Answer:
= 3289.8 m / s
Explanation:
This exercise can be solved using the definition of momentum
I = ∫ F dt
Let's replace and calculate
I = ∫ (at - bt²) dt
We integrate
I = a t² / 2 - b t³ / 3
We evaluate between the lower limits I=0 for t = 0 s and higher I=I for t = 2.74 ms
I = a (2,74² / 2- 0) - b (2,74³ / 3 -0)
I = a 3,754 - b 6,857
We substitute the values of a and b
I = 1500 3,754 - 20 6,857
I = 5,631 - 137.14
I = 5493.9 N s
Now let's use the relationship between momentum and momentum
I = Δp = m - m v₀o
I = m - 0
= I / m
= 5493.9 /1.67
= 3289.8 m / s
Energy is calculated as power*time, so give the wattage of 1200 W (equivalent to 1200 Joules/second) and time of 30 seconds, multiplying these gives 36000 J or 36 kJ of electrical energy.
If electrical charge or current is needed: Power = voltage * current, so given the power of 1200 watts and voltage of 120 V, current is 1200 W / 120 V = 10 Amperes. Charge is calculated by multiplying 10 A*30 s = 300 C.
The force of friction is 19.1 N
Explanation:
According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:
where
is the net force
m is the mass
a is the acceleration
The bag is moving at constant speed, so its acceleration is zero:
Therefore the net force is zero as well:
Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:
where
is the horizontal component of the applied force, with
F = 22.5 N
is the force of friction
And solving for , we find
Learn more about friction:
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Answer:
0.22m/s
Explanation:
The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.
momentum of an object = mass* velocity
Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;
Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;
Solve for x.
