The equation 
(option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:
Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:
When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:
Therefore, the equation represents the horizontal momentum (option 3).
Learn more about linear momentum here:
I hope it helps you!
Answer:
53.64 m/s
Explanation:
Applying,
a = (v-u)/t............. Equation 1
Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
make u the subject of the equation
u = v-at............. Equation 2
From the question,
Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)
Substitute these values into equation 2
u = 0-(-5.364×10)
u = 0+53.64
u = 53.64 m/s
It would be B
Explanation:
Because if you're not measuring in inches you want to go the next one down other than inches which would be millimeters!(: hope this helps.
The answer is c: <span>1960 J
</span>Potential Energy :
<span>PE = m x g x h = 40*9.8*5=1960
</span>
Answer:
1.84 kJ (kilojoules)
Explanation:
A specific heat of 0.46 J/g Cº means that it takes 0.46 Joules of energy to raise the temperature of 1 gram of iron by 1 Cº.
If we want to heat 50 g of iron from 20° C to 100° C, we can make the following calculation:
Heat = (specific heat)*(mass)*(temp change)
Heat = (0.46 J/g Cº)*(50g)*(100° C - 20° C)
[Note how the units cancel to yield just Joules]
Heat = 1840 Joules, or 1.84 kJ
[Note that the number is positive: Energy is added to the system. If we used cold iron to cool 50g of 100° C water, the temperature change would be (Final - Initial) or (20° C - 100° C). The number is -1.84 kJ: the negative means heat was removed from the system (the iron).
