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Anna007 [38]
3 years ago
8

"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its

speed when it passes the equilibrium position."
Physics
2 answers:
AleksandrR [38]3 years ago
6 0

Answer:

haha lol

Explanation:

sorry but i dont know this

prisoha [69]3 years ago
4 0

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

x=Acos\alpha \\

we can express the velocity by the derivative of the displacement,

Hence

V=-Awsin\alpha \\

at equilibrium, the velocity becomes

V=wA\\w=2\pi f

Hence if we substitute values we arrive at

V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s

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zlopas [31]

Answer:

b

Explanation:

8 0
3 years ago
What do we mean by the observable universe?
mote1985 [20]
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4 0
3 years ago
A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r
kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = K.E=\frac{1}{2}I\omega^2

where,

I is the moment of inertia = mr²

on substituting the values, we get

K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2

or

K.E = 0.0075 J

3 0
3 years ago
A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the ma
Nana76 [90]

Answer:

30.63 m

Explanation:

From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

Maximum height (h) =.?

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

t = 2.5 s

Thus, the time (t) taken to reach the maximum height is 2.5 s

Finally, we shall determine the maximum height reached by the ball as follow:

Time (t) taken to reach the maximum height = 2.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

h = ½gt²

h = ½ × 9.8 × 2.5²

h = 4.9 × 6.25

h = 30.625 ≈ 30.63 m

Therefore, the maximum height reached by the cannon ball is 30.63 m

3 0
3 years ago
Any two application of gravity
Brrunno [24]

Answer:

Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.

Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

5 0
3 years ago
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