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balu736 [363]
3 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]3 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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Balance the equation-<br> Al+Mn02 ———-&gt; Mn + Al2O3
joja [24]

Answer:

                                                                                                         

Explanation:

3 0
3 years ago
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
3 years ago
A football is thrown vertically with an initial velocity of 33m/s. Calculate the velocity of the ball when it’s height increased
tatyana61 [14]

The final velocity becomes 31.48 m/s

<u>Explanation:</u>

Given:

Initial velocity, u = 33 m/s

Height, h = 5m

Final velocity, v = ?

According to Newton's law:

v² - u² = 2gh

where,

g is the acceleration due to gravity and

g = 9.8 m/s²

On substituting the values we get:

v^2 - (33)^2 = 2 X -9.8 X 5\\\\v^2 - 1089 = -98\\\\v^2 = 991\\\\v = 31.48 m/s

Therefore, the final velocity becomes 31.48 m/s

5 0
3 years ago
Please help on this one?
bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0
3 years ago
From a hot air balloon that is at rest at a certain height, a projectile is launched horizontally at 30m / s, how fast will it h
Yuki888 [10]

Answer:

A. 50 m/s

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 4 s

Find: v

v = at + v₀

v = (10 m/s²) (4 s) + 0 m/s

v = 40 m/s

In the x direction, the velocity is constant at 30 m/s.

The overall speed is:

v² = (30 m/s)² + (40 m/s)²

v = 50 m/s

3 0
3 years ago
Read 2 more answers
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