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3 years ago

How long will it take an object to hit the ground if it is dropped from a hight of 176.4 meters

Physics

1 answer:

Oliga [24]3 years ago

5 0

There's a short handy formula for that.

If the object is just dropped and not tossed, and it's not affected by air resistance on the way down, then the distance it falls in T seconds is

D = (1/2) (gravity) (T²)

For this problem . . .

176.4 m = (1/2) (9.8 m/s²) (T²)

Divide each side by (4.9 m/s²) :

T² = (176.4 m) / (4.9 m/s²)

T² = (36 s²)

Take the square root of each side:

<em>T = 6 seconds</em>

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3 years ago

Un alumno menciona que al abrir la ventana de su casa sintió cómo el frío ingresaba a su cuerpo. Menciona cuál es la verdadera r

stepan [7]

Answer:

My believe the answer is

A.) or B.)

Explanation:

Here is why I think A is the answer.

If we use the process of elimination, it would look like this,

a) Porque el aire tiene una temperatura menor que la de su cuerpo; por eso se propaga más rápido.

<em>This makes sense because we all know in winter the weather is very cold and freezing.</em>

b) Porque la temperatura de su cuerpo, siente el aire frio que entra por la ventana.

<em>I feel like this answer is the question, but it could also be an answer, sorry, I'm a little uncertain.</em>

c) Porque el calor de su cuerpo se propaga al medio ambiente, al ser la temperatura del niño mayor que la del aire exterior.

<em>This answer has nothing to do with the question, plus it is very false, our body heat is not enough to overcome the very cold temperature from outside.</em>

d) Porque la temperatura del aire es igual a la temperatura del cuerpo.

<em>This is false because again our body heat is not even compared to the freezing cold temperatures from the winter.</em>

<em />

<h2>Well, have a nice rest of the day!</h2><h3>ba baiii!</h3>

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3 years ago

The next four questions refer to the situation below.

Anna11 [10]

Answer:

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}, t_{out} = $\frac{D}{v_s +v_r}$

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

$v_{sg 1} = v_{sr} + v_{rg}$

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

$v_{sg1}$ = D / $t_{out}$

D = v_{sg1} t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

$v_{sg 2} = v_{sr} - v_{rg}$

$v_{sg 2}$ = D / $t_{in}$

D = v_{sg 2} t_{in}

with the distance is the same we can equalize

$v_{sg1} t_{out} = v_{sg2} t_{in}$

t_{out} = t_{in}

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}

This must be the answer since the return time is known. If you want to delete this time

t_{in}= D / $v_{sg2}$

we substitute

t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

t_{out} = $\frac{D}{v_s +v_r}$

7 0

2 years ago

I'm not really sure how to go about creating the equation, can anyone help me?

AlexFokin [52]

The displacement vector (SI units) is
$\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}$

The speed is a scalar quantity. Its magnitude is
$v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}$

Answer: At√(t⁴ - 12t³ + 36t² + 1)

3 0

3 years ago

How long will it take an object to hit the ground if it is dropped from a hight of 176.4 meters ​

How long will it take an object to hit the ground if it is dropped from a hight of 176.4 meters