The thickness of a $1 bill is 0.11 mm. If you have a stack of $1 bills 450 m tall, how much money do you have?

Answers

Answer 1
Answer:

Answer:

4090909

Explanation:

Thickness of one bill = 0.11 mm

Total thickness = 450 m

No of $1 bills = total thickness / thickness of one bill

No of $1 bills = 450 / 0.11 × 10^-3

= 4090909

Answer 2
Answer:

Answer:

You will have 4.5 million dollar

Explanation:

The thickness of a $1 bill is 0.11 mm

So we have

              1 $ = 0.1 mm

              0.1 mm = 1 $

              0.0001 m =  1 $

              1 m = 10000 $

           450 m = 450 x 10000 = 4500000 $

So you will have 4.5 million dollar


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A series circuit contains a 20-Ω resistor, a 200-mH inductor, a 10-μF capacitor, and an ac power source. At what frequency should the power source drive the circuit in order to have maximum power transferred from the driving source?

Answers

Answer:

f = 113 Hz

Explanation:

In order to have maximum power transferred from the driving source, as the RMS voltage doesn´t depends on frequency, the current I must be maximum.

This condition is met when the circuit behaves a purely resistive, as the impedance is at a minimum.

Such condition is known as resonance, and it satisfies the following equation:

XL = XC ⇒ ω₀ * L = 1 / ω₀*C, where ω₀, is the angular frequency at resonance.

Solving for ω₀, we have:

ω₀ = 1/√LC = 1/√200*10⁻3 H* 10⁻6F = 707 rad/sec

As we need to find the frequency (in cycles/sec), we need to convert from angular frequency to frequency, as follows:

ω₀ = 2*π*f₀ ⇒ f₀ = ω₀ / 2*π = 707 rad/sec / 2*π rad = 113 Hz

Hercules X-1 is a pulsating X-ray source. The X-rays from this source sometimes completely disappear for 6 hours every 1.7 days because the neutron star has a 1.7-day orbital period around its companion star, and it is eclipsed for ____ hours once every orbital period.

Answers

Answer:

06 Hours

Explanation:

As per the details given in the question it self, the neutron star X-1 is revolving around its companion star. The orbital period is 1.7 years which means it will complete the revolution in 1.7 years. During the movement in the orbit we will be able to detect the x-rays except for the time when it goes behind the companion star and eclipsed by it as seen from Earth.

Since the x-rays disappear completely for around 6 hours. This clearly means that eclipse period is 06 hours.

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct

Answers

Answer:

4.25m/s

E. None of the option is correct

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

Hot air enters a rectangular duct (20cm wide, 25cm high, and 5m long) at 100 kPa and 60 degrees C at an average velocity of 5 m/s. While air flows the duct, it gets cool down (loses energy) so that air leave the duct at 54 degrees C. Determine the rate of heat loss from the air under steady condition

Answers

Answer:

1.57 kW

Explanation:

The rate of heat loss is given by:

q = Gm * Cp * (tfin - ti)

Where

q: rate of heat loss

Gm: mass flow

Cp: specific heat at constant pressure

The Cp of air is:

Cp = 1 kJ/(kg*K)

The mass flow is the volumetric flow divided by the specific volume

Gm = Gv / v

The volumetric flow is the air speed multiplied by the cruss section of the duct.

Gv = s * h * w (I name speed s because I have already used v)

The specific volume is obtained from the gas state equation:

p * v = R * T

60 C is 333 K

The gas constant for air is 287 J/(kg*K)

Then:

v = (R * T)/p

v = (287 * 333) / 100000 = 0.955 m^3/kg

Then, the mass flow is

Gm = s * h * w / v

And rthe heat loss is of:

q = s * h * w * Cp * (tfin - ti) / v

q = 5 * 0.25 * 0.2 * 1 * (54 - 60) / 0.955 = -1.57 kW (negative because it is a loss)

What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *1 point
45 N
450 N
450 kg
10N

Answers

Answer:

450N

Explanation:

weight= m*g

weight=45*10

weight=450N

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °

Answers

Answer:\theta =45.73^(\circ)

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

T\cos\theta =mg-----1

T\sin \theta =(mv^2)/(r)------2

Divide 2 & 1

tan\theta =(v^2)/(rg)

tan\theta =(4.65^2)/(2.15* 9.8)

tan\theta =1.026

\theta =45.73^(\circ)