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Montano1993 [528]
3 years ago
6

Which would fall with greater acceleration in a vacuum a leaf or a stone?

Physics
2 answers:
Sergeu [11.5K]3 years ago
3 0
Because its a vacuum, there's no air resistance, they will fall at same time
Applying gravity acceleration rule g=9.8m/s which is taken as 10m/s sometimes.
Mkey [24]3 years ago
3 0

Answer:

They will both have the same acceleration

Explanation:

Because its a vacuum, which ordinarily means it will be closed (unless otherwise stated), there will be no opposing forces (like air interference) to each of there acceleration due to gravity - each of them will have an acceleration due to gravity of 9.8 m/s2 (or approximately 10 m/s2) and will hit the "ground" at the same time if "dropped" at the same time.

Here, the masses will not be considered because there are no obstructions/opposing forces.

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The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
A mass of 3 slugs (this is the English unit of mass, a pound is a force) is attached to a vertical spring with a spring constant
motikmotik

Answer:

equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin  √{α² - ω²} t )

Explanation:

Given data

mass = 3 slugs = 3 * 32.14 = 96.52 lbs

constant k = 9 lbs/ft

Beta = 6lbs * s/ft

mass is pulled =  1 ft below

to find out

equation of motion for the mass

solution

we know that The mass is pulled 1 ft below so

we will apply here differential equation of free motion i.e

dx²/dt² + 2 α dx/dt + ω² x =0     ........................1

here 2 α  =  Beta / mass

so 2 α  = 6 / 96.52

α  = 0.031

α²  = 0.000961         ...............2

and

ω² = k/mass

ω² = 9 /96.52

ω² = 0.093     ..................3

we can say that from equation 2 and 3 that α² - ω²  = -0.092239

this is less than zero

so differential equation is

x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin  √{α² - ω²} t )

equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin  √{α² - ω²} t )

3 0
3 years ago
Which of the following is an example in which kinetic energy is converted to potential energy?
Ivan

Answer:

C. A spring is stretched.

Explanation:

Kinetic energy is the energy of motion, potential energy is stored energy. The motion of stretching the spring is kinetic, the energy it has before it's released is potential.

3 0
4 years ago
The angle of reflection does not equal the angle of incidence?<br> True or False
leonid [27]

False, it does. Hope this helped!

3 0
3 years ago
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.15Hz , and the acc
Minchanka [31]

Answer:

See answer below

Explanation:

Hi there,

I'm a bit late with this response, but I attached the work below.

Good luck with physics, have fun!

thanks,

4 0
3 years ago
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