.3%5C%20kg)
Initial speed of the cart, ![u_1=1.2\ m/s](https://tex.z-dn.net/?f=u_1%3D1.2%5C%20m%2Fs)
Mass of the larger cart, ![m_2 = 2\ kg](https://tex.z-dn.net/?f=m_2%20%3D%202%5C%20kg)
Initial speed of the larger cart, ![u_2=0](https://tex.z-dn.net/?f=u_2%3D0)
After the collision,
Final speed of the smaller cart,
(as its recolis)
To find,
The speed of the large cart after collision.
Solution,
Let
is the speed of the large cart after collision. It can be calculated using conservation of momentum as :
![m_1u_1+m_2u_2=m_1v_1+m_2v_2](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2)
![m_1u_1+m_2u_2-m_1v_1=m_2v_2](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2-m_1v_1%3Dm_2v_2)
![v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}](https://tex.z-dn.net/?f=v_2%3D%5Cdfrac%7Bm_1u_1%2Bm_2u_2-m_1v_1%7D%7Bm_2%7D)
![v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}](https://tex.z-dn.net/?f=v_2%3D%5Cdfrac%7B0.3%5Ctimes%201.2%2B0-0.3%5Ctimes%20%28-0.81%29%7D%7B2%7D)
![v_2=0.301\ m/s](https://tex.z-dn.net/?f=v_2%3D0.301%5C%20m%2Fs)
So, the speed of the large cart after collision is 0.301 m/s.
Answer:
Rice
Explanation:
Because I can't control eating lots of rice