A 55 kg person falling with a velocity of 0.6

In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.

Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

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

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

$g=9.8 m/s^2$ (acceleration of gravity)

Therefore the period of this pendulum is

$T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s$

#LearnWithBrainly

3 0

3 years ago

9. Since current is the rate at which charge is flowing, if the current in the circuit decreases, what does that mean about the

TiliK225 [7]

Answer:

Electric current is defined as the rate of flow of electric charge in a circuit from point one point to another. This is carried by electrically charged particles within the circuit. Current is represented by the symbol I and its unit measured in Amperes. It is therefore related to the voltage and resistance of the circuit. If the current in the circuit reduces, the rate at which the charge and current on the capacitor reduces also proportionally in an exponential manner.

Explanation:

Since a decrease in the flow of current in the circuit is observed, the implication for the rate at which the charge and voltage on the capacitor is also an exponential decrease in the rate of flow with time. This is because the electric current is directly proportional to the electric charge and the time.

5 0

3 years ago

HELP PLEASE!!! 30+ points!!

GarryVolchara [31]

1) 9.26 cm

Explanation:

The focal length of a plane mirror is virtually infinite. Considering the lens equation,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where f is the focal length, p is the object distance, q the image distance. If we replace f with infinity, we get

$q=-p$

The magnification equation states that

$y' = -\frac{q}{p}y$

where y is the size of the object and y' the size of the image. Substituting q=-p, we get

$y'=y$

this means that the image produced by a plane mirror is always:

- Upright (y' is positive)

- The same size as the object

In this case, we have a book of height 9.26 cm (y=9.26 cm). This means that the magnitude of the size of the image (y') will be 9.26 cm as well.

2) 22.7 cm

As we said before, due to the infinite focal length of a plane mirror,

$q=-p$

this means that the image produced by a plane mirror is always:

- Virtual (because q is negative)

- At the same distance from the mirror as the object

In this case, we have a book placed at 22.7 cm from the mirror (p=22.7 cm). This means that the magnitude of the distance of the image from the mirror (q) will be 22.7 cm as well.

3) 1.60 m/s

We said previously that the image produced by a plane mirror is always at the same distance from the mirror as the real object. This implies that whenever we move the object toward/away from the mirror, the distance p will alway remain equal to the distance q. But this also means that the object and the distance are moving toward/away from the mirror at the same speed.

Therefore, since in this case the person is moving away from the mirror at 1.60 m/s, the image will also move away at a speed of 1.60 m/s.

4 0

3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is: