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liberstina [14]
3 years ago
7

The accompanying map shows the route traveled by a school bus.

Physics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

C

Explanation:

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Earths gravitational pull just got 3 times stronger what happens to your weight?
djverab [1.8K]
You get 3 times heavier
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3 years ago
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How does jet stream form
barxatty [35]
Even though the wind "tries" to flow from high pressure to low pressure, the turning of the Earth causes the air flow to turn to the right (in the Northern Hemisphere), so the jet stream flows around the air masses, rather than directly from one to the other.
7 0
3 years ago
Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70
IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2

here \omega _{o}^{2}= k/m = (6.30)/(0.135) = 46.67 N/m kg

F / mA = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

\omega ^{2}= \omega _{o}^{2}\pm F/m

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π =  8.53 /6.28  or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

8 0
3 years ago
Assume the space shuttle's main engines produce 764,576 newtons of thrust, and the shuttle has a mass of 78,018 kg. Why does the
Nady [450]

Weight of anything = (mass) x (gravity in the place where the thing is)

Weight of anything on Earth = (mass) x (9.81 m/s²)

Weight of the shuttle = (78,018 kg) x (9.81 m/s²)

Weight of the shuttle, on Earth = 765,357 Newtons

Thrust of main engines = 764,576 Newtons

Are you starting to see the problem yet ?

The weight of the whole thing standing on the launch pad is 751 Newtons more than the maximum thrust of the main engines, and the engines can't lift it !  Even with all throttles wide open, the main engines alone would need about 175 <em>more</em> pounds of thrust to budge that load off the ground.  Even with the pedal to the metal, with flame and smoke belching out and covering the whole launch complex, the shuttle would just sit there and never leave the pad.

Well, no.  That's not exactly what would happen.  As the fuel in the main monster fuel tank is burned, the weight decreases.  So it would actually happen like this:  After the man announced "Zero !  We have ignition !  All engine running !", the ship would just sit there on the pad ... at first.  It would go nowhere and not even wiggle, <em>UNTIL</em> the first 175 pounds of fuel got burned without accomplishing anything.  The ship would then be 175 pounds lighter.  At that point, the weight would be exactly equal to the thrust of the main engines, and the vertical forces on the ship would be balanced.  Then, as MORE fuel continued to be wasted and the weight continued to decrease, the main engines could just begin to lift the ship off the pad.

So the correct answer is <em>choice-D</em> .  It tells the whole story, quicker than I can tell it.

4 0
3 years ago
A solid, uniform disk of radius 0.250 m and mass 45.2 kg rolls down a ramp of length 5.40 m that makes an angle of 17.0° with th
serious [3.7K]

Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

Length of the ramp, l = 5.4 m

Angle made by the ramp with horizontal, \theta=17^{\circ}

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

h is the height of the ramp

sin\theta=\dfrac{h}{5.4}

h=sin(17)\times 5.4=1.57\ m

v is the speed of the disk's center

I is the moment of inertia of the disk,

I=\dfrac{1}{2}mr^2

\omega=\dfrac{v}{r}

mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{1}{2}mr^2\times (\dfrac{v}{r})^2

gh=\dfrac{1}{2}v^2+\dfrac{1}{4}v^2

gh=\dfrac{3}{4}v^2

9.8\times 1.57=\dfrac{3}{4}v^2

v = 4.52 m/s

(b) At the bottom of the ramp, the angular speed of the disk is given by :

\omega=\dfrac{v}{r}

\omega=\dfrac{4.52}{0.25}

\omega=18.08\ rad/s

Hence, this is the required solution.

4 0
3 years ago
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