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Helen [10]
3 years ago
10

he adventurous robot M.A.N.D.I. is orbiting Saturn’s moon Dione. She wants to cause an impact with themoon to kick up some of th

e surface dust so that she can do a spectral analysis of it. She tosses a steel ball bearing inthe opposite direction of her orbital velocity with just the right impulse to make the ball come to a dead stop. That’sthe back story. The actual problem starts here: The ball, starting with zero velocity, falls straight down to the surfaceof the moon. If the moon has a radius of5.61×103m and a mass of1.10×1021kg, and if the ball bearing starts at analtitude of2.73×103m above the surface of the moon, how fast will it be going when it hits the surface? Note thatthe gravitational constantG= 6.67408×10−11N m2/kg2.
Physics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

v = 2.928 10³ m / s

Explanation:

For this exercise we use Newton's second law where the force is the gravitational pull force

         F = ma

         a = F / m

Acceleration is

        a = dv / dt

        a = dv / dr dr / dt

        a = dv / dr v

        v dv = a dr

We substitute

       v dv = a dr

       ∫ v dv = 1 / m G m M ∫ 1 / r² dr

We integrate

       ½ v² = G M (-1 / r)

We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon

       v² = 2G M (- 1 / R +2.73 10³+ 1 / R)

         

We calculate

       v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61  - 10⁻³ /(5.61 + 2.73))

       v² = 14.6828 10⁷ (0.1783 -0.1199)

       v = √8.5748 10⁶

       v = 2.928 10³ m / s

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Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De
My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0
3 years ago
A 40.0 kg box is placed on a 2.00 m tall shelf. If the box falls off the shelf, what is its kinetic energy as it strikes the gro
RUDIKE [14]
M = 40 Kg , g=9.8 m/s² , h = 2 m

PE = m g h

PE = (40) (9.8) (2)

PE = 784 J

KE = PE

½m v² = m g h

½ v² = g h

½ v² = (9.8) (2)

½ v² = 19.6

v² = 19.6×2

v² = 39.2

V = √39.2

V = 6.26 m/s

KE = ½mv²

KE = ½(40) (6.26)²

KE =783.8 J
6 0
2 years ago
How much force must you apply to give the go cart an acceleration of 0.9m/s^2
qaws [65]
It's dependent on the mass. You can fimd the force needed using the formula F = ma. Where F is force, m is mass of the cart and a is the acceleration (0.9m/s^2). The heavier it is the more force you are going to need. Remember unit of force is N
5 0
3 years ago
if an object's density is decreasing but its mass stays constant, what must be true of the object's volume?
astra-53 [7]
<h2>Answer</h2>

The volume will be <u>increased</u>

<h2>Explanation</h2>

Look at the formula

Density = Mass/Volume

If the mass increase the density will be increased because of their direct relationship. But in the case of constant mass. The volume increase the density will decrease because there is an inverse relationship exists between them. In the inverse relationship, the two objects perform differently. for example, if you are going to inflate the balloon, its volume will increase but density decrease

8 0
3 years ago
The density for potassium is 0.856 g/cm3. What would be the mass of a 40 cm3 piece of potassium?
SCORPION-xisa [38]

Answer:

38.52g

Explanation:

3 0
3 years ago
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