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2 years ago

It is easier to climb up a slanted slope than a vertical slope

2 answers:

8 0

Yes it is easier to clime a gentle slope than a vertical or more steep slope. This is because gravity is always pulling objects down so when you climb upward gravity weighs you down, you have to use effort to move yourself in the opposite direction that gravity is pulling you that is why it is harder to climb up a slope. When you climb a gentle slope you still are using energy but it takes longer to reach the same height on a gentle slope versus a vertical slope. On a vertical slope you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope. When climbing a gentle slope you don't rise as rapidly so gravity doesn't take as much of a toll on you and you have a chance to recover.

Bas_tet [7]2 years ago

7 0

Answer:

Yes, it is easier to climb a slanted slope than a vertical or more steep slope.

Explanation:

On a vertical slope, you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope

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iogann1982 [59]

Different atomic masses

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3 years ago

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When you voice the vowel sound in "hat," you narrow the opening where your throat opens into the cavity of your mouth so that yo

xz_007 [3.2K]

Answer: 0.1 m, 0.0583 m

Explanation:

We are given that:

Frequency for throat= 800 Hz

Frequency for mouth= 1500 Hz

Sound speed= 350 m/s

We have to find the corresponding lengths.

We have

f= $\frac{v}{4L}$

or L= $\frac{v}{4f}$

For the throat= L= $\frac{350}{4*800\\}$ = 0.1 m

For the mouth= L= $\frac{350}{4*1500}$ = 0.0583 m

3 0

3 years ago

Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

5 0

2 years ago

carlis lives 100 m away from his friend home what is his average speed if he reaches his friends home in 50s

Savatey [412]

100m ÷ 50s = 2m/s
Just some simple divison.

8 0

3 years ago

Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface

tatyana61 [14]

Answer:

Option 3: -48 cm

Explanation:

We are given:

refractive index; n = 1.5

radius of curvature; r2 = 24 cm

Formula for the focal length is given as;

1/f = (n - 1) × [(1/r1) - (1/r2)]

As r1 tends to infinity, 1/r1 = 0

Thus,we now have;

1/f = (n - 1) × (-1/r2)

Plugging in the relevant values;

1/f = (1.5 - 1) × (-1/24)

1/f = -0.02083333333

f = -1/0.02083333333

f = -48 cm

3 0

2 years ago

It is easier to climb up a slanted slope than a vertical slope​

It is easier to climb up a slanted slope than a vertical slope