Is it 1314000 Or something
Doug needs to change the <em>exercises </em>he is doing. Your body begins to become accustomed to what you do, and if you do the same thing every single time - you will stop making progress, and they'll simply become easy.
However, if you walk into the gym and your body is expecting you to just go over and bench press some, go rock some curls, head over to the squat rack then go home - but instead you start out with some dips, then go do lunges, then attack your triceps instead of biceps, your body will be shocked!
When you change it up, your body will react. The goal is to be comfortable being uncomfortable. Also, one of the most overlooked things in working out is the intensity. If you bring it everything you got and go to the last gut wretching rep, you'll see results.
However, if you just leisurely do things with 4 minute rests in between - you won't see true progress.
Hopefully, this helps! =)
Answer:
W = 20 J
Explanation:
Given that,
The mass of a loaded sack, m = 1000 g = 1 kg
It falls down from the floor of a lorry 200 cm high, h = 2 m
We need to find the work done by the gravity. The work done by an object under the action of gravity is given by :
W = mgh
Substitute all the values,
W = 1 × 10 × 2
= 20 J
Hence, the required work done by gravity is equal to 20 J.
Answer:
F = 20.4 i ^
Explanation:
This exercise can be solved using the ratio of momentum and amount of movement.
I = F t = Dp
Since force and amount of movement are vector quantities, each axis must be worked separately.
X axis
Let's look for speed
cos 45 = vₓ / v
vₓ = v cos 45
vₓ = 8 cos 45
vₓ = 5,657 m / s
We write the moment
Before the crash p₀ = m vₓ
After the shock
= -m vₓ
The variation of the moment Δp = mvₓ - (-mvₓ) = 2 m vₓ
The impulse on the x axis Fₓ t = Δp
Fₓ = 2 m vₓ / t
Fx = 2 0.450 5.657 / 0.250
Fx = 20.4 N
We perform the same calculation on the y axis
sin 45 = vy / v
vy = v sin 45
vy = 8 sin 45
vy = 5,657 m / s
We calculate the initial momentum po = m 
Final moment
= m
Variations moment Δp = m
- m
= 0
Force in the Y-axis
= 0
Therefore the total force is
F = fx i ^ + Fyj ^
F = Fx i ^
F = 20.4 i ^
Answer:
Velocity=14[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy.
In the attached image we can see the illustration of the ball falling from the height of 20 meters, at this time the potential energy will have the following value.
![Ep=m*g*h\\where:\\m=3[kg]\\h=20[m]\\](https://tex.z-dn.net/?f=Ep%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%3D3%5Bkg%5D%5C%5Ch%3D20%5Bm%5D%5C%5C)
![Ep=3*9.81*20\\Ep=588.6[J]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A20%5C%5CEp%3D588.6%5BJ%5D)
When the ball passes through half of the distance (10m) its potential energy will have decreased by half as shown below.
![Ep=3*9.81*10\\Ep=294.3[m]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A10%5C%5CEp%3D294.3%5Bm%5D)
If we know that potential energy is transformed into kinetic energy, we can find the value of speed.
![Ek=\frac{1}{2} *m*v^{2} \\therefore\\v=\sqrt{\frac{Ek*2}{m} } \\v=\sqrt{\frac{294.3*2}{3} } \\\\v=14[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Ctherefore%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BEk%2A2%7D%7Bm%7D%20%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B294.3%2A2%7D%7B3%7D%20%7D%20%5C%5C%5C%5Cv%3D14%5Bm%2Fs%5D)