Which area on the line has both some kinetic energy and some gravitational energy?

Wall-E the robot is resting when he randomly explodes into two pieces that fly off in opposite directions. His head has a mass o

Brut [27]

Answer:

<em>The body flies off to the left at 9.1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2\qquad\qquad[1]$

Wall-E robot is initially at rest, its two parts together. His head has a mass of m1=0.75 kg and his body has a mass of m2=6.2 kg. Both parts have initial speeds of zero v1=v2=0.

After the explosion, his head flies off to the right at v1'=75 m/s. We are required to find the speed of his body v2'. Solving [1] for v2':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}$

Substituting values:

$\displaystyle v'_2=\frac{0.75*0+6.2*0-0.75*75}{6.2}$

$\displaystyle v'_2=-9.1 \ m/s$

The body flies off to the left at 9.1 m/s

3 0

3 years ago

What will happen to the fringe spacing if the wavelength of the light is decreased?.

REY [17]

Answer:

The formula is the form -

m λ = d sin Θ

As the wavelength λ is decreased sin Θ will also decrease.

One can see from the derivation that as the wavelength, being considered,

is decreased the dispersion will also decrease.

3 0

3 years ago

a student throws a coin vertically downward frok the top of a building. the coin leaves the throwers hand with a speed of 15.0m/

Elina [12.6K]

Answer:

Final speed after 2 seconds = 34.6 m/s

Explanation:

Given:

Initial speed of coin (u) = 15 m/s

Time taken = 2 seconds

Find:

Final speed after 2 seconds

Computation:

Gravitational acceleration of earth = 9.8 m/s²

Using first equation of motion;

v = u + at

or

v = u + gt

where,

v = final velocity

u = initial velocity

g = Gravitational acceleration

t = time taken

v = 15 + 9.8(2)

v = 15 + 19.6

Final speed after 2 seconds = 34.6 m/s

8 0

3 years ago

A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the lad

pantera1 [17]

Answer: 45.3°

Explanation:

Given,

Length of ladder = l

Weight of ladder = w

Coefficient of friction = μs = 0.495

Smallest angle the ladder makes = θ

If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,

N1 = mg and

N2 = μmg

Moment at a point A in the clockwise direction is

N2 Lsinθ - mg.(L/2).cosθ = 0

μmgLsinθ - mg.(L/2).cosθ = 0

μmgLsinθ = mg.(L/2).cosθ

μsinθ = cosθ/2

sin θ / cos θ = 1 / 2μ

Tan θ = 1 / 2μ

Substituting the value of μ = 0.495, we have

Tan θ = 1 / 2 * 0.495

Tan θ = 1 / 0.99

Tan θ = 1.01

θ = tan^-1(1.01)

θ = 45.3°

3 0

3 years ago

a boy pulls 23 kg box with a 100 N force at 39 above a horizontal surface if the coefficient of kinetic friciton between the box

tatuchka [14]

Answer:

Total work done is 2606.08 J.

Explanation:

Given :

Mass of box , m = 23 kg .

Force applied , F = 100 N .

Angle from horizon , $\theta=39^o$ .

Coefficient of kinetic friction , $\mu=0.16$ .

Distance travelled by box , d = 34 m .

Now ,

Total work done = work done by boy + work done by friction.

$W=Fdcos\theta+f_sdcos\theta=Fd-\mu(mg) \\\\W=100\times cos39^o\times 34-0.16\times 23 \times 9.8 \\\\W=77.71\times 34-36.06=2606.08\ J.$

Hence , this is the required solution.

6 0

3 years ago