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3 years ago

Beating a carpet with a carpet beater.

1 answer:

8 0

I would say a short person with muscles considering they are closer to the ground, but they may not be able to build up as much force in such a short time compared to the tall person.

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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

How many orbitals are allowed in a subshell if the angular momentum quantum number for electrons in that subshell is 3?

svp [43]

Answer:

7 orbitals are allowed in a sub shell if the angular momentum quantum number for electrons in that sub shell is 3.

Explanation:

We that different values of m for a given l provide the total number of ways in which a given s, p,d and f sub shells in presence of magnetic field can be arranged in space along x, y ,z- axis or total number of orbitals into which a given subshell can be divided.

Range for given l lies between -l to +l .

The possible values of m are -3 , -2 , -1 , 0 , 1 ,2 , 3 .

Total number of orbitals are 7.

4 0

3 years ago

Sports managers have the potential to earn more than a million dollars per year. True false

andriy [413]

Answer:

true

Explanation:

hope this helped!

4 0

3 years ago

A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\

zloy xaker [14]

Answer:

t = 1.77 s

Explanation:

The equation of a traveling wave is

y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

v = λ f

Where the frequency and period are related

f = 1 / T

we substitute

v = λ / T

let's develop the initial equation

y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

2π /λ = 1.65

λ = 2π / 1.65

λ = 3.81 m

2π / T = 4.64

T = 2π / 4.64

T = 1.35 s

we seek the speed of the wave

v = 3.81 / 1.35

v = 2.82 m / s

Since this speed is constant, we use the uniformly moving ratios

v = d / t

t = d / v

t = 5 / 2.82

t = 1.77 s

3 0

3 years ago

Which of the following is a correct formula of speed

satela [25.4K]

Answer:s= d/t

Explanation:

7 0

3 years ago