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3 years ago

What do these compounds have in common H2O cl2 NH3

Physics

1 answer:

nasty-shy [4]3 years ago

7 0

Answer:

All the compounds are covalent compounds . This means that they are formed by the sharing of pair of electrons.

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Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only

Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

$v=u+at$

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

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

$\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2$

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

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

$\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2$

$\Rightarrow 16\times t^2=576$

$\Rightarrow t^2=\frac{576}{16}$

$\Rightarrow t^2=36$

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

8 0

2 years ago

Lillle is running. She increases her initial speed of 30 km/h to 40 km/h so she

Alex777 [14]

Answer

$200km {h}^{ - 2}$

Explanation

$Acceleration = \frac{final \: \: \: velocity - initial \: \: velocity }{time} \\ = \frac{(40 - 30)km {h}^{ - 1} }{0.05h} \\ = \frac{10}{0.05} \\ = 200km {h}^{ - 2}$

Hope this helps you.

Let me know if you have any other questions :-):-)

6 0

2 years ago

Read 2 more answers

A worker drops a wrench from the top of a tower 80m tall. What is the velocity when the

OlgaM077 [116]

Answer:

39.6 m/s

Explanation:

Taking down to be positive:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

Δy = 80 m

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (80 m)

v = 39.6 m/s

3 0

3 years ago

Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c

mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

$V = \frac{k*q}{r}$

where k = $\frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2$

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

$V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)$

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

$d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}$

Replacing by the values in (1) we have:

$V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q) = 0 V$

which is equal to the option d).

6 0

3 years ago

Un objeto que es empujado sobre una superficie plana, tiene una masa de 80 kg y dicha fuerza tiene un valor de 300 N. Calcule la

Gennadij [26K]

Responder:

A. Ff = 300 N N = 784,8 N

Explicación:

Dado

Masa del cuerpo = 80 kg

Fuerza de movimiento Fm = 300N

Dado que el cuerpo no está acelerando, la fuerza de fricción (Ff) es igual a la fuerza de movimiento que actúa sobre el cuerpo, ya que la fuerza de fricción es una fuerza de oposición, es decir, Fm = Ff

Dado que Fm = 300N, Ff = 300N

La reacción normal que actúa en el cuerpo es igual al peso.

N = W = mg

g es la aceleración debida a la gravedad

g = 9,8 m / s

N = mg

N = 80 (9,81)

N = 784,8N

Por tanto, la fuerza normal que actúa sobre el cuerpo es 784,8 N

3 0

2 years ago