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3 years ago

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You're driving at 50 mph and then speed up to 70 mph. how much more "force" does the engine in your car have to produce to maint

ain the new speed (as compared to the force at 50 mph)? explain.

1 answer:

Fittoniya [83]3 years ago

6 0

You drove 20 more mph than what you did before. The more force is 20 because you drove 20 more mph.

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A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

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3 years ago

Kenitic Energy (KE)<br> 1.doubled

astra-53 [7]

Answer:

mass

Explanation:

This energy of motion is what we call kinetic energy. ... In fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses.

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2 years ago

The position-time equation for a cheetah chasing an antelope is:

pochemuha

Answer:

x = 1.6 + 1.7 t^2 omitting signs

a) at t = 0 x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0 V = 0

c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)

d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)

4 0

2 years ago

A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear

Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

$f_o = f_s\frac {(v+vs)}{v}$ and $f_o = f_s\frac {(v-vs)}{v}$ when moving towards observer

With $f_o$ of 76 then taking speed in air as 343 m/s we have

$76 = f_s\times\frac {(343-vs)}{343}$

$f_s=\frac {343\times 76}{343-v_s}$

Similarly, with $f_o$ of 65 we have

$65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}$

Now

$f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}$

v_s=27.76 m/s

Substituting the above into any of the first two equations then we obtain

$f_s=70.07 Hz$

4 0

3 years ago

A gas sample is confined within a chamber that has a movable piston. A small load is placed on the piston; and the system is all

timurjin [86]

Explanation:

It is given that,

Total weight of the piston, W = F = 70 N

Area of the piston, $a=5\times 10^{-4}\ m^2$

Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{70\ N}{5\times 10^{-4}\ m^2}$

$P=1.4\times 10^5\ Pa$

We know that the atmospheric pressure is given by :

$P_o=1.013\times 10^5\ Pa$

So, the pressure is given by :

$p=P+P_o$

$p=1.4\times 10^5+1.013\times 10^5$

$p=2.41\times 10^5\ Pa$

Hence, this is the required solution.

7 0

3 years ago