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3 years ago

7

a ball is dropped and falls with an acceleration of 9.8m/s^2 downward. it hits the ground with a velocity of 49m/s downward. how

long did it take the ball to fall to the ground

1 answer:

yaroslaw [1]3 years ago

8 0

The answer below...........

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A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet

puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0

3 years ago

For this trajectory, what would the vertical component of acceleration for the module be at time tm=t0−σ=325s? Recall that accel

Karo-lina-s [1.5K]

I think its half of pie so you divid that by 3.14 .

4 0

3 years ago

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What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes

mr Goodwill [35]

The current is defined as the amount of charge Q that passes through a given point of a wire in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Since I=500 A and the time interval is
$\Delta t=4.0 min=240 s$
the charge is
$Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C$

One electron has a charge of $q=1.6 \cdot 10^{-19}C$ , therefore the number of electrons that pass a point in the wire during 4 minutes is
$N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23}$ electrons

3 0

3 years ago

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover

givi [52]

Answer:

Acceleration due to gravity, $a=4.61\times 10^{-14}\ m/s^2$

Explanation:

It is given that,

Mass of Pluto, $m=1.4\times 10^{22}\ kg$

Distance between Neptune and Pluto, $r=4.5\times 10^{12}\ m$

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}$

$a=4.61\times 10^{-14}\ m/s^2$

So, the acceleration due to gravity at Neptune due to Pluto is $4.61\times 10^{-14}\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

The acceleration due to gravity on Mars is less than that on Earth. On Mars, a person will weigh than on Earth.

pantera1 [17]

On mars people would way less.
An example of this is that if I weighed 700 pounds (I don't by the way) I would then weigh 500 pounds or less.

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3 years ago

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