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True [87]
3 years ago
10

In case 1, a force f is pushing perpendicular on an object a distance l/2 from the rotation axis. in case 2 the same force is pu

shing at an angle of 30 degrees a distance l from the axis. 1 in which case is the torque due to the force about the rotation axis biggest?
Physics
1 answer:
Serggg [28]3 years ago
6 0
In case 1, the torque is given by the product between the force and the arm:
\tau_1 = F \cdot  \frac{L}{2}

In case 2, the torque is given by the product between the component of the force perpendicular to the arm and the arm itself, so we have:
\tau_2 = F \cos 30^{\circ} L=F  \frac{ \sqrt{3} }{2} L =  \sqrt{3} F  \frac{L}{2}= \sqrt{3} \tau_1

and since \sqrt{3} is larger than 1, than the torque in case 2 is larger than the torque in case 1.
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2. An 873 kg dragster accelerates at a rate of 44.6 m/s during a race.
Ira Lisetskai [31]

Answer:

<h3>38,673.9N</h3>

Explanation:

According to newton's second law:

Force = mass * acceleration

Given

Mass = 873kg

acceleration = 44.66m/s²

Magnitude of the force is expressed as;

F = ma

F = 873 * 44.6

F = 38,673.9N

<em>Hence the magnitude of the net force exerted on the dragster during this time is 38,673.9N</em>

8 0
2 years ago
Why does the current is reduced as electrons move through a conductor
hammer [34]
Because the electrons collide with the particles inside the conductor so are therefore slowed down seen as current is the rate of flow of electrons
3 0
3 years ago
Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb
murzikaleks [220]

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        x_{cm} = \frac{1}{M} \sum  x_{i} m_{i}i

        x_{cm} = \frac{1}{m+m} ( 0 + l m)

        x_{cm} = \frac{1}{2m}  lm

        x_{cm} = \frac{1}{2}

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

4 0
3 years ago
Read 2 more answers
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

a_c =w^2r = \frac{v^2}{r}

in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

a_c = \frac{8^2}{28.2}\ m/s^2

a_c = 2.27\ m/s^2

The answer is the last option

3 0
3 years ago
What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and
Andrei [34K]

Answer:

A. 8.29\times 10^{-18}\ J

Explanation:

Given that:

p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

k = Boltzmann constant = 9\times 10^{9}\ Nm^2/C^2

r = distance between the two carbon nuclei = 1.00 nm = 1.00\times 10^{-9}\ m

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

U = \dfrac{kQq}{r}

So, the potential energy between the two nuclei of carbon is as below:

U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J

Hence, the energy stored between two nuclei of carbon is 8.29\times 10^{-18}\ J.

8 0
2 years ago
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