A student wishes to conduct an investigation on heat transfer that demonstrates convection, Which

A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i

Effectus [21]

Answer:

s = 1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}

• s = $\sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}$

s = $\sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}$

s = 1.7 m

7 0

2 years ago

What is the kinetic energy of a 0.5 kg puppy that is running 1.5m/s

bija089 [108]

Answer: Well the answer is KE = 5.625E-7 i just don't know the units for it...

Hope this helps....... Stay safe and have a Merry Christmas!!!!!!!!!! :D

3 0

2 years ago

Describe J.J Thomson's plum pudding model,and explain how it compares to the modern day atomic theory.

juin [17]

The correct answer to this open question is the following.

The J.J Thomson's plum pudding model establishes that when he researched the atom, he found that electrons are positively charged. Because he observed that electrons were embedded like plums. Electrons are negative. However, compared to the modern-day atomic theory, scientist Ernest Rutherford established that the space between an atom is empty, the nucleus of the atom is positive, and negative electrons have fixed motion paths.

8 0

2 years ago

Read 2 more answers

A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t

scoray [572]

Answer:

a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

let's use subscript 1 for the ladder and 2 for the firefighter

∑ τ = 0

-W₁ x₁ - W₂ x₂ + N₁ y = 0

N₁ = $\frac{W_1 x_1 + W_2 x_2}{y}$ (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

cos 60 = x₁ / d₁

x₁ = d₁ cos 60

x₁ = 7.5 cos 60

x₁ = 3.75 m

for the firefighter d₂ = 4 m

cos 60 = x₂ / d₂

x₂ = d₂ cos 60

x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

sin 60 = y / d₃

y = d₃ sin 60

y = 15 sin 60

y = 13 m

we look for the Normal by substituting in equation 1

N₂ = $\frac{500 \ 3.75 \ + 800 \ 2}{13}$

N₂ = 267.3 N

now let's use the translational equilibrium relations

X axis

F₁ - N₂ = 0

F₁ = N₂

F₁ = 267.3 N

Axis y

N₁ - W₁ -W₂ = 0

N₁ = W₁ + W₂

N₁ = 500 + 800

N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

x₂ = 9 cos 60

x₂ = 4.5 m

we substitute in 1

N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}

N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

fr = F₁ = N₂

fr = 421.15 N

friction force has the expression

fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

μ = fr / N₁

μ = 421.15 / 1300

μ = 0.324

8 0

2 years ago

Caregivers who are observed to be abusive or neglectful have been associated with the __________ type of attachment. A. secure B

Whitepunk [10]

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4 0

2 years ago