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Anarel [89]
3 years ago
12

The physical quantity represented as rate of change of change in position in a

Physics
1 answer:
Oduvanchick [21]3 years ago
3 0

Answer:

B

rate of change of its position with respect to a frame of reference, and is a function of time.

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4 years ago
Leash a block of 0.5 kg down the entire length of an inclined plane forming a name of 37 ° with the horizontal. Friction with th
castortr0y [4]
The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height

The potential energy of a spring is given by:
V = 0.5 * k * x²

k spring constant
x compression of the spring

If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:

m * g * h = 0.5 * k * x²

m = 0.5 kg
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h = 2.5m * sin 37° = 1,5 m
x = 0,6 m

Solve for k.

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6 0
3 years ago
Kate walks up to an automated teller machine and pushes a button. The pressure exerted by her finger is transformed into electri
lozanna [386]
The answer is Transduction .
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What type of system would allow light to enter and exit, but would keep any
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A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl
raketka [301]
The wavelength of the note is \lambda = 39.1 cm = 0.391 m. Since the speed of the wave is the speed of sound, c=344 m/s, the frequency of the note is
f= \frac{c}{\lambda}=879.8 Hz

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }
where \mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
L= \frac{1}{2f}  \sqrt{ \frac{T}{\mu} } =0.31 m
8 0
3 years ago
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