A substance that is present in a solution in a smaller amount and is dissolved by the solvent is called the solute.

A runner of mass 53.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cen

const2013 [10]

Answer:

0.336 rad/s

Explanation:

$\omega_1$ = Angular speed of the turntable = -0.2 rad/s

R = Radius of turntable = 2.9 m

I = Moment of inertia of turntable = $76\ kgm^2$

M = Mass of turn table = 53 kg

$v_1$ = Magnitude of the runner's velocity relative to the earth = 3.6 m/s

As the momentum in the system is conserved we have

$Mv_1R+I\omega_1=(I + MR^2)\omega_2\\\Rightarrow \omega_2=\dfrac{Mv_1R+I\omega_1}{I + MR^2}\\\Rightarrow \omega_2=\dfrac{53\times 3.6-76\times 0.2}{76+53\times 2.9^2}\\\Rightarrow \omega_2=0.336\ rad/s$

The angular velocity of the system if the runner comes to rest relative to the turntable which is the required answer is 0.336 rad/s

4 0

3 years ago

Telephone and electrical lines are allowed to sag between poles so that the tension will not be too great when something hits or

navik [9.2K]

Answer:

Tension in the wire= 1837.5 N

Explanation:

The answer along with solution can be found in the image attached.

7 0

3 years ago

A group of hikers hear an echo 2.6 s after they shout. The temperature is 20 ◦C. How far away is the mountain that reflected the

Vadim26 [7]

Answer:

$x = 445.9\,m$

Explanation:

The velocity of sound in air at 20 °C is 343 meters per second. The distance of mountain can be derived by the following expression:

$2\cdot x = v_{air}\cdot \Delta t$

$x = \frac{v_{air}\cdot \Delta t}{2}$

$x = \frac{(343\,\frac{m}{s} )\cdot (2.6\,s)}{2}$

$x = 445.9\,m$

4 0

3 years ago

Can a 20 N force and 40 N force ever produce a resultant with magnitude of 27 N?

SVEN [57.7K]

Sure ,Let's find angle between forces

Vectors be A and B and resultant be R

$\\ \sf\longmapsto R^2=A^2+B^2+2ABcos\theta$

$\\ \sf\longmapsto 27^2=20^2+40^2+2(20)(40)cos\theta$

$\\ \sf\longmapsto 729=400+1600+1600cos\theta$

$\\ \sf\longmapsto 729=2000+1600cos\theta$

$\\ \sf\longmapsto 1600cos\theta=-271$

$\\ \sf\longmapsto cos\theta=-0.169$

$\\ \sf\longmapsto \theta=cos^{-1}(-0.169)$

$\\ \sf\longmapsto \theta=80.2°$

7 0

2 years ago

An MRI technician moves his hand from a region of very low (B=0) magnetic field strength into an MRI scanner’s 3.50 T field with

sattari [20]

Answer:

The change in the magnetic flux in the ring is 10.99 Wb.

Explanation:

Given that,

An MRI technician moves his hand from a region of very low (B=0) magnetic field strength into an MRI scanner’s 3.50 T field with his fingers pointing in the direction of the field.

Diameter of the ring, d = 2 cm

Radius, r = 1 cm

It takes 0.4 s to move it into the field. We need to find the change in magnetic flux in the ring. Magnetic flux is given by :

$\phi=BA\\\\\phi=B\pi r^2\\\\\phi=3.5\pi \times (1)^2\\\\\phi=10.99\ Wb$

So, the change in the magnetic flux in the ring is 10.99 Wb. Hence, this is the required solution.

4 0

3 years ago