How much does a crate weigh (force) if it is 0.25 meters wide and 1.5 meters long and it
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Answer:
The velocity of the police car relative to earth is
Given:
time for flash generation of the inter galactic police car, t = 1.2 s
time between flashes as measured from earth, t' = 2.2 s
Solution:
Utilising Einstein's equation for time dilation to calculate the velocity of the police car, the equation is given by:
(1)
where, c = speed of light in vacuum =
re arranging eqn (1) for velocity, v:
(2)
Now, from eqn (2)
The rod has a mass of m = 4.4 kg and a length of L = 0.4 m.
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s
Answer:
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)
Explanation:
For this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where q is the charge inside the surface.
In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product
∫ E dA = q_{int}/ ε₀
The area of a sphere is
A = 4π r²
- The electric field for a distance r < R_int
The charge inside is zero, so the electric field
E = 0 r <R_in t
- The field for a radio inside the shell
Let's use the concept of density
ρ = Q / V
q = ρ (4/3 π r³)
dq = ρ 4π r² dr
We substitute in the Gaussian equation
E ∫ dA = ρ 4π r² dr / ε₀
E 4π r² = ρ 4π/ε₀ r³ / 3
E = ρ / 3ε₀ r
We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E
E- 0 = ρ / 3ε₀ (r –R_int)
Density is
ρ = q / 4/3 π (R_out³ - R_int³)
Where R <r
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)