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MissTica
2 years ago
13

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

mark. What is the mass of the meter stick?
Physics
2 answers:
Airida [17]2 years ago
6 0

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

BlackZzzverrR [31]2 years ago
6 0

Answer:

127 g

Explanation:

Let m be the mass of the metre stick.

50 g mass is placed on 90 cm mark is balance at 61.3 cm mark. Take moments about the 61.3 cm mark.

Anticlockwise torque = clock wise torque

m x (61.3 - 50) = 50 x (90 - 61.3)

m x 11.3 = 50 x 28.7

m = 127 g

Thus, the mass of the mere stick is 127 g.

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What visible signs indicate a precipitation reaction when two solutions are mixed?
Illusion [34]

Formation of an insoluble solid

Explanation:

One of the remarkable visible signs that indicates a precipitation reaction when two solutions are mixed is the formation of an insoluble solid. The insoluble solid formed is the precipitate.

  • Precipitates usually forms in single replacement reactions and double replacement or double decomposition reactions.
  • They form when two soluble compounds react. One of the product is an insoluble solid in the solution called the precipitate.
  • The solubility table helps to predict whether precipitates forms in a reaction.

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6 0
2 years ago
A Porsche 944 Turbo has a rated engine power of 217hp . 30% of the power is lost in the drive train, and 70% reaches the wheels.
scZoUnD [109]

Explanation:

(a)  It is given that two-third of weight is over the drive wheels. So, mathematically, w = \frac{2}{3}mg.

Hence, maximum force is expressed as follows.

                F_{max} = \mu_{s} \times w

           m \times a_{max} = \mu_{s} (\frac{2}{3} mg)

Hence, the maximum acceleration is calculated as follows.

             a_{max} = \frac{2}{3} \mu_{s} \times g

                          = \frac{2}{3} \times 1.00 \times 9.8 m/s^{2}

                          = 6.53 m/s^{2}

Hence, the maximum acceleration of the Porsche on a concrete surface where μs = 1 is 6.53 m/s^{2}.

(b)  Since, 30% of the power is lost in the drive train. So, the new power is 70% of P_{max}.

That is,   new power = 0.7 \times P_{max}

Now, the expression for power in terms of force and velocity is as follows.

                      P = F_{max} \nu

              0.7 P_{max} = ma_{max} \nu

Therefore, speed of the Porsche at maximum power output is as follows.

            \nu = 0.7 \times \frac{P_{max}}{ma_{max}}

                      = 0.7 \times \frac{217 hp \times \frac{746 W}{1 hp}}{1500 kg \times 6.53 m/s^{2}}

                      = 11.568 m/s

                      = 11.57 m/s

Therefore, speed of the Porsche at maximum power output is 11.57 m/s.

(c)   The time taken will be calculated as follows.

             time = \frac{\text{velocity}}{\text{acceleration}}

                     = \frac{11.57 m/s}{6.53 m/s^{2}}

                     = 1.77 s

Therefore, the Porsche takes 1.77 sec until it reaches the maximum power output.

6 0
2 years ago
1) What is the highest atomic number element a red dwarf star can produce in its core? a. Carbon b. Oxygen c. Helium d. Iron
bonufazy [111]

Answer:

1) c. Helium

2) Iron

3) False.

Explanation:

1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.

2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.

3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.

3 0
2 years ago
Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p
makvit [3.9K]
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space). 

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
5 0
2 years ago
three neutral metal cans mounted on isulating stands are touching a negatively charge ballon is brought near can a can b is then
Nimfa-mama [501]
Charge on can A is positive. 
Charge on can C is negative.  
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon. 
 As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
7 0
2 years ago
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