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Vikentia [17]
2 years ago
15

A clock has an hour hand of length 2.4 cm and a minute hand of length 3.8 cm. (a) Calculate the position and velocity of the hou

r hand at noon. (b) Calculate the position and velocity of the minute hand at 12:15.
Physics
1 answer:
9966 [12]2 years ago
6 0

Answer:

  1. At 12:00 the hour hand it's located at 0 degrees and its tip travels with a velocity of 1.26 cm/hour, approximately.
  2. At 12:15 the minute hand it's located at 90 degrees and its tip travels with a velocity of 23.88 cm/hour, approximately.

Explanation:

Position is given by an angle, in this case starting from the 12:00 mark and opening clockwise.

  1. At noon, the the hour hand is located at the 12:00 mark, so its position is 0 degrees; and it has an angular velocity of (2π)/(12 hour) = π/(6 hour). So its tip travels with linear velocity of (2.4 cm) * π/(6 hour) ≅ 1.26 cm/hour.
  1. At 12:15, the minute hand is located at the 3 hour mark, so its position is 90 degrees or π/2; and it has an angular velocity of (2π)/(1 hour) = 2π/hour. So its tip travels with linear velocity of (3.8 cm) * 2π/hour ≅ 23.88 cm/hour.
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An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48
kramer

Answer:

Er = 108 [J]

Explanation:

To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

Where:

Er = remaining energy [J]

Er = 108 [J]

3 0
2 years ago
What is the magnitude of the force a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away?
sweet [91]
The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:

F = kq1q2/r^2

where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters

Using direct substitution, the force F is determined to be 1920 Newtons.</span>
7 0
2 years ago
Is it possible for two pieces of the same metal to have different recrystallization temperatures? Is it possible for recrystalli
Butoxors [25]

Answer:

Explained

Explanation:

Two pieces of the same metal can have different recrystallization temperatures if the pieces  have been cold worked to different amounts. The piece of work cold worked to greater extend will have more internal energy to drive the recrystalline process and lower recrystallization temperature.

Yes, its possible that recrystallization to take place in some regions of a part before it does in other regions of the same part if the work has been unevenly strained or if the part have different thickness at different sections.

6 0
2 years ago
13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?
sergeinik [125]

The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

Explanation:

The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of  mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

Q = mc del T

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

Q = 20*C*80

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

5 0
2 years ago
A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an
Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

\omega = 2\pi f

\omega = 2\pi (2.5)

\omega = 5\pi rad/s

The angular displacement is given as the form:

\theta (t) = \theta_0 cos(\omega t)

In the equlibrium we have to t=0, \theta(t) = \theta_0 and in the given position we have to

\theta(t) = \theta_0 cos(5\pi t)

Derived the expression we will have the equivalent to angular velocity

\frac{d\theta}{dt} = 2.7rad/s

Replacing,

\theta_0(sin(5\pi t))5\pi = 2.7

Finally

\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°

Therefore the maximum angular displacement is 9.848°

6 0
2 years ago
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