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cupoosta [38]
3 years ago
14

A pendulum completes 30 cycles every minute. What is the frequency of the pendulum

Physics
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

1 cycle every 2 seconds if that is what it is asking. So you have to do 2 x 30 and you will get 60 so ur awnser is 60

Explanation:

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Mrs. Davis has asked Kyle to retrieve the compound mixture from the chemical cabinet. Which of the following should Kyle retriev
nignag [31]

Answer:

water

Explanation:

coz out of the choices water is the only one that is not an element.

water formula is H2O

3 0
3 years ago
A ___ is when too much water moves into an area
Marina86 [1]
A. Flood.
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6 0
3 years ago
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Line segment gj is a diameter of circle l. angle k measures (4x 6)°. circle l is inscribed with triangle g j k. line segment g j
choli [55]

The value of x in the given right triangle in a semicircle is determined as 21.

<h3>What is the measure of a triangle in a semicircle?</h3>

The triangle in a semicircle is always a right angle triangle.

From the figure shown, we can say that the triangle  G J K is right triangle and m<K = 90degrees.

Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:

4x + 6 = 90

4x = 90 - 6

4x = 84

x = 21

Thus, the value of x in the given right triangle in a semicircle is determined as 21.

Learn more about right angle here: brainly.com/question/64787

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8 0
2 years ago
The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started
bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J

The change in electrostatic potential energy \Delta U, of one point charge q is defined as the product of the charge and the potential difference.

\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V

5 0
3 years ago
A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an
Vanyuwa [196]

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁  Where T₁ is temperature of hot source  and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

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output work = 20.25 KJ.

= 20 KJ .

6 0
3 years ago
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