Which of the following is an example of static friction?

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th

Igoryamba

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

5 0

2 years ago

An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider compl

horsena [70]

Answer:

A = 2,8333 s

Explanation:

El periodo es definido como el tiene que toma de dar una oscilación.

En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.

T = t/n

calculemos

A = 34,0/ 12,0

A = 2,8333 s

6 0

2 years ago

When a planet's orbit takes it closest to the Sun, its called:________.

andriy [413]

Answer:

perihelion

Explanation:

The point at which a planet is closest to the sun is called perihelion. The farthest point is called aphelion

3 0

2 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration ( $a_c$ ) and the tangential acceleration ( $a_t$ ):

$a=\sqrt{a_c^2+a_t^2}$

We know that:

a = 4.4g

$a_c = 3.2 g$

So, we can find the tangential acceleration:

$a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2$

The angular acceleration is related to the tangential acceleration by

$\alpha = \frac{a_t}{r}$

where r = 10.9 m is the length of the centrifuge. Substituting,

$\alpha = \frac{29.6}{10.9}=2.7 rad/s^2$

Learn more about centripetal and angular acceleration here:

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#LearnwithBrainly

8 0

3 years ago

Give a strong idea to protect , purify and the beautify of the Dal lake

Sholpan [36]

Answer:

Water in dal lake is test for any heavy metals and pollutant, sewage and drainage system are also monitored for the same.

Explanation:

Dal lake is located in Srinagar that is the state capital of Kashmir and is known for recreation and tourism purposes. The area covers about 18 km sq. and forms a part of natural wetlands.

The lake is prone to pollution and has recently undergone restoration measures. To address the problems of eutrophication algae and large-scale microplankton have been removed from the water.

The government of India has taken various measures to check the pollution by setting up a committee to monitor the proper use of allotted funds.

3 0

3 years ago