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3 years ago

On what factors, current sensitivity voltage sensitivity ofa galvanometer depend?

Physics

2 answers:

Nataly_w [17]3 years ago

6 0

That depends upon several factors

Number of turns of coil
Current flowing through the galvanometer
Torsion constant of suspension fibre
Strength of magnetic field
Restoring force.

Alexeev081 [22]3 years ago

5 0

<h3> it depends on numbers of turns coils area restoring force per unit twist and magnetic fields</h3><h2 /><h2 /><h2>hope it helps</h2>

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Pls help I will give brainlist and don’t give me a link I can’t open them

dezoksy [38]

Answer:

200

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3 years ago

Find the net work done by friction on the body of a snake slithering in a complete circle of radius. The coefficient of friction

Tamiku [17]

Answer:

C. 0 J

Explanation:

When an object moves in a circle, friction provides the centripetal force. Hence, it is acting towards the centre of the circle. The displacement of the object is tangential to the circle at the point of interest. Hence, the angle between the frictional force and the displacement is 90°.

Work done is given by

where <em>F</em> is the force, <em>d</em> is the displacement and <em>θ</em> is the angle between them.

<em>W</em> = <em>Fd</em> cos 90° = 0 J

Hence, the work done by friction is 0 J.

7 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

4 years ago

How does the width of the floodplain change over time?

serg [7]

Floodplains are landscapes shaped by running water. As streams and their larger forms, rivers, flow across the surface of land, they transport eroded rock and other material. (Erosion is the gradual wearing away of Earth surfaces through the action of wind and water.) At points along that journey, when their flow slows, the material they carry is dropped to create what are termed depositional landforms. Among these landforms are deltas and floodplains.

Read more: http://www.scienceclarified.com/landforms/Faults-to-Mountains/Floodplain.html#ixzz7BNHuUb00

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2 years ago

On what factors, current sensitivity voltage sensitivity ofa galvanometer depend?​

On what factors, current sensitivity voltage sensitivity ofa galvanometer depend?