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3 years ago

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A car is driving directly north on the freeway at a speed of 91 km/h and a truck is leaving the freeway driving 56 km/h in a dir

ection that is 28° west of north. What is the velocity of the truck relative to the car? Please report the answer with a positive angle.

1 answer:

Sladkaya [172]3 years ago

3 0

Answer:

v = 49.69 km/hr

Explanation:

velcoity of truck with respect to car is given as = velocity of truck - velocity of car

velcoity of truck with respect to car is given as = (-54 sin(28)i +54cos(28)j - 91km/h

= -24.35i - 43.32j

magnitude of velocity is

$v = \sqrt{ 24.35^2+43.32^2}$

v = 49.69 km/hr

the direction is

$tan^{-}[\frac{ 43.32}{24.35}]$

= 60.65 degree towrd south of west

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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2 years ago

How do the causes of surface and deepwater currents differ

zepelin [54]

The temperature is colder, and the water pressure is higher.

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3 years ago

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

JulijaS [17]

Total amount of energy would remain constant according to law of conservation of energy. i.e., 50 Joules

In short, Your Answer would be Option C) <span>50 Joules because as energy converts from one form to another, it cannot be created or destroyed during the conversion.
</span>
Hope this helps!

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3 years ago

Read 2 more answers