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balu736 [363]
4 years ago
13

In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v

olume of gold in the phone were equal to the volume of a 10-centimeter cube of gold. The density of gold is 19,300 kg/m3.. Could such a phone be casually passed around a table from hand to hand? What is the weight of the phone? 1 kg of mass is about 2.2 lb.. . 1. Yes; it weighs about 9 lbs. . 2. No; it weighs about 90 lbs. . 3. No; it weighs about 45 lbs. . 4. Yes; it weighs about 4.5 lbs.
Physics
1 answer:
dangina [55]4 years ago
7 0
Volume of gold in the phone = 10 cm^3
                                              = 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
                                        = 0.193 kg 
So
0.193 kg = 0.193 * 2.2 pounds
               = 0.43 pounds
I think there is something wrong with the options given in the question.</span>
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A titanium bicycle frame displaces 0.314 l of water and has a mass of 1.41 kg.part what is the density of the titanium in g/cm3
nlexa [21]

Answer;

= 4.49 g/cm³

Explanation;

Density = Mass / Volume  

mass = 1.41 kg,

volume = 0.314 L

but; 1 kg = 1000 g   and 1 L = 1000 mL

Thus;

Density = 1.41 kg/0.314L  

Density = 4.49 kg/L (1000g/ 1kg)(1L/1000cm^3)  

Density = 4.49 g/cm^3

4 0
3 years ago
What is the magnitude of the force required to stretch a 20 cm-long spring, with a spring constant of 100 N/m, to a length of 21
GalinKa [24]

The correct answer to the question is: 1 N.

EXPLANATION:

As per the question, the spring constant or the force constant of the spring is given as k = 100 N/m.

The original length of the spring L = 20 cm.

The stretched length of the spring L'= 21 cm.

Hence, the change in length will be-

                              ∆L = L' - L

                                    = 21 cm - 20 cm

                                    = 1 cm

                                    = 0.01 m

We are asked to calculate the magnitude of force acting on the spring .

From Hooke's  law, we know that the restoring force that acts on the spring is proportional to the distance .

Mathematically it can be written as -

                F = - kx.

Here, k is the force constant.

         x is the change in length due to compression or elongation.

The negative sign is due to the fact that it is opposite to the applied force.


Hence, the applied force on the spring is calculated as -

            F = kx

               = k × ∆L

               = 100 N/ m × 0.01 m

               = 1 N.

Hence, the force acting on the spring is 1 N.


                                                   

8 0
3 years ago
Please help me !!!!!
riadik2000 [5.3K]
C. 70 inches is the answer
6 0
3 years ago
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
frozen [14]

Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

==> A = 3.142 x 0.002^{2} = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

4 0
3 years ago
A piece of aluminum requires 4,000 J of energy to change
photoshop1234 [79]

Answer:

Mass, m = 105.58 g

Explanation:

We have,

Heat required in aluminium to change the temperature from 68°C to 110°C. It is required to find the mass of aluminium.

Concept used : Specific heat capacity

Solution,

The heat required to raise the temperature is given by :

Q=mc\Delta T

c is specific heat capacity, for Aluminium, c = 0.902 J/g-°C

m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{4000\ J}{0.902\ J/g^{\circ} C\times (110-68)^{\circ} C}}\\\\m=105.58\ g

So, the mass of aluminium is 105.58 grams.

8 0
3 years ago
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