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Grace [21]
3 years ago
14

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?​
Physics
1 answer:
zalisa [80]3 years ago
7 0

Answer:

I got a question Can someone tell me what colors is the American Samoan Flag

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A force of 100 newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the
valentina_108 [34]


work is distance * force so 15*100=1500

and to find time you know power = diastance * force / time

so 25=15*100/t

25=1500/t

25/1500=t

.016=time


5 0
3 years ago
A transverse wave on a long horizontal rope with a
frosja888 [35]

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and  speed by the equation

f=\frac{v}{\lambda}

where

f is the frequency

v is the speed of the wave

\lambda is the wavelength

For the wave in this problem,

v = 2 m/s

\lambda=8 m

So the frequency is

f=\frac{2}{8}=0.25 Hz

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

T=\frac{1}{f}=\frac{1}{0.25}=4 s

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

t=\frac{T}{2}=\frac{4}{2}=2 s

4 0
3 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
Is friction and pushing similar ????
olya-2409 [2.1K]

Answer:

Yes, they are.

Explanation:

3 0
3 years ago
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Two point charges are separated by a distance of 1 meter. How much would the force change if one charge was 4x larger?
jekas [21]

Answer:

the force would increase 4 times more

Explanation

more force results more mass or acceleration

6 0
2 years ago
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