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natta225 [31]
1 year ago
12

A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period o

f the motion
Physics
1 answer:
Jobisdone [24]1 year ago
7 0

Answer:

Explanation:

The amplitude is extraneous information for this question.

Period (T) is simply a measure of time (how long it takes to complete a full cycle).

The frequency (f), is the reciprocal (how many cycles it can complete per unit of time).

The frequency reported is 12Hz.  Hz is a unit meaning "cycles per second"

Thus 12cycles = 1second.

To find the number of seconds per cycle, simply create the other unit fraction and simplify:

T=\frac{1second}{12cycles}\\T=0.8\bar{3}\frac{seconds}{cycle}\\

T=0.83seconds

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A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a
9966 [12]

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination \theta = 15°

\sin \theta = \frac{AB}{AC}

\sin 15 = \frac{5900}{x}

x = \frac{5900}{0.2588}

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

7 0
2 years ago
Which use combustion to release thermal energy?
julia-pushkina [17]
Heat of combustion.<span> The calorific value is the total energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon or other organic molecule reacting with oxygen to form carbon dioxide and water and release heat.</span>
8 0
2 years ago
Read 2 more answers
How does the gravitational force between two objects change if the distance
Anestetic [448]

Answer:

The Gravitational Force is reduced 4 times

Explanation:

The equation of Gravitational force follows:

F = (G*m1*m2)/r^2

Assume that G*m1*m2 = 1 and r = 1:

F = 1/1^2 = 1 N

Multiply the radius by 2

F = 1/2^2 = 1/4 N

So doubling the distance reduces the force 4 times.

7 0
3 years ago
(ASAP) would it be 125 m/s2 to calculate for her speeding up?
serg [7]

Answer:

0\:\mathrm{ m/s^2}

Explanation:

Recall the formula for acceleration:

\displaystyle\\a=\frac{v_f-v_i}{\Delta t}, where v_f is final velocity, v_i is initial velocity, and \Delta t is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0
2 years ago
A convex mirror, like the passenger-side rearview mirror on a car, has a focal length of -3.0 m . An object is 6.0 m from the mi
Artist 52 [7]

A) -2.0 m

Look at the ray diagram attached in the picture, where:

p identifies the location of the object

q identifies the location of the image

F identifies the focus of the mirror

Each tick represents 1 m

We have

p = 6.0 m is the distance of the object from the mirror

f = -3.0 m is the focal length

From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so

q = -2.0 m

This can also be verified by using the mirror equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-3.0 m}-\frac{1}{6.0 m}=-\frac{3}{6.0 cm}\\q = \frac{-6.0 cm}{3}=-2.0 cm

B) Upright and virtual

As we see from the picture, the image is upright, since it has same orientation as the object.

Also, we notice that the image is on the other side of the mirror, compared to the object. For a mirror,

- An image is said to be real if it is on the same side of the object

- An image is said to be virtual if it is on the opposite side of the mirror

Therefore, this means that the image is virtual.

8 0
2 years ago
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