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2 years ago

A 73.9 kg weight-watcher wishes to climb a

Physics

1 answer:

Tresset [83]2 years ago

7 0

The height to which the weight-watcher must climb to work off the equivalent 991 (food) Calories is 0.59 Km

<h3>How to determine the energy. </h3>

1 food calorie = 103 calories

Therefore,

991 food calories = 991 × 103

991 food calories = 102073 calories

Multiply by 4.2 to express in joule (J)

991 food calories = 102073 × 4.2

991 food calories = 428706.6 J

<h3>How to determine the height </h3>

Energy (E) = 428706.6 J
Mass (m) = 73.9 kg
Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

E = mgh

Divide both side by mg

h = E / mg

h = 428706.6 / (73.9 × 9.8)

h = 591.95 m

Divide by 1000 to express in km

h = 591.95 / 1000

h = 0.59 Km

Learn more about energy:

brainly.com/question/10703928

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A cell phone is released from the top with the speed of 10ms what is the speed 3s after?

sergeinik [125]

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

6 0

3 years ago

The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin

alekssr [168]

Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = $200\mu m = 200\times 10^{- 6}\ m$

Gauge Pressure inside, $P_{in} = 25\ mmHg$

Blood Pressure outside, $P_{o} = 10\ mmHg$

Now,

Change in pressure, $\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa$

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

$\Delta P = \frac{4\pi T}{R}$

$T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m$

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

3 0

3 years ago

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

Triss [41]

The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

$F = q(v \times B)$

$F = qvB \: sin \: θ$

$F = 1.6 \times 10 ^{ - 19} \times 5.02 \times 10 ^{6} \times 0.180 \times \: sin \: 60°$

$= 1.25 \times 10 ^{ - 13} \: N$

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

brainly.com/question/23096032

#SPJ4

4 0

1 year ago

a distant galaxy is studied with a radio telescope x ray telescope and optical light telescope. the images form which set of mus

aivan3 [116]

Answer:

The modern instruments or we can say the different levels of telescopes are used to explore and study the distant galaxies. i.e the Hubble telescope is out there providing the data regarding the different properties of the celestial entities which in other case is not visible to the human naked eye.

Explanation:

Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.

As, these telescopes includes the most recent "Hubble telescope", which is out there inside the space to explore the universe and more over the galaxies by subjecting them with x-rays and then provide us with a very rough but valid results to study the distant galaxies.

6 0

3 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$