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BlackZzzverrR [31]
1 year ago
14

A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li

fted if one pound of fat contains approximately 3500 Calories. If the weight can be lifted once every two seconds, how long would this activity take? 1 Cal=4184 J.
Physics
1 answer:
statuscvo [17]1 year ago
3 0

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required = \frac{(14.644 x 10^6)}{784}

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
2 years ago
somewhere between the earth and the moon is a point where the gravitational attraction of the earth is canceled by the gravitati
mote1985 [20]
<span>It's pretty easy problem once you set it up.

Earth------------P--------------Moon

"P" is where the gravitational forces from both bodies are acting equally on a mass m

Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon

You are correct to use that equation. If the gravitational forces are equal then

GMearth*m/Rep² = Gm*Mmoon/Rpm²

Mearth/Mmoon = Rep² / Rpm²

Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be

Rpm = Rem - Rep

Mearth / Mmoon = Rep² / (Rem - Rep)²

Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²

Everything is done now. The most complicated part now is the algebra, so bear with me as we solve for Rep. I may skip some obvious or too-long-to-type steps.

81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0

We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem

Obviously, point P cannot be 9/8 of the way to the moon because it'll be beyond the moon. Therefore, the logical answer would be 9/10 the way to the moon or B.

Edit: The great thing about this idealized 2-body problem, James, is that it is disguised as a problem where you need to know a lot of values but in reality, a lot of them cancel out once you do the math. Funny thing is, I never saw this problem in physics during Freshman year. I saw it orbital mechanics in my junior year in Aerospace Engineering. </span> sylent_reality · 8 years ago
8 0
2 years ago
During metamorphism, what is the major effect of chemically active fluids?
xz_007 [3.2K]

Answer:

Option b

Explanation:

Metamorphism is the process where the variation of the geological texture resulting from the different arrangement of the minerals or the variation of minerals in protoliths, i.e., pre- existing rocks take place such that there occurs no change in state of the protolith, i.e., it does not melt into magma.

The change takes place as a result of the presence of chemically active fluids, heat and pressure.

There is a reaction between the chemically active fluid and the rock through which it passes and promotes the movement of the dissolved ions of silicate and promotes the growth of the mineral grains.

7 0
2 years ago
What's the momentum of a 3.5-kg boulder rolling down hill at 5 m/s
ICE Princess25 [194]
P = mv 
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Hope this helps!
6 0
2 years ago
A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
Makovka662 [10]

Answer:

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

Explanation:

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v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k

a=5 i+2t j - 6\ t^2k

Therefore the acceleration function a will be

a=5 i+2t j - 6\ t^2k

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k  m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

a=\sqrt{5^2+4^2+24^2}\ m/s^2

a= 24.83 m/s²

The direction of the acceleration a is given as

a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

5 0
2 years ago
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