Physics, 21.09.2019 02:00 princeton23124

A4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same constant braking force, and are brought to rest. which object travels the greater distance before stopping? a 4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same constant braking force, and are brought to rest. which object travels the greater distance before stopping? the 1.0-kg object the 4.0-kg object both objects travel the same distance. it is impossible to know without knowing how long each force acts.

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Answer from: ghari112345

Since $\rm s_1 \;and\;s_2$ are same, therefore both the objects covers the same distance.

Given :

Mass of both the object -- $\rm m_1 = 4\;Kg\;and\; m_2 = 1\;Kg$

Initial velocity of both the object -- $\rm u_1 = 2\;m/sec \; and \; u_2 = 4\;m/sec$

Final velocity of both the object -- $\rm v_1 = 2\;m/sec \; and \; v_2 = 4\;m/sec$

Solution :

We know that,

F = ma

$\rm a= \dfrac{F}{m}$ ---- (1)

We know that,

$\rm v^2 = u^2+2as$ --- (2)

Where,

final velocity - v,

initial velocity - u,

a is acceleration,

and s is displacement.

For object 1, from equation (1) and (2),

$\rm s_1 = -\dfrac{u_1^2m_1}{2F}$

$\rm s_1 = -\dfrac{2^2\times 4 }{2F} = -\dfrac{8}{F}$

For object 2, from equation (1) and (2),

$\rm s_2 = -\dfrac{u_2^2m_2}{2F}$

$\rm s_2 = -\dfrac{4^2 \times 1}{2F} = - \dfrac {8}{F}$

Since $\rm s_1 \;and\;s_2$ are same, therefore both the objects covers the same distance.

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Answer from: theo39

Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
F=ma

from which we find an expression for the acceleration:
$a= \frac{F}{m}$ (1)

Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
v_f^2 - v_i^2 = 2 a S

(2)
where

is the final speed of the object
v_i

is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing v_f

(since the final velocity of the two objects is zero), we find
$-v_i^2 = 2 \frac{F}{m}S$
$S=- \frac{v_i^2 m}{2F}$
where we can ignore the negative sign (because the force F will bring another negative sign).

For the first object, we have
$S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m]$
And for the second object we have
$S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m]$

And since the braking force applied to the two objects is the same, the two objects cover the same distance.

Answer from: Quest

you'd use the equation that establishes a relationship between energy and wavelength, which looks like this

e=h⋅cλ, where

λ - the wavelength of the photon; c - the speed of light, usually given as 3⋅108m/s; h - planck's constant, equal to 6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m

Answer from: Quest

Ithink it is a. but i'm not 100 percent sure.

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