Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.
where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
![(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]](https://tex.z-dn.net/?f=%281000%2A10%29%2B%282000%2A0%29%3D%281000%2B2000%29%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D3.33%5Bm%2Fs%5D)
<span>Electric field repulsive for objects of like charge and attractive for opposite type of charges and for a magnet you can say that like poles repel and unlike attracts so D makes sense</span>
Answer:
theres an decrease in temperature because 252,000 is more than 42,000. so its colder and not as hot as 252,000.
Explanation:
Answer:
28.2 m/s
Explanation:
The range of a projectile launched from the ground is given by:
where
v is the initial speed
g = 9.8 m/s^2 is the acceleration of gravity
is the angle at which the projectile is thrown
In this problem we have
d = 81.1 m is the range
is the angle
Solving for v, we find the speed of the projectile:
Answer:
-0.9045 j
Explanation:
Here K= 270 N/m
distance 1 = 4.60 mm=0.0046 m
when pen is compressed ;
distance 2 = 6.70 mm + distance1 = 0.0067 m+0.0046 m= 0.0113 m
The pen store in itself two types of PE
Which is Initial PEi = 1/2 ×K× [
=0.5×270×0.0046 =0.621 j
and Final PEf = 1/2 × K × 
= 0.5×270×0.0113=1.5255 j
When the pen gets ready - it goes ahead distance d2 and is then pressed back distance d1 for writing.
Total PE when ready = 0.621-1.5255= - 0.9045 j
so, work = -0.9045 j
