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frozen [14]
3 years ago
14

Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown

made of pure (100%) gold? The density of water is rhow=1.00 grams per cubic centimeter. The density of gold is rhog=19.32 grams per cubic centimeter.
Physics
1 answer:
9966 [12]3 years ago
4 0

Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

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Answer:

2 m = E / c^2      where m is mass of electron

E = h v     where v is the frequency ( nu) of the incident photon

E = h c / y      where y is the incident wavelength (lambda)

2 m = h / (c y)

y = h / (2 m c)      wavelength required

y = 6.62 * 10E-34 / (2 * 9.1 * 10E-31 * 3 * 10E8)  m

y = 3.31 / 27.3 E-11 m

y = 1.21 E -12 m   = .0121 Angstrom units

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3 years ago
What is the role of the neutral wire
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The neutral wire is often confused with ground wire, but in reality, they serve two distinct purposes. Neutral wires carry currents back to power source to better control and regulate voltage. Its overall purpose is to serve as a path to return energy.

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2 years ago
Consider three planets. All have the same mass as Earth, but with different radii (from largest to smallest: Planet 1, 2, 3). Fo
LuckyWell [14K]

Answer:

option C

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given,

mass of the three planet is same

radius of the planets are

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v = \sqrt{\dfrac{2GM}{R}}

G is the gravitational constant

M is the mass of the planet

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Hence planet 3 has the smallest radius so the escape velocity of the third planet will be maximum.

The correct answer is option C

3 0
3 years ago
On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s
dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

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4 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

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substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
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