Answer:
Magnitude of impulse, |J| = 4 kg-m/s
Explanation:
It is given that,
Mass of cart 1, 
Mass of cart 2,
Initial speed of cart 1,
Initial speed of cart 2,
(stationary)
The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

V = 1 m/s
The magnitude of the impulse exerted by one cart on the other is given by:


J = -4 kg-m/s
or
|J| = 4 kg-m/s
So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.
We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

Where
A = Cross-sectional Area
Y = Young's modulus
= Coefficient of linear expansion for steel
= Temperature Raise
Our values are given as,




Replacing we have,


Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N
Answer:
A) x _electron = 0.66 10² m
, B) x _Eart = 1.13 10² m
, C) d_sphere = 1.37 10⁻² mm
Explanation:
A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of
Let's use proportions rule
x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5
x _electron = 0.66 10⁵ mm = 0.66 10² m
B) the radii of the Earth and the sun are
= 6.37 10⁶ m
tex]R_{Sum}[/tex] = 6.96 10⁸ m
Distance = 1.5 10¹¹ m
x_Earth = 1.5 10¹¹ / 6.96 10⁸ 1.5
x _Eart = 1.13 10² m
C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions
d_sphere = 1.5 / 6.96 10⁸ 6.37 10⁶
d_sphere = 1.37 10⁻² mm