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alexandr1967 [171]

2 years ago

3

A toy car of mass 6 kg, moving in a straight path, experiences a net force given by the function F = −3t. At time t = 0, the car

has a velocity of 4 m/s in the positive direction and is located +8 m from the origin.
The car will come instantaneously to rest at time t equal to ________.

2 answers:

Mademuasel [1]2 years ago

7 0

Answer:

The car will come instantaneously to rest at time t equal to <u>4 seconds</u>.

Explanation:

"The acceleration varies linearly with time

As here the accelartion varies with respect to time so we can not simply apply Newtons Law"

Using the informations:

Area = F*t/2 = -3t²/2

I = Qf - Qi (Qf = 0, the velocity is zero)

So I = -Qi = -mv

I = -6*4 = -24

-24 = -3t²/2

t = 4s

Nutka1998 [239]2 years ago

7 0

Answer:

t = 4 s

Explanation:

We all know that:

I = F*dt.

And the impulse is the variation of the linear momentum.( I = dP)

Now,

Area = F*t/2 = -3t²/2

I = Pf - Pi (Pf = 0, the velocity is zero)

So,

I = -Pi = -mv

I = -6*4 = -24

-24 = -3t²/2

48 = 3t²

16 = t²

t = 4 s

∴ The car will come instantaneously to rest at time t equal to 4 s

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Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

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where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

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∑τ = Iα ------(2)

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Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

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∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

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2 years ago

Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cl

kozerog [31]

Answer:

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Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.

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An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t

marissa [1.9K]

The energy transferred to the spring is given by:
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Let's convert the data into the SI units:
$k=52.9 N/cm = 5290 N/m$
$x=40.0 cm=0.4 m$

so now we can use these data inside the equation ,to find the energy transferred to the spring:
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2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

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Answer:

(a). The initial velocity is 28.58m/s

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Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

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$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

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$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

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2 years ago