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Serjik [45]
3 years ago
9

a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of

the spring from its unstrained length is -0.115 m, what is the acceleration a of the object
Physics
1 answer:
NeTakaya3 years ago
8 0

Answer:

The acceleration of the object is 5.57\ m/s^2.

Explanation:

Given that,

Mass of the block, m = 2.6 kg

Spring constant of the spring, k = 126 N/m

At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.

When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :

kx=ma

a is acceleration of the object

a=\dfrac{kx}{m}\\\\a=\dfrac{126\times 0.115}{2.6}\\\\a=5.57\ m/s^2

So, the acceleration of the object is 5.57\ m/s^2.

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Explanation:

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processed foods and certain dairy products are acid forming.

4 0
3 years ago
Calculate the speed with which the moon orbits the earth given the distance from earth to moon as R = 3.84 · 108 m. (Astronomers
marshall27 [118]
This sounds pretty easy, in fact. The orbital motion can be assumed to be circular and with constant speed. Then, the period is the time to do one revolution. The distance is the length of a revolution. That is 2*pi*R, where R is the distance between the Moon and the Earth (the respective centers to be precise). In summary, it's like a simple motion with constant speed:

v = 2*pi*R/T,

you have R in m and T is days, which multiplied by 86,400 s/day gives T in seconds.

Then v = 2*pi*3.84*10^8/(27.3*86,400) = 1,022.9 m/s ~ 1 km/s (about 3 times the speed of sound :)

For the Earth around the Sun, it would be v = 2*pi*149.5*10^9/(365*86,400)~ 29.8 km/s!

I know it's not in the problem, but it's interesting to know how fast the Earth moves around the Sun! And yet we do not feel it (that's one of the reasons some ancient people thought crazy the Earth not being at the center, there would be such strong winds!)
3 0
3 years ago
Two men decide to use their cars to pull a truck stuck in the mud. They attach ropes and one pulls with a force of 615 N at an a
stiv31 [10]

Answer:

1398.12 N

Explanation:

We define the x-axis in the direction parallel to the movement of the truck  on and the y-axis in the direction perpendicular to it.

x-components  of the ropes forces

T₁x = 615N*cos31°=527.1579 N  :Tension in direction x of the rope of the car 1

T₂x= 961 N*cos25°=870.96 N  :Tension in direction x of the rope of the car 2

Net forward force exerted on the truck in the direction it is headed (Fnx)

Fnx = T₁x  + T₂x

Fnx = 527.1579 N  + 870.96 N

Fnx = 1398.12 N

4 0
4 years ago
The count rate of a radioactive source decreases from 1600 counts per minute to 400 counts per minute in 12 hours. What is the h
kirill115 [55]

Answer:

t_{1/2}=6 h

Explanation:

Let's use the decay equation.

A=A_{0}e^{-\lambda t}

Where:

  • A is the activity at t time
  • A₀ is the initial activity
  • λ is the decay constant

We know that \lambda=\frac{ln(2)}{t_{1/2}}

So we have:

\lambda=\frac{ln(A/A_{0})}{t}

\frac{ln(2)}{t_{1/2}}=\frac{ln(A/A_{0})}{t}

t_{1/2}=\frac{t*ln(2)}{ln(A/A_{0})}

t_{1/2}=6 h

Therefore, the half-life of the source is 6 hours.

I hope it helps you!

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3 years ago
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