Help please, i am lost.

A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest.

Rudik [331]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as

$m_1 = m$

$m_2 = M$

$v_{1i} =v_0$

$v_{2i} = 0$

Initial momentum will be

$P_i = m_iv_{1i}+m_2v_{2i}$

$P_i = mv_0$

After collision

$v_{1f} = v_{2f} = \frac{v_0}{3}$

Final momentum

$P_f = (m_1+m_2)(\frac{v_0}{3})$

$P_f = (m+M)(\frac{v_0}{3})$

From conservation of momentum

$P_f = P_i$

Replacing,

$(m+M)(\frac{v_0}{3})=mv_0$

$(m+M)\frac{1}{3} = m$

$m+M=3m$

$M=3m-m$

$M=2m$

3 0

3 years ago

A thin rod of length d on a frictionless surface is pivoted about one end

motikmotik

The magnitude of the angular momentum of the rod immediately after the collision is given as (pf + pi) * d. (Option A).

<h3>What is Angular Momentum?</h3>

Angular Momentum may be defined or described as the vector quantity of the rotation of a body, which is arrived at by multiplying its moment of inertia by its angular velocity.

The formula for Angular Momentum is given as:

L = mvr; Where

L = Angular Momentum

m = mass

v = velocity; and

r = radius.

Learn more about Angular Momentum at:
brainly.com/question/4126751

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3 0

2 years ago

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

3 years ago

Which are examples of perfectly in elastic collisions

sweet [91]

Typical examples of inelastic collision are between cars, airlines, trains, etc.
For instance, when two trains collide, the kinetic energy of each train is transformed into heat, which explains why, most of the times, there is a fire after a collision. However, the momentum of the two trains that are involved in the collision remains unaffected. So, the trains collide with all their speed, maintaining their momentum, yet their kinetic energy is transformed into heat energy.
Another way to explain a train or a car collision is this: when the two trains or cars collide, they stick together while slowing down. They slow down because their kinetic energy is gradually lost. Still, they collide because they conserve their momentum.

8 0

3 years ago

Consider two wires made of copper. They have the same diameter but one of them has three times the length of the other. Choose t

boyakko [2]

Answer:

a. Both wires have the same resistivity

Explanation:

For the resistance of a wire , following formula holds good .

R = ρ l / S , R is resistance , l is length , S is cross sectional area and ρ is resistivity of the material that the wire is made of. Resistance is dependent on length and cross sectional area but resistivity does not depend upon length or cross sectional area . It only depends upon the type of material.

If we replace copper wire with aluminium wire , then resistivity will change .

Hence , since the wire remains made of copper , resistivity will not change.

6 0

3 years ago