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DochEvi [55]
2 years ago
10

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what

does she have to do with the mass of the smiley face to keep the mobile in perfect balance? Note that she may have to change masses of other objects to keep the entire structure balanced
Physics
1 answer:
charle [14.2K]2 years ago
4 0

Answer:

She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

Explanation:

mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.

Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

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How many significant figures are in 0.0069
posledela
Two significant figures, the 6 and the 9
7 0
2 years ago
A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0
2 years ago
If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it
Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

      \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

Explanation:

From the question we are told that

   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is a =  27.0 \  m/s

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

        a = \frac{u *  \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}

Here \frac{\Delta m}{\Delta  t }  is the rate at which gas is being ejected with respect to time

Substituting values

      27 = \frac{2000 *  \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}

=>   170100 -27* \frac{\Delta m}{\Delta t} = 2000 *  \frac{\Delta m}{\Delta t}

=>   170100  = 2027 *  \frac{\Delta m}{\Delta t}

=>   \frac{\Delta m}{\Delta t}   = \frac{170100}{2027}

=>   \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

     

3 0
2 years ago
Alonzo sprints for 500 meters along a straight track during a race. After crossing the finish line, he to walks back along the t
Nata [24]
450 meters, straight ahead is the answer.
6 0
2 years ago
Read 2 more answers
A 50 g tennis ball travels at a velocity of 15 m/s, hits a basketball with a mass of 600 g that is stationary on a frictionless
statuscvo [17]

Answer:

1.75 m/s

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(50 g) (15 m/s) + (600 g) (0 m/s) = (50 g) (-6 m/s) + (600 g) v

v = 1.75 m/s

8 0
2 years ago
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