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Musya8 [376]
4 years ago
11

A light spring with force constant 3.80 n/m is compressed by 7.20 cm as it is held between a 0.300-kg block on the left and a 0.

600-kg block on the right, both resting on a horizontal surface. the spring exerts a force on each block, tending to push the blocks apart. the blocks are simultaneously released from rest. find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is 0, 0.067, and 0.429. (let the coordinate system be positive to the right and negative to the left. indicate the direction with the sign of your answer. assume that the coefficient of static friction is the same as the coefficient of kinetic friction. if the block does not move, enter 0.)
Physics
1 answer:
jeka944 years ago
3 0
Some dogs may inherit a susceptibility to epilepsy.
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What is 300k fahrenheit ?
Yakvenalex [24]

Answer:

80.33F

Explanation:

(300-273.15)*9/5+32=80.33

7 0
3 years ago
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The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec
Elanso [62]
On Earth, the period of a pendulum is given by:
T_{earth}=2\pi  \sqrt{ \frac{L}{g_{earth} }
where L is the length of the pendulum and g_{earth}=9.81~m/s^2 is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }
where g_{mars}=3.71~m/s^2 is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} }  =  \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.
6 0
4 years ago
What is the amount of nickel, which has an average atomic mass of 58.7 amu, present in 525 g nickel
Wittaler [7]

Answer:

bhgyg 78 gt78gyug

Explanation:

4 0
3 years ago
Read 2 more answers
The distribution of total body protein in healthy adult men is approximately Normal, with mean 12.3 kg and standard deviation 0.
Sophie [7]

Answer:

P(12.25\leq x \leq 12.35 ) = 0.9876

Explanation:

given,

mean (μ) = 12.3 Kg

standard deviation (σ ) = 0.1

random sample = 25

probability between 12.25 and 12.35 kg

P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{\sqrt{n}}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{\sqrt{n}}}\leq z)

P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{\sqrt{25}}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{\sqrt{25}}}\leq z)

P(12.25\leq x \leq 12.35 ) = P(\dfrac{12.35-12.3}{\dfrac{0.1}{5}}\leq z)- P(\dfrac{12.25-12.3}{\dfrac{0.1}{5}}\leq z)

P(12.25\leq x \leq 12.35 ) = P(\dfrac{5 (12.35-12.3)}{0.1}\leq z)- P(\dfrac{5(12.25-12.3)}{0.1}\leq z)

P(12.25\leq x \leq 12.35 ) = P(\dfrac{2.5\leq z)- P(-2.5\leq z)

using z-table

P(12.25\leq x \leq 12.35 ) = 0.9938 - 0.0062

P(12.25\leq x \leq 12.35 ) = 0.9876

4 0
3 years ago
What is one way to take advantage of creating even more kinetic energy on this particular technologically advanced field
ch4aika [34]

Incomplete question. However, I provided a brief about Kinetic energy generation.

<u>Explanation:</u>

Interestingly, Kinetic energy in simple terms refers to the energy possessed by a body in motion.

It is often calculated using the formula E =  \frac{1}{2}MV^{2}

A good example of creating even more kinetic energy is a hand crank toy car that moves after you wind it a little, when the car moves it is generating another measure of K.E.

7 0
3 years ago
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