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Elodia [21]
3 years ago
12

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:
pochemuha3 years ago
6 0

Answer:

m = 63 grams

Explanation:

ω = 10 cycles/s(2π radians/cycle) = 20π rad/s

ω = √(k/m)

m = k/ω² = 250/(20π)² = 0.06332... kg

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According to the Law of Universal Gravitation, which scenario would generate the most
ehidna [41]

Answer:

im not 100 percent but its probably 2 turtles 20m apart

3 0
3 years ago
Answer the question please
kirill [66]

Answer:

i think is third

Explanation:

7 0
4 years ago
A housefly walking across a clean surface can accumulate a significant positive or negative charge. In one experiment, the large
Lisa [10]

Answer:

4.56*10^8 \text{    electrons.}

Explanation:

Since the fly accumulated a positive charge of +73pC, it must have lost an equal number of negative charge of -73pC to the surface (because the housefly was neutral to begin with).

Therefore, to answer our question we have to ask ourselves <em>how many electrons combine to make -73pC of charge? </em>

The answer is since one electron carries a charge of -1.6*10^{-19}C, the number n of electrons that make up -73pC (-73*10^{-12}C) are

n= \dfrac{-73*10^{-12}C}{-1.6*10^{-19}C/e}

\boxed{n= 4.56*10^8e.}

Thus, the housefly lost about 456 million electrons to the surface!

4 0
4 years ago
A .005kg projectile leaves a 1500kg launcher with a velocity of 750 m/s. What is the recoil velocity of the projectile
docker41 [41]

Answer:

The recoil velocity of the projectile is 0.0025m/s

Explanation:

Given:

Mass of the projectile =0.005kg

Mass of the launcher = 1500kg

Velocity =  750 m/s.

To Find:

The recoil velocity of the projectile = ?

Solution:

The recoil velocity is the obtained by dividing the "recoil momentum"  by the "mass of the recoil body".  The recoil momentum is equal to the momentum of the other body. The momentum of the other body is equal to it mass times its velocity.

Lets find the recoil momentum,

Recoil momentum = mass of the projectile X velocity

Recoil momentum =0.005 \times 750

Recoil momentum = 3.75

Now Recoil Velocity,

Recoil Velocity = \frac{\text { Recoil Momentum}}{\text {Mass of the launcher}}

Recoil Velocity = \frac{ 3.75}{1500}

Recoil Velocity = 0.0025m/s

7 0
4 years ago
Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o
Mila [183]

Answer:

a) x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) x = 4.47 cm

c) x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm

c) The center of the mass of the beads realtive to the center of bead is:

x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm

6 0
3 years ago
Read 2 more answers
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