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STALIN [3.7K]
3 years ago
7

A 0.18-kg ball on a stick is whirled on a vertical circle at a constant speed. When the ball is at the three o’clock position, t

he stick's tension is 19 N. Find the tension in the stick when the ball is (a) at the twelve o’clock and (b) at the six o’clock positions.
Physics
1 answer:
Stells [14]3 years ago
7 0

Answer:

a) 17 N

b) 21 N

Explanation:

At the 3 o'clock position, the sum of the forces towards the center is:

∑F = ma

T = m v² / r

19 = m v² / r

At the 12 o'clock position, the sum of the forces towards the center is:

∑F = ma

T + mg = m v² / r

T + (0.18)(9.8) = 19

T = 17.2 N

At the 6 o'clock position, the sum of the forces towards the center is:

∑F = ma

T − mg = m v² / r

T − (0.18)(9.8) = 19

T = 20.8 N

Rounding to two significant figures, the tensions are 17 N and 21 N.

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An atom undergoes nuclear decay, but its atomic number is not changed.
andreev551 [17]

Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

  • Gamma rays have zero atomic and mass numbers.
  • When they cause decay, they cause no change to the mass and atomic numbers.
  • They simply produce gamma rays during such reactions and these rays are very energetic.
5 0
3 years ago
The length of a 100 mm bar of metal increases by 0.3 mm when subjected to a temperature rise of 100°C. The coefficient of linear
Juli2301 [7.4K]

Answer:

α = 3×10^-5 K^-1

Explanation:

let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.

then, the coefficient of linear expansion is given by:

α = ΔL/(ΔT×L)

   = (0.3×10^-3)/(100)(100×10^-3)

   = 3×10^-5 K^-1

Therefore, the coefficient of linear expansion is 3×10^-5 K^-1

5 0
3 years ago
Label the following descriptions with (M) for Mass or (W) for Weight. ____measured with a spring scale ____measured in Newtons _
Valentin [98]
1. mass
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3. weight
4. weight
5. mass
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8 0
3 years ago
A teapot with a surface area of 700 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic
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Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

7 0
3 years ago
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