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Sliva [168]
3 years ago
8

A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs

all day. Given that there are 86,400 seconds / day, how much energy, in Joules, would you burn in climbing stairs all day?
Physics
1 answer:
defon3 years ago
8 0

Answer:

2940.1 joules  would you burn in climbing stairs all day.

Explanation:

Work = W = F\times d

going up stairs  would be against force of gravity

W = mgh

where h is the  height

the question is not complete because we need speed or distance

h =  v \times t

so assuming 1 step per second

h = 86,400 steps  \times 7inchs/step \times 0.0254 m/inch

h = 15362 m

so from this    

W = 800 N \times 15362

   = 12289600 J  

that means YOU  need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal  

Kcal =  \frac{W}{(J/Kcal)}

       = \frac{12289600}{4180}

       = 2940.1 Kcal

if you ran faster you would use more energy  2 steps per second would mean 5880 Kcal.

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The milky way is often considered to be an intermediately wound, barred spiral, which would be type ________ according to hubble
lora16 [44]
SBb type spiral. Its a type B because its not too tightly wound but its still too tight to be a type C
4 0
3 years ago
A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found
kati45 [8]

Answer:

1)  \rm q_2 is<u> positive.</u>

<u></u>

2) \rm q_2=4.56\times 10^{-10}\ C.

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., \rm q_2 is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

  • Mass of the balloon, m = 0.00275 kg.
  • Charge on the balloon, \rm q_1 = -3.50\times 10^{-8}\ C.
  • Distance between the rod and the balloon, d = 0.0640 m.
  • Acceleration due to gravity, \rm g = 9.81\ m/s^2.

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, \rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.

The magnitude of the electrostatic force on the balloon due to the rod is given by

\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.

\rm \dfrac{1}{4\pi \epsilon_o} is the Coulomb's constant.

For the elecric force and the weight to be balanced,

\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.

3 0
3 years ago
A 50n box is lifted 2 meters in 3 seconds.
anygoal [31]

F=G=50 N

L=F*d=50*2=100 J

is the work

P=L/t=100/3=33.33 W

4 0
3 years ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
3 years ago
IP The x and y components of a vector r⃗ are rx = 16 m and ry = -8.5 m , respectively
fiasKO [112]

as it is given that

r_x = 16 m

r_y = -8.5 m

now we will have

\vec r = 16 \hat i - 8.5 \hat j

now the magnitude of this vector is given as

|r| = \sqrt{16^2 + 8.5^2}

|r| = 18 m

now to find the direction we can use

tan\theta = \frac{r_y}{r_x}

tan\theta = \frac{-8.5}{16}

\theta = tan^{-1}(-0.53)

\theta = - 28^0

4 0
3 years ago
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