What is the molecular geometry or of the molecular molecule

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

3 years ago

The convection zone is the outer most layer of the suns blank hot gases rise and then blank hear the loops form by singing cool

lora16 [44]

Answer:

Explanation:

The sun is made up of 6 parts. Namely:

The core
The radiation zone
The convection zone
The photosphere
The chromosphere and
The corona

The convection area is just above the radiation zone. As materials from the suns core are heated, they rise above the radiation zone towards the EDGE of the convection area then sinks back again into the radiative zone for more heat.

The radiative zone is 12.6 million Fahrenheit hot and is just above the core.

The core of the son is not solid but plasma whose motion is like gas. Its temperature stands at 48 million Fahrenheit

Cheers

5 0

3 years ago

HELP ASAP

Sladkaya [172]

Answer:
I think the answer is
C) iron nails are attracted towards all materials

8 0

2 years ago

Explain the uses of Electrophorus <br>

Maslowich

Answer:

Explanation:

Comment

Interesting creature.

These creatures are the only survivors from an era long past, where nothing else has survived. They are eels capable of discharging a large electrical current. A pack of them (they can hunt in groups) can kill just about anything.

Tribes use and tame them to guard the tribe's tame Electrophorus. They are bread and maintained in schools. While the charge is huge, once discharged, these creatures take a very long time to recharge, so they are not entirely invincible.

8 0

2 years ago

) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th

Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

Now, we will substitute 0 for both $z_{1}$ and $z_{2}$ , 0 for w, 334.9 kJ/kg for $h_{1}$ , 2726.5 kJ/kg for $h_{2}$ , 5 m/s for $V_{1}$ and 220 m/s for $V_{2}$ .

Putting the given values into the above formula as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

$1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0$

q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

6 0

3 years ago