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2 years ago

14

a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis

placement of the bus. (Answer is 4.7x10^2m)

1 answer:

KatRina [158]2 years ago

7 0

The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.

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You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

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Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

$W=-Fd$

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

$F=\mu mg$

where

$\mu$ is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

$W=-\mu mg d$ (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

$-\frac{1}{2}mu^2 = -\mu mg d$

And solving for $\mu$ , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

$\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767$

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2 years ago

What is the approximate pressure of a storage cylinder of recovered r-410a that does not contain any non-condensable impurities

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Answer:

173psig

Explanation:

The storage cylinder of recovered R-410A is mixture of difluoromethane and pentafluoroethane which is used as a refrigerant in air conditioning application. The refrigeration sector has low temperatures for installation. The pressure of cylinder at 80 F will be 173 psig. The pure refrigerants have inside a container have saturation temperature which is equal to ambient temperature.

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7. A local sign company needs to install a new billboard. The signpost is 30 m tall, and the ladder truck is parked 24 m away fr

wolverine [178]

<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

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Jules is conducting an experiment involving friction. He is measuring the temperature of various objects and surfaces after quic

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Answer:

The correct prediction will be:

The temperature of the surface of the ball bearing when rubbed over glass will be the least.

The incorrect prediction:

The tennis ball over the linoleum floor will have no friction, as the temperatures will not change.

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Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

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

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

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

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

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)

aₓ shows displacement in horizontal direction

From (3)

∑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>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

2 years ago