3 years ago

Formation of hydrogen bonds requires hydrogen atoms and what else?

Physics

1 answer:

ohaa [14]3 years ago

8 0

Answer:

either a Nitrogen atom, Oxygen atom, or a Flourine atom

Explanation:

The atom has to be more electronegative than hydrogen for the bond to form.

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Convert 5.7 miles to km

gayaneshka [121]

5.7 kilometers is equal to 3.5418157957528034 miles

4 0

3 years ago

Read 2 more answers

A wave of water moving up a river, initiated by tidal action and normal resonances within a river estuary, is called a:

IgorC [24]

Answer:

friction solar system

Explanation:

6 0

3 years ago

Ahmad was moving to the south with v= 10 km/hr ,and Mohammed was moving with half of Ahmad's speed to the North .Write the vecto

Alexxx [7]

Answer:

5 km/hr to the South

Explanation:

VM = VA/2

VM = 10km/2hr =5km/hr

8 0

3 years ago

A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o

Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

$\delta = \frac{PL}{AE}$

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

$P = 68*10^3 N$

E = 200GPa

$A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2$

Replacing we have,

$\delta = \frac{PL}{AE}$

$\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}$

$\delta = 0.001932m$

$\delta = 1.93mm$

Therefore the change in length is 1.93mm

7 0

3 years ago

An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?

Ostrovityanka [42]

Answer:

4.02 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s²

Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

T = 2 × 35 × Sine 35 / 10

T = 70 × 0.5736 / 10

T = 7 × 0.5736

T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

4 0

3 years ago