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3 years ago

Would you be doing more work going up the stairs twice as fast?

2 answers:

7 0

If you use the stairs at normal speed, you will survive longer, and go up higher.
When going up the stairs twice as fast, you lose energy more quickly, but height would remain the same.

BabaBlast [244]3 years ago

5 0

Yes you would because if you go at a slower rate you don't feel the work that you're doing but if you go fast then you can tell that you put work to it and you can tell but how fast your heart is beating

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Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,

adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

x_cm = 1 /M ∑ $x_{i}$ m_{i}

y_cm = 1 /M ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

x4=?

y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

M = 2.00 + 2.20 + 3.40

M = 7.6 kg

0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

x4 = -6.8 / 6

x4 = -1,133 m

Axis y

0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

y4 = -11/6

y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0

3 years ago

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

4 years ago

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

lisabon 2012 [21]

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

6 0

3 years ago

A radio signal has a frequency of 1.023 x 108 HZ. If the speed of the signal in air is 2.997 x 108m/s, what is the wavelength of

Sladkaya [172]

Answer:

2.93 m (which agrees with answer "C" on the list)

Explanation:

Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:

Wavelength = 2.997 10^8 / 1.023 10^8 = 2.93 m

5 0

3 years ago

If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th

Trava [24]

Answer:

- path differnce = 2.18*10^-6

- 1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

$path\ difference\ = dsin\theta$ (1)