Diego rivera's mural for the lobby of the rca building was destroyed because

What is the maximum m2 value that the system can be stationary at?

Ne4ueva [31]

Answer: 37.5 kg in 3 s.f.

Explanation:

7 0

3 years ago

Que carro ese que está en la foto

Taya2010 [7]

Travis Scott!3&;8284$28&:!;&29395

7 0

3 years ago

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Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

DiKsa [7]

Am not sure
Have a nice day

6 0

3 years ago

(III) An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable breaks when the e

Iteru [2.4K]

Answer: 12Mg/h

Explanation:

Let the spring is compressed by a distance x,before the lift stops,then

Mg(h+x)= 1/2 kx^2 ............... 1

Kx - Mg = M ( 5g ) ............ 2

Make x the subject in equation 2

Kx = 5Mg + Mg

Kx = 6Mg

x = 6Mg/k ............ 3

Put equation 3 into 1

Mg ( h + x ) = 1/2 kx^2

Mgh + Mgx = 1/2kx^2

Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2

Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2

h =18Mg/k - 6Mg/h

K = 12Mg/h

8 0

3 years ago

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago