Would a decrease in force cause a decrease in pressure

A simple harmonic oscillator has an amplitude of 3. 50 cm and a maximum speed of 26. 0 cm/s. What is its speed when the displace

Marianna [84]

Answer:

1.) A simple harmonic oscillator has an amplitude of 3.50 cm and a maximum speed of 26.0 cm/s. What is its speed when the displacement is 1.75 cm? 2.) Both pendulum A and B are 3.0 m long. The period of A is T. Pendulum A is twice as heavy as pendulum B. What is the period of B? 3.) The time for one cycle of a periodic process is called the _ ? 4.) In simple harmonic motion, the acceleration is proportional to? 5.) The position of a mass that is oscillating on a spring is given by x= (18.3 cm) cos [(2.35 s-1)t]. What is the frequency of this motion?

Explanation:

3 0

2 years ago

Nancy immediately remembers she had promised to return her grandmother's call from earlier this morning. What type of memory is

Solnce55 [7]

I believe its Procedural Memory

6 0

3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

3 years ago

(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0

3 years ago

Volcanoes that have fast flowing liquid lava will be narrow like a river

vodomira [7]

Answer:

Shield volcanoes, the third type of volcano, are built almost entirely of fluid lava flows. Flow after flow pours out in all directions from a central summit vent, or group of vents, building a broad, gently sloping cone of flat, domical shape, with a profile much like that of a warrior's shield.

Explanation:

8 0

3 years ago