What is a voltage source

If the hiker starts climbing at an elevation of 350 ft, what will their change in gravitational potential energy be, in joules,

Kitty [74]

Answer:

352,088.37888Joules

Explanation:

Complete question;

A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.

A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)

Chane in potential energy is expressed as;

ΔGPH = mgΔH

m is the mass of the hiker

g is the acceleration due to gravity;

ΔH is the change in height

Given

m = 53kg

g = 9.8m/s²

ΔH = 2574-350 = 2224ft

since 1ft = 0.3048m

2224ft = (2224*0.3048)m = 677.8752m

Required

Gravitational potential energy

Substitute the values into the formula;

ΔGPH = mgΔH

ΔGPH = 53(9.8)(677.8752)

ΔGPH = 352,088.37888Joules

Hence the gravitational potential energy is 352,088.37888Joules

7 0

2 years ago

A student places blocks on a 100cm long see-saw as shown/

Hunter-Best [27]

Answer:

Part 1)

$\tau_1 = 5 \times (0.50) = 2.5 N m$

Part 2)

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Part 3)

$\tau_3 = 1.4 N m$

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

$\tau_1 = 5 \times (0.50) = 2.5 N m$

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Now torque due to weight of the scale is given as

$\tau_3 = 7(0.20)$

$\tau_3 = 1.4 N m$

now torque on left side of scale is given as

$\tau_{left} = \tau_1 + \tau_3$

$\tau_{left} = 2.5 + 1.4 = 3.9 N m$

Torque on right Side is given as

$\tau_{right} = \tau_2 = 4.2 Nm$

Since torque on right side is more so here it will turn and slip over it

8 0

3 years ago

Select the correct answer.

lutik1710 [3]

I’m sorry i haven’t found the answer to this

8 0

3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago

A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0

2 years ago

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