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2 years ago

How many kisses is 50 hectokisses? What would you be if you earn a megabuck year?

Physics

2 answers:

never [62]2 years ago

6 0

Answer:

you would be a millionaire.

Explanation:

Given that,

Kisses = 50 hecto kisses

We know that,

We represent of hecto from $10^{2}$

$1\ hecto = 10^{2}$

If you earn a mega buck year.

We represent of mega from $10^{6}$

$1\ megabuck = 10^{6}\ buck$

So, The kisses is

$50\ hecto kisses=50\times10^{2}\ kisses$

Hence, you would be a millionaire.

Keith_Richards [23]2 years ago

4 0

Answer:

1.) 6 hectokisses 50 times

2.) Millionaire

Explanation:

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A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte

Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

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2 years ago

A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

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2 years ago

The speed of sound in air changes with the temperature. When the temperature T is 32 degrees Fahrenheit, the speed S of sound is

dezoksy [38]

Answer and Explanation:

A. We have temperature t = 32

Speed of sound, s = 1087.5

As t increases by 1⁰f speed increases by 1.2

So that

S = 1088.6

T= 33⁰f

We have 2 equations

1087.5 = k(32) + c

1088.6 = k(33) + c

Subtracting both equations

(33-32)k = 1088.6-1087.5

K = 1.1

b.). S = kT + c

1087.5 = 32(1.1) + c

Such that

C = 1052.3

Therefore

S = 1.1(t) + 1052.3

C.). S = 1.1t + 1052.3

We make t subject of the formula

T = s/1.1 - 1052.3/1.1

T = 0.90(s) - 956.3

D. This means that We have temperature to rise by 0.90 whenever speed is increased

8 0

2 years ago

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull

Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E) 351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

work done = force x distance x cos θ

since the rope is parallel to the ramp the angle between the rope and

the ramp θ will be 0

work done = 72 x 5.2 x cos 0

work done by the rope = 374.4 J

(B) calculate the work done by gravity

the work done by gravity = weight of carton x distance x cos θ

The weight of the carton = force exerted by the mass of the carton = m x g

the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

work done by gravity = 128 x 5.2 x cos 120

work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

work done by the normal force = force x distance x cos θ

the angle between the normal force and the ramp is 90 degrees

work done by the normal force = Fn x distance x cos θ

work done by the normal force = Fn x 5.2 x cos 90

work done by the normal force = Fn x 5.2 x 0

work done by the normal force = 0 J

(D) what is the net work done ?

The net work done is the addition of the work done by the rope, gravitational force and the normal force

net work done = 374.4 - 332.8 + 0 = 41.6 J

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

work done by the rope= force x distance x cos θ
the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

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2 years ago

A runner has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?

antiseptic1488 [7]

I could be wrong, but I'm pretty sure it's 144kg.

8 0

2 years ago

Read 2 more answers