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3 years ago

Brainliest..... plz help!!!!!!!!!

1 answer:

4 0

Tamam hocam nasılsınız hocam proje konularını atabilirmisiniz ne kadar çok seviyorum çok öpüyorum iyi akşamlar yarın görüşürüz iyi geceler Allah rahatlık var bir şey var ya bir insan değilim ki bir şey olmaz sen kaç gibi müsait olur musun lütfen bana yardımcı olabilir misin bir insan bir

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Liquid water has surface tension, has a high specific heat, and functions effectively as a solvent for polar covalent and ionica

MrRissso [65]

Answer:

The property to form polar covalent bond and hydrogen bonds

Explanation:

Liquid water exhibits the property of surface tension, functions as solvent for ionic and polar covalent bonded molecules and have high specific heat.

All these properties can be credited to the property of water to form hydrogen bonds with liquid water molecules where the hydrogen bonding results when hydrogen atom of a water molecule is attracted to a more electro negative oxygen atom of other water molecule in the neighborhood while polar covalent bond formation results in the the water molecule as it is a polar covalent molecule as it contains hydrogen atom and an electronegative oxygen atom which results in the charge distribution in the molecule.

3 0

3 years ago

A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc

Ray Of Light [21]

Answer:

A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.

Explanation:

A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.

If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

vₓ = 16.6 m/s i

In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s

Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:

$v =\sqrt{(16.6m/s)^{2}+ (-17.2m/s)^{2}} = 23.9 m/s$

v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ = $\frac{-17.2}{16.6} =-1.036$

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0

4 years ago

A 62 pound box is on an incline. Determine the minimum force P that would result in the box starting to slide up the incline. (i

Anon25 [30]

Answer:

F > W * sin(α)

Explanation:

The force needed for the box to start sliding up depends on the incline (α).

The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.

These forces can be decomposed on their normal and tangential (to the slide plane) components.

The weight will be split into

Wn = W * cos(α) (in normal direction)

Wt = W * sin(α) (in tangential direction)

The normal reaction will be alligned with the normal axis, and will be equal to -Wn

N = -W* cos(α) (in normal direction)

To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger

F > |-W * sin(α)| (in tangential direction)

8 0

4 years ago

brainliest. Which layer of the atmosphere has the air we breathe? Mesosphere Stratosphere Thermosphere Troposphere

Nat2105 [25]

Answer:

mesosphere

Explanation:

3 0

4 years ago

Read 2 more answers

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago