Answer:
I = 20 i ^ N s
Explanation:
For this problem let's use the Impulse equation
I = Δp = m 
- v₀
The impulse and the velocity are vector quantities, let's calculate on each axis, let's decompose the velocity
cos 60 = vₓ / v
vₓ = v cos 60
sin60 = 
/ v
= v sin60
vₓ = 10 cos 60
= 10 sin60
vₓ = 5.0 m / s
= 8.66 m / s
Let's calculate the impulse on each axis
X axis
Iₓ = m 
- m vₓ₀
How the ball bounces
= - vₓ₀ = vₓ
Iₓ = 2 m vₓ
Iₓ = 2 2 5
Iₓ = 20 N s
Y axis
= m 
- m vyo
On the axis and the ball does not change direction so
= vyo
= 0
The total momentum is
I = Iₓ i ^ + j ^
I = 20 i ^ N s
Answer:
At the dimer-dimer interface there might be acting non-covalent forces (van der waals, Hidrogene bridges, hydrophobic forces)
At the monomer-monomer interface there might be covalent forces acting (disulfide bridges).
Explanation:
On the SDS-PAGE application works by disrupting non-covalent bonds in the proteins, and so denaturing them. Therefore, the disulfide bridges won´t be disrupted, so the monomers will remain bounded.
Answer:
a) a = 91.4 m / s², b) t = 0.175 s, c)
Explanation:
a) This is a kinematics exercise
v² = vox ² + 2a (x-xo)
a = v² - 0/2 (x-0)
let's calculate
a = 16² / 2 1.4
a = 91.4 m / s²
b) the shooting time
v = vox + a t
t = v-vox / a
t = 16 / 91.4
t = 0.175 s
c) let's use Newton's second law
F = ma
F = 7.9 91.4
F = 733 N
Answer:
The motion of the ball relative to the ground is stationary
The motion of the ball relative to the wagon is backwards
Explanation:
To describe the motion of the ball relative to the ground, we note that
Assuming the ball is perfectly round and rotate freely, then we have
Force on the ball due to motion of the wagon = 0 N,
Then by the law of motion, an object will remain at rest when no force is applied to it
Therefore, apart from rotation of the ball, it will remain no displacement relative to the ground.
The motion of the ball relative to the wagon
Relative to the wagon, the ball appears to be moving in the opposite direction to the wagon, that is backwards.
