Pube Goldberg machine is a complicated contraption designed to do a very simple task, like the one shown above

A mass hanging from a spring is set in motion and its ensuing velocity is given by v (t )equals 2 pi cosine pi t for tgreater th

lianna [129]

Answer:

2(maximum), -2(minimum), -2(maximum).

Explanation:

V(t)= 2πcos πt--------------------------------------------------------------------------------(1).

Therefore, there is a need to integrate v(t) to get S(t).

S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).

Applying the condition given, we have s(0)= 0.

S(0)= 2sin ×π(0) + C.

Which means that; 0+C= 0. That is; C=0.

S(t)= 2 sin πt.

The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.

Take note that; sin(π/2) = sin(5π/2) = 1 .

Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.

Therefore, we have that 2 maximum; -2 minimum and -2 maximum.

7 0

2 years ago

Sending food to a country that has been hit by a hurricane is an example of which foreign policy tool?

sasho [114]

International Disaster Management or FEMA

8 0

3 years ago

Read 2 more answers

The mass of the Sun is 1. 99 × 1030 kg. Jupiter is 7. 79 × 108 km away from the Sun and has a mass of 1. 90 × 1027 kg. The gravi

german

The gravitational force is s type of force that has the ability to attract any two objects having mass. The gravitational force will be $4.16\times10^{23}$ .

<h3>What is the gravitational force?</h3>

The gravitational force is s type of force that has the ability to attract any two objects with mass. Gravitational force tries to pull two masses towards each other.

$F= G\frac{m_1m_2}{r^{2} }$

Given,

mass of the sun ( $m_1$ )= $1.99\times10^{23}$ kg

mass of Jupiter( $m_2$ )= $7.79\times10^{8}$ kg

distance between the sun and Jupiter (r)= $1.90\times10^{27}$ m

$F= G\frac{m_1m_2}{r^{2} }\\\\\\F=4.16\times10^{23}\times\frac{1.99\times10^{23}\times7.79\times10^{8}}{({1.90\times10^{27})^2} }$

$F= 4.16\times(10)^{23}$ Newton

Hence the gravitational force between the sun and Jupiter will be $4.16\times10^{23}$

To learn more about gravitational force refer to the link:

brainly.com/question/24783651

4 0

2 years ago

1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and

saul85 [17]

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

Centripetal acceleration = $\frac{v^{2} }{r}$

v is the velocity of the body

r is the radius of the circle

putting in the parameters:

Centripetal acceleration = $\frac{4^{2} }{0.75}$

Centripetal acceleration = 10.67m/s²

Centripetal force = m $\frac{v^{2} }{r}$

m is the mass

Centripetal force = mass x centripetal acceleration

= 3 x 10.67

= 32N

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

4 0

3 years ago

Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

$f=\frac{qB}{2\pi m}$

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

$m=9.31\cdot 10^{-31} kg$ is the mass of the electron

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

7)

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.