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Gnoma [55]
3 years ago
9

A mass on a spring with k=88.7 N/m oscillates 15 times in 9.24s. what is the objects mass? unit=kg?

Physics
1 answer:
sweet [91]3 years ago
3 0

The mass on the spring is 0.86 kg

Explanation:

The period of a mass-spring system is given by the equation

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

In this problem, we have:

k = 88.7 N/m is the spring constant

The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

T=\frac{9.24}{15}=0.62 s

Now we can re-arrange the first equation  to solve for the mass:

m=k(\frac{T}{2\pi})^2=(88.7)(\frac{0.62}{2\pi})^2=0.86 kg

Learn more about period:

brainly.com/question/5438962

#LearnwithBrainly

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A net force of 16 N causes a mass to accelerate at a rate of 5 m/s^2. Determine the mass.
Gnom [1K]

Answer:

So mass of the object will be 3.2 kg

Explanation:

We have net force on the object F = 16 N

Acceleration of the object a=5m/sec^2

We have top find the mass of the object

From newton law of motion we know that force is given by

F = ma , here m is mass and a is acceleration

So 16=5\times m

m=3.2kg

So mass of the object will be 3.2 kg

3 0
3 years ago
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
Name TWO WEAKNESSES of the model pictured below
Nikolay [14]

Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare,  and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

7 0
2 years ago
The answer answer it it la
ivolga24 [154]

Answer:

c) a tube light

Explanation:

a solar panel converts light energy into electricity

a tube light converts electricity into light

6 0
3 years ago
What do astronomers use in addition to parallax to find the actual distance of stars that are close to Earth?
IgorLugansk [536]

Answer:

trigonometry (guessing)

Explanation:

ellipse: is the shape of an orbit : looks like an oval

periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth

parallax is triangulation. like how gps works. looking at a star one day and then looking at it again 6 months later, an astronomer can see a difference in the viewing angle for the star. With trigonometry, the different angles yield a distance. This technique works for stars within about 400 light years of earth

https://science.howstuffworks.com/question224.htm

By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars

1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022

https://www.space.com/30417-parallax.html

alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years

blossoms.mit.edu

.

7 0
2 years ago
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