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Delvig [45]
3 years ago
10

An object of mass 8kg is attached to massless string of length 2m and swum with a tangential velocity of 3 what is the tension o

n the string
Physics
1 answer:
Paul [167]3 years ago
6 0

Answer:

36 N

Explanation:

If the object of mass, m = 8 kg is swung in a horizontal circle of radius, r = 2m = length of string with tangential velocity v = 3 m/s, the tension in the string is the centripetal force which is T = mv²/r

= 8 kg × (3 m/s)²/2 m

= 4 kg × 9 m/s²

= 36 N

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How are the ice cubes and water similar?<br> How are they different?
KATRIN_1 [288]

Answer:

As the liquid cools down, the amount of potential energy is reduced and the molecules start to move slower. When the water temperature reaches around 0°C, the molecules stick together and form a solid – ice

The "stuff" (molecules) in water is more tightly packed than in ice, so water has greater density than ice. Don't let the fact that ice is a solid fool you! As water freezes it expands. So, ice has more volume (it takes up more space, but has less density) than water

Explanation:

3 0
2 years ago
Read 2 more answers
At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl
bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0
3 years ago
Phosphorus-32, a radioactive isotope of phosphorus-31 (atomic number 15), undergoes a form of radioactive decay whereby a neutro
dimulka [17.4K]

Answer:

The product of the decay its Sulfur-32

Explanation:

Phosphorus-32 ( lets write it _{15}^{32}P, where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:

_{Z}^{A}X \to _{Z+1}^{A}Y + e^- + \bar{\nu_e}

where e^- its the electron, and \bar{\nu_e} the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).

So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:

_{15}^{32}P \to _{15+1}^{32}Y + e^- + \bar{\nu_e}.

_{15}^{32}P \to _{16}^{32}Y + e^- + \bar{\nu_e}.

The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its

_{15}^{32}P \to _{16}^{32}S + e^- + \bar{\nu_e}.

and the product of such decay its Sulfur-32

5 0
4 years ago
Density of a substance ratio​
Inga [223]

Answer:

Density of a substance is the ratio of mass of the substance to its volume.

6 0
2 years ago
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The forces acting on a child sitting in a seat are described by the free-body diagram shown. What is the net force acting on the
r-ruslan [8.4K]

The net force is 0 N.

To find the total net force, all you need to do is add the forces.

So 100 + -100 = 0

Most of these questions are trick questions, but if you know Newton's laws of forces, then you should know that all forces should cancel out with each other.

Hope this helps!!!

3 0
3 years ago
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