Two tubes and two other what total for this

Rearrange the equation = KE -1 2 - my? to solve for v. Show your work.

vesna_86 [32]

Answer:

See below

Explanation:

KE = 1/2 m v^2 multiply both sides by 2

2 (KE) = mv^2 divide both sides by m

2(KE) / m = v^2 sqrt both sides

√ [(2KE)/m ] = v

3 0

2 years ago

Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.

balu736 [363]

Answer:

1. False

2. True

3. False

4. True

Explanation:

Conservation of Momentum

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

$p_a=m_a.v_a$

The total momentum of both pucks at the initial condition is

$p_1=m_a.v_a+m_b.v_b$

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

$m_a=m$

$m_b=2m$

We are given

$v_a=6\ m/s\\v_b=2\ m/s$

The total initial momentum is

$p_1=6m+2(2m)=10m$

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

$p_2=3m.v'$

Equating the initial and final momentum

$10m=3m.v'$

Solving for v'

$v'=10/3\ m/s=3.33\ m/s$

1. Compute the initial kinetic energy:

$\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2$

$K_1=18m+4m=22m$

The final kinetic energy is

$\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2$

$K_2=16.63m$

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

$p_1=10m$

$p_2=3m.v'=3m(10/3)=10m$

Proven

5 0

3 years ago

Catherine gently pushes the tip of her finger against the eraser of her pencil and the pencil does not move. Which of the follow

Sonbull [250]

The best illustration that represents the interaction is D

8 0

3 years ago

Read 2 more answers

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to

SpyIntel [72]

This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s

3 0

3 years ago

Students in physics lab times 10 oscillations of the pendulum and get 11.50 seconds how long is a pendulum in centimeters?

allochka39001 [22]

Answer: 36.5

Explanation:

7 0

3 years ago

Two tubes and two other what total for this​

Two tubes and two other what total for this