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2 years ago

10

A dart in a toy dart gun is pressed 4 cm against a spring. if it has 1.5 J of stored potential energy what is the spring constan

t?

1 answer:

Sphinxa [80]2 years ago

8 0

U = 0.5k(x^2)
1.5 J = 0.5k(0.04m^2)
k = 1875 N/m

Check my math i might be wrong

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When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?

Oduvanchick [21]

First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
$a = r \times ( \frac{1}{n1} - \frac{1}{n2} )$
We insert the values

$a = 2.18 \times {10}^{ - 18} \times ( \frac{1}{ {1}^{2} } - \frac{1}{ {6}^{2} } )$
$= 2.12 \times {10}^{ - 18}$
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
$wavelength = \frac{h \times c}{a}$
Finally we insert the values
$\frac{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{2.12 \times {10}^{ - 18} } = 9.376 \times {10}^{ - 8}$
Which is the same as 93.8 nm

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3 years ago

Technician A says that the evacuation process will remove dirt and debris from the refrigerant system. Technician B says that th

GuDViN [60]

Answer: Technician B is right.

Explanation:

Evacuation process is used in refrigeration systems to remove moisture, air and non-profit condensable gases in order to achieve maximum function of the system.

vacuum pump is used to draw the sealed AC system into a vacuum. Evacuation of a refrigerant system also helps to maintain pressure, this is so as pulling a vacuum on the system is simply removing matter (mostly air and nitrogen) from inside the system so that the pressure inside drops below atmospheric pressure.

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3 years ago

A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The

Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

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3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

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3 years ago

The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to

katrin2010 [14]

Answer:D- 0.2528

Explanation:

T=(Rp/Rd)^3/2

T=(9378/23459)^3/2

T=0.2528

The answer is D

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3 years ago