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3 years ago

Read the information and study the image.

Physics

1 answer:

erastova [34]3 years ago

4 0

Answer:

It is impossible to detect underground water from the surface. Dowsing practitioners refuse to explain their secrets.

Explanation:

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If the force of gravity changes what will happen and what would cause it to change?

FromTheMoon [43]

We will fall to pluto.

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3 years ago

You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling a

Sonja [21]

Answer:

L=55.9m

Explanation:

The equation for the period of a simple pendulum is:

$T=2\pi\sqrt{\frac{L}{g}}$

In our case what we know is the period and the acceleration of gravity, and we need to know the length of the pendulum, so we can write:

$L=(\frac{T}{2\pi})^2g$

Which for our values is:

$L=(\frac{15s}{2\pi})^2(9.81m/s^2)=55.9m$

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3 years ago

A testing instrument that's used to measure electrical signals in a circuit and display them as waveforms on a screen is called

disa [49]

A testing instrument that's used to measure electrical signals
in a circuit and display them as waveforms on a screen is called
an oscilloscope.

8 0

3 years ago

Which of the following would result if thermal energy is added to an object?

seropon [69]

Answer:

is there supposed to be a pic and abcd options?

8 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago