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4 years ago

The frequency of a simple pendulum that makes 120 complete oscillations in 3 minutes is:

Physics

1 answer:

nekit [7.7K]4 years ago

5 0

Answer: 0.67 Hz

Explanation:

Given that:

Number of oscillations = 120

Time taken for oscillations = 3 minutes

Since standard unit of time is seconds, convert 3 minutes to seconds

(If 1 minute = 60 seconds

3 minutes = 3 x 60 = 180 seconds)

Then, recall that frequency is the number of complete cycles or oscillations in one second, and its unit is Hertz.

Thus, frequency = (number of oscillations / time)

F = 120 / 180 seconds

F = 0.67 Hz

Thus, the frequency of a simple pendulum is 0.67Hz

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Why are units of measurement useful?

Zanzabum

Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories. Not only is measurement important in science and the chemical industry, it is also essential in farming, engineering, construction, manufacturing, commerce, and numerous other occupations and activities.

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4 years ago

Read 2 more answers

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis.At

timurjin [86]

Answer:

a) The initial velocity of the green car is -13 m/s

b) The acceleration of the green car is - 2.25 m/s²

Explanation:

The equation for the position of objects moving in a straight line with constant acceleration is as follows:

x = x0 + v0·t + 1/2·a·t²

where:

x = position

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

If the velocity is constant, then a = 0 and x = x0 + v·t

a)The initial position of the red and green car is 0 m and 220 m respectively. We know that at 44.5 m the cars pass each other if the red car has a constant velocity of 20 km/h. So let´s find how much time it takes the cars to pass each other in this case:

The position of the red car is:

x = x0 + v·t

then:

0.0445 km = 0 km + 20 km/h · t

t = 0.0445 km/ 20 km/h = 8.0 s

We also know that if the red car has a velocity of 40 km/h, both cars pass each other at 76.6 m. So let´s find the time it takes the cars to reach that position using the equation for the red car:

0.0766 km = 0 km + 40 km/h · t

t = 0.0766 km / 40 km/h = 6.9 s

The position of the green car at t= 6.9 s and t = 8.0 s must be the same as the red car because both cars pass each other at those times.

Then, for the green car:

x = x0 + v0·t + 1/2·a·t²

0.0445 km = 0.220 km + v0 · 8.0 s + 1/2·a· (8.0 s)²

and

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·a· (6.9 s)²

Now we have a system of two equations with two unknowns.

Solving for "a" in the first equation

0.0445 km - 0.220 km - v0 · 8.0 s = 32 s²·a

(-0.176 km - v0 · 8.0 s) / 32 s² = a

Replacing a = (-0.176 km - v0 · 8.0 s) / 32 s² in the second equation and solving for v0:

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·((-0.176 km - v0 · 8.0 s)/32 s²)·(6.9 s)²

-0.143 km = v0 · 6.9 s - 0.74(0.176 km + v0 · 8.0 s)

-0.143 km = v0 · 6.9 s - 0.130 km - v0 · 5.9 s

-0.143 km + 0.130 km = v0 · 6.9 s - v0 · 5.9 s

-0.013 km = 1 s · v0

v0 = -13 m/s

b) The acceleration of the green car is:

a = (-0.176 km - v0 · 8.0 s) / 32 s²

a = (-0.176 km - (-0.013 km/s) · 8.0 s) / 32 s² = -2.25 m/s²

3 0

3 years ago

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

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3 years ago

Read 2 more answers

What is energy?

Anuta_ua [19.1K]

Energy is the ability to do work

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3 years ago

A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

likoan [24]

Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

3 0

3 years ago