Answer:
B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s
Explanation:
The momentum of an object is given by the product between its mass (m) and its velocity (v):
Let's apply this formula to calculate the initial momentum and final momentum of the ball:
- initial momentum:
- Final momentum:
So, the correct answer is
B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s
Condensation does not effect the temperature of the area, condensation is the product the temperature difference between an object and the area it is in
Answer:
V = (5.8cm/s)i, (4.7cm/s)j
Explanation:
Given :
r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^
To obtain the average velocity (V)
V = (r2 - r1) / (t2 - t1)
To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above
r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j
r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j
r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j
r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j
r2 = (16.1cm)i + (9.4cm)j
V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0
V = 11.6i / 2 ; 9.4j / 2
V = (5.8cm/s)i, (4.7cm/s)j
Explanation:
Given:
Acceleration of car = 10 m/s
Distance travelled in 5 sec = 150 m
To find:
Distance travelled in the next 5 seconds
Concept:
There are 2 ways to app this kind of questions .
Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.
Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.
Calculation:
v² = u² + 2as
=> ( u + at)² = u² + 2as
=> u² + 2uat + a²t² = u² + 2as
=> 2uat + (at)² = 2as
=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)
=> 100u = 3000 - 2500
=> 100u = 500
=> u = 5 m/s
Distance travelled in 10 seconds :
s = ut + ½at²
=> s = (5 × 10) + ½(10)(10)²
=> s = 50 + 500
=> s = 550 m
Distance travelled in the 2nd half will be :
d = 550 - 150
=> d = 400 m
So final answer is :
There are a lot of same examples that you may have worked before, where the mass on a spring uses a classics when it comes to mechanics. So in this system, always put in your mind that there is an enormous quantum standard that one can use in the equation. It should be 2.10x10 raise to a negative sixth. J.
