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My name is Ann [436]

2 years ago

6

A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t

he spring exerts on the car is given by the equation F=−Kx2, where K=5000 Nm2. The potential energy stored in the spring when the car is pushed against it is most nearly:________________

1 answer:

dybincka [34]2 years ago

5 0

Answer:

1.56 J

Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

F = - k x

k = 5000 N/m

Potential energy stored = ?

energy stored in the spring

$PE = \dfrac{1}{2}kx^2$

$PE = \dfrac{1}{2}\times 5000\times 0.025^2$

PE = 1.56 J

Hence, the potential energy stored in the car is equal to 1.56 J.

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A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em

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Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

$P = \ro g h = 1000 * 10 * 20 = 200000 Pa$

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

$P_1V_1 = P_2V_2$

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

$V_1 = Ah$

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

$P_1Ah_1 = P_2Ah_2$

$h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m$

b) If the temperatures changes, we can still reuse the ideal gas equation above:

$\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}$

$h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m$

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2 years ago

Any kind of wave spreads out after passing through a small enough gap in a barrier. This phenomenon is known as _________. a) di

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Answer:

a) diffraction

Explanation:

Diffraction occurs when waves pass through small openings, around obstacles or sharp edges.When an opaque object is between the point of light and a screen, the border between the shaded and illuminated regions on the screen is not defined. A careful inspection of the scrubber shows that a small amount of light is diverted to the shaded region. The region outside the shadow contains bright and dark altered bands, where the intensity of the first band is brighter than the region of uniform illumination.

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The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and cons

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Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

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2 years ago

Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show

Ksenya-84 [330]

Answer:

Show attached picture

Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call $R_M$ its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

$R_M = 1000 R$ (1)

Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

$V_M = V_R$

Using Ohm's law, $V=RI$ , we can rewrite the previous equation as:

$R_M I_M = R I_R$

where

$I_M$ is the current in the meter

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Using (1), this equation becomes

$(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}$

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2 years ago

The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the

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Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

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$E = \dfrac{kq}{r^2}$

k = 9 x 10⁹ N.m²/C²

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$q= \dfrac{Er^2}{k}$

now,

$q= \dfrac{180000\times 0.015^2}{9\times 10^9}$

q = 4.5 x 10⁻⁹ C

q = 4.5 nC

the charge on the object is equal to 4.5 nC

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