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ss7ja [257]
2 years ago
12

Wonder Woman and Superman fly to an altitude of 1510 km, carrying between them a chest full of jewels that they intend to put in

to orbit around Earth. They want to make this tempting treasure inaccessible to their evil enemies who are trying to gain possession of it, yet keep it available for themselves for future use when they retire and settle down. But perhaps the time to retire is now! They accidentally drop the chest, which leaves their weary hands at rest, and discover that they are no longer capable of catching it as it falls into the Pacific Ocean.
At what speed does the chest impact the surface of the water? Ignore air resistance, although in the real world it would make a world of difference. The radius and mass of Earth are 6370 km and 5.98 × 1024 kg, respectively.
Physics
1 answer:
blagie [28]2 years ago
4 0

Answer:

10061.56 m/s

Explanation:

Gravitation potential of a body in orbit from the center of the earth is given as

Pg = -GM/R

Where G is the gravitational constant 6.67x10^-11 N-m^2kg^-2

M is the mass of the earth = 5.98x10^24 kg

R is the distance from that point to the center of the earth = r + Re

Where r is the distance above earth surface, Re is the earth's radius.

R = 1510 km + 6370 km = 7880 km

Pg = -(6.67x10^-11 x 5.98x10^24)/7880x10^3

Pg = -50617512.69 J/kg

The negative sign means that the gravitational potential is higher away from earth than it is at the earth's surface (it shows convention).

This indicates the kinetic energy per kilogram that the chest of jewel will fall with to earth.

For the jewel chest, the velocity V will be

0.5v^2 = 50617512.69

V^2 = 101235025.4

V = 10061.56 m/s

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If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the induc
mel-nik [20]

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

5 0
2 years ago
A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and
victus00 [196]

Answer:

a)  t=195.948N.m

b)  \phi=13.6 \textdegree

Explanation:

From the question we are told that:

Density \rho=1.225kg/m^2

Velocity of wind v=14m/s

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient \mu=2.05

a)

Generally the equation for Force is mathematically given by

F=\frac{1}{2}\muA\rhov^2

F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2

F=163.29

Therefore Torque

t=F*r*sin\theta

t=163.29*1.2*sin90

t=195.948N.m

b)

Generally the equation for torque due to weight is mathematically given by

t=d*Mg*sin90

Where

d=sin \phi

Therefore

t=sin \phi*Mg*sin90

195.948=833sin \phi

\phi=sin^{-1}\frac{195.948}{833}

\phi=13.6 \textdegree

5 0
2 years ago
A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a
Naily [24]

Answer:

V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

a=\dfrac{P}{M+Rt}-g

We know that

a=\dfrac{dV}{dt}

\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g

\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt

V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT

V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT

This is the velocity of bucket at the instance when it become empty.

6 0
2 years ago
Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir
Svet_ta [14]
Let's ask this question step by step:
 Part A) 
 a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
 ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 Part (c)
 (a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
 (a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
 (a + b) b = 10 + 36
 (a + b) b = 46
 Part (d)
 comp (ba) = (a.b) / lbl
 a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
 comp (ba) = 26 / root (20)
 answer
 2k
 26
 46
 26 / root (20)
3 0
2 years ago
To ensure the greatest nutritional benefits, dietary guidelines suggest which of the following?
oee [108]
The answer would be D. Eating nutrient-dense foods
3 0
2 years ago
Read 2 more answers
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