Answer: turns to gas
Explanation:
when a liquid gets heated up the chemicals start to boil/evaporate turning the liquid into a gas
Answer:
v₁ = 3.5 m/s
v₂ = 6.4 m/s
Explanation:
We have the following data:
m₁ = mass of trailing car = 400 kg
m₂ = mass of leading car = 400 kg
u₁ = initial speed of trailing car = 6.4 m/s
u₂ = initial speed of leading car = 3.5 m/s
v₁ = final speed of trailing car = ?
v₂ = final speed of leading car = ?
The final speed of the leading car is given by the following formula:
![v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B2m_1%7D%7Bm_1%2Bm_2%7Du_1-%5Cfrac%7Bm_1-m_2%7D%7Bm_1%2Bm_2%7Du_2%5C%5C%5C%5Cv_2%3D%5Cfrac%7B%282%29%28400%5C%20kg%29%7D%7B400%5C%20kg%2B400%5C%20kg%7D%286.4%5C%20m%2Fs%29-%5Cfrac%7B400%5C%20kg-400%5C%20kg%7D%7B400%5C%20kg%20%2B%20400%5C%20kg%7D%283.5%5C%20m%2Fs%29)
<u>v₂ = 6.4 m/s</u>
The final speed of the leading car is given by the following formula:
![v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7Bm_1-m_2%7D%7Bm_1%2Bm_2%7Du_1%2B%5Cfrac%7B2m_2%7D%7Bm_1%2Bm_2%7Du_2%5C%5C%5C%5Cv_1%3D%5Cfrac%7B400%5C%20kg-400%5C%20kg%7D%7B400%5C%20kg%20%2B%20400%5C%20kg%7D%286.4%5C%20m%2Fs%29%2B%5Cfrac%7B%282%29%28400%5C%20kg%29%7D%7B400%5C%20kg%2B400%5C%20kg%7D%283.5%5C%20m%2Fs%29)
<u>v₁ = 3.5 m/s</u>
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
Answer:
360 N
Explanation:
m = 30kg u = 2 m/s a = -2m/s/s
Since the object has an initial velocity of 2 m/s and acceleration of -2 m/s/s
the object will come to rest in 1 second but the force applied in that one second can be calculated by:
F = ma
F = 30 * -2
F = -60 N (the negative sign tells us that the force is acting downwards)
Now, calculating the force applied on the box due to gravity
letting g = -10m/s/s
F = ma
F = 30 * -10
F = -300 N (the negative sign tells us that the force is acting downwards)
Now, calculating the total downward force:
-300 + (-60) = -360 N
<em></em>
<em>Hence, a downward force of 360 N is being applied on the box and since the box did not disconnect from the rope, the rope applied the same amount of force in the opposite direction</em>
Therefore tension on the force = <u>360 N</u>
Utilize the formula: ![V _{f} = V _{i} + a\Delta t](https://tex.z-dn.net/?f=V%20_%7Bf%7D%20%3D%20V%20_%7Bi%7D%20%2B%20a%5CDelta%20t)
= Final Velocity (86 m/s)
= Initial Velocity (0 m/s)
a = acceleration (m/s²)
t = Time (100 seconds)
As a result,
86 m/s = 0 + (a)(100 seconds)
Using algebra, divide 86 m/s by 100 seconds:
86 m/s = 100a
a = 0.86 m/s²
Rounded to one decimal place: 0.9 m/s²
Let me know if you have any questions!