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3 years ago

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An object of mass m is placed on Spring A, which is compressed by distance x. The spring is released and the velocity of the pus

hed object as it leaves the spring is measured. The experiment is repeated with the same object on Spring B, which is compressed by the same distance x. The object travels faster when pushed by Spring B. What can be concluded?
A .Spring A had a higher elastic potential energy than Spring B.
B. Spring A's spring constant is higher than Spring B's.
C. Spring A's spring constant is lower than Spring B's.
D. Spring A has more mass than Spring B.

1 answer:

ss7ja [257]3 years ago

3 0

Answer:

C

Explanation:

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A woman drives 200 miles west and then turns south and drives 375 miles. Her trip takes her 11.75 hrs. What is the woman's avera

FromTheMoon [43]

Answer:

The woman's average velocity during the trip is 36.2 miles/hour.

Explanation:

Velocity can be define as the displacement of an object per time. It is a vector quantity, and measured in m/s.

i.e velocity = $\frac{displacement}{time}$

From the given question,

Displacement = $\sqrt{200^{2} + 375^{2} }$

= $\sqrt{40000+140625}$

= $\sqrt{180625}$

= 425

The displacement of woman is 425 miles.

velocity = $\frac{425}{11.75}$

= 36.1702 miles/hour

The woman's average velocity during the trip is 36.2 miles/hour.

7 0

3 years ago

An air pump does 5,600 J of work to launch a water bottle rocket into the air. If the air pump applies 150 N of force to the roc

Alchen [17]

Answer: 53 m

Explanation:

Work = Force × Displacement

W= F s cosΘ

Where, s is the displacement and F is the force. Θ is the angle between force and displacement

F cosθ is the component of force acting in the horizontal direction.

The air pump applies F = 150 N force at an angle Θ = 45°

W = 5600 J

⇒s = W/F cosθ

$\Rightarrow s =\frac{5600 J}{150\times cos 45^o}\approx 53 m$

Thus, the horizontal distance the water bottle rocket travels is 53 m

8 0

3 years ago

Read 2 more answers

PLZZZ HELP ME ASAP!!!!

Leokris [45]

Answer:

D?

Explanation:

not a 100 percent sure

3 0

3 years ago

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A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

So

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

$\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

$\Delta \eta = 647 \pi$

$\Delta \eta = 2032.9$

4 0

4 years ago

If you are looking straight ahead an object that is behind you is in your cone of vision.

Dimas [21]

False
You cannot see behind you, therefore it is not in your cone of vision.

5 0

3 years ago

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