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3 years ago

To measure the potential drop across a resistor, a voltmeter is connected in ____ with the resistor.

Physics

1 answer:

slava [35]3 years ago

5 0

A voltmeter is connected in PARALLEL with the resistor.

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lorasvet [3.4K]

Answer:

A) Option 1 is the correct answer.

B) Option 4 is the correct answer.

Explanation:

A) Weight of liquid = 7 N

Volume of liquid = 1 L = 0.001 m³

Specific weight = $\frac{7}{0.001}=7000N/m^3$

Density = $\frac{7000}{9.81}=713.5kg/m^3$

Specific gravity = $\frac{713.5}{1000}=0.7135$

Option 1 is the correct answer.

B) The Stokes(St) is the cgs physical unit for kinematic viscosity, named after George Gabriel Stokes.

We have

1 St = 10⁻⁴ m²/s

Option 4 is the correct answer.

7 0

3 years ago

Could anyone help with number 9?

Alisiya [41]

The answer would be A

3 0

3 years ago

A car is moving at a speed of 50 km/hour. The driver takes her foot off of the accelerator and the car coasts to a stop. Why

harina [27]

Removing the foot from the accelerator means that no ac

8 0

3 years ago

Read 2 more answers

The net force acting on an object equals the applied force plus the force of friction.

Georgia [21]

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

3 0

3 years ago

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

4 years ago