Answer:
I = -12070 N.s
Explanation:
Given that:
mass of 1.0 metric tons = 1000 kg
initial speed = 72mi/hr = 32.19 m/s
final speed = 45mi/hr = 20.12 m/s
Impulse = change in momentum
I = mv-mu
I = m(v-u)
I = 10³(20.12 - 32.19)
I = 10³(-12.07)
I = -12070 N.s
The negative sign signifies that the impulse way is to the opposite direction.
This question involves the concepts of Newton's Law of Gravitation and mass.
The force on Procyon A from Procyon B will be "equal" to the force on Procyon B from Procyon A, which has a value of "3.75 x 10²⁶ N".
Applying Newton's Law of Gravitation, we can find the force on Procyon A from Procyon B, which is equal to the force on Procyon B from Procyon A:

where,
F = force = ?
G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = mass of Procyon A = 3 x 10³⁰ kg
m₂ = mass of Procyon B = (2.5)(3 x 10³⁰ kg) = 7.5 x 10³⁰ kg
r = distance between them = 2 x 10¹² m
Therefore,

<u>F = 3.75 x 10²⁶ N</u>
Learn more about Newton's Law of Gravitation here:
brainly.com/question/17931361?referrer=searchResults
Answer:
a)Yes will deform plastically
b) Will NOT experience necking
Explanation:
Given:
- Applied Force F = 850 lb
- Diameter of wire D = 0.15 in
- Yield Strength Y=45,000 psi
- Ultimate Tensile strength U = 55,000 psi
Find:
a) Whether there will be plastic deformation
b) Whether there will be necking.
Solution:
Assuming a constant Force F, the stress in the wire will be:
stress = F / Area
Area = pi*D^2 / 4
Area = pi*0.15^2 / 4 = 0.0176715 in^2
stress = 850 / 0.0176715
stress = 48,100.16 psi
Yield Strength < Applied stress > Ultimate Tensile strength
45,000 < 48,100 < 55,000
Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.
The angular momentum is defined as,

Acording to this text we know for conservation of angular momentum that

Where
is initial momentum
is the final momentum
How there is a difference between the stick mass and the bug mass, we define that
Mass of the bug= m
Mass of the stick=10m
At the point 0 we have that,

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity
vector from the point of reference (O), and ve is the velocity
At the end with the collition we have

Substituting




Applying conservative energy equation we have


Replacing the values and solving

Substituting
l=\frac{13}{0.54(9.8)}
