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3 years ago

13

Once you're in heaven, do you get stuck wearing the clothes you were buried in for eternity?

2 answers:

Zolol [24]3 years ago

5 0

Answer:

It depends on the faith

Explanation:

Some you show up naked, some you show up clothed, and others you are reincarnated as something

vaieri [72.5K]3 years ago

3 0

Answer:

Well, I think it actually depends from faith to faith... 'do u go to heaven or hell' is a much more heavier question to ask...

Explanation:

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You can increase the rate solute dissolves in solvent by______

V125BC [204]

Crushing particles of solute. The more surface area of the solute exposed to the solvent the faster the solute will dissolve and react with the solvent

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4 years ago

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A 2.80 kg mass is dropped from a

inn [45]

Answer:

PE = 82.32J

Explanation:

PE = m*g*h

PE = 2.80kg*9.8m/s²*3m

PE = 82.32J

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3 years ago

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Assume your car reaches a speed of 21.7 m/s at a steady rate for 5.05 s after the light turns green. (a) What distance have you

boyakko [2]

Answer:

The distance and average speed are 54.79 m and 10.85 m.

Explanation:

Given that,

Speed = 21.7 m/s

Time = 5.05 s

(a). We need to calculate the distance

Firstly we will find the acceleration

Using equation of motion

$v = u+at$

$a = \dfrac{v-u}{t}$

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

$a = \dfrac{21.7-0}{5.05}$

$a = 4.297 m/s^2$

Now, using equation of motion again

For distance,

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

$s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2$

$s=54.79\ m$

The distance is 54.79 m.

(b). We need to calculate the average speed during this time

$v_{avg}=\dfrac{D}{T}$

Where, D = total distance

T = time

Put the value into the formula

$v_{avg}=\dfrac{54.79}{5.05}$

$v_{avg}=10.85\ m/s$

Hence, The distance and average speed are 54.79 m and 10.85 m.

3 0

3 years ago

1. A group of students were trying to find the greatest

deff fn [24]

Answer:

D th

Explanation:

D B. The force applied to the ball is a balanced force.

8 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago