All categories

3 years ago

The pointer of an analog meter is connected to a

Physics

2 answers:

dangina [55]3 years ago

8 0

C. coil suspended by bearings.
<span>but im not 100% sure</span>

Rom4ik [11]3 years ago

5 0

Answer: the correct answer is c coil suspended by bearings.

Explanation: this is so that whenever the coil detect a parameter to be measured say current voltage or resistance it energizes and the bearings moves accordingly to indicate the corresponding reading

You might be interested in

What airplanes did the ffa ground earlier this year after several crashes?

frez [133]

The FFA is the Future Farmers of America. It has no authority in aeronautical matters.

8 0

4 years ago

If an airplane is flying directly north at 300.0 km/h, and a crosswind is hitting the airplane at 50.0 km/h from the east, what

Rashid [163]

Answer:

magnitude = 304.14 km/h

direction: $9.46^o$ West of North

Explanation:

The final plane's vector velocity will be the result of the vector addition of one pointing North of length 300 km/h, another one pointing West of length 50 km/h.

To find the magnitude of the final velocity vector (speed) we need to apply the Pythagorean theorem in a right angle triangle with sides: 300 and 50, and find its hypotenuse:

$|v|=\sqrt{300^2+50^2}=\sqrt{92500} = 304.14$ km/h

The actual direction of the plane is calculated using trigonometry, in particular with the arctan function, since the tangent of the angle can be written as:

$tan(\theta)=\frac{50}{300} = \frac{1}{6} \\\theta = arctan(\frac{1}{6} ) = 9.46^o$

So the resultant velocity vector of the plane has magnitude = 304.14 km/h,

and it points $9.46^o$ West of the North direction.

3 0

3 years ago

Which statement is true about the thermal energy of an object? Choose the correct answer. 1). Thermal energy is the internal pot

Brums [2.3K]

This thermal energy flows as heat within the box and floor, ultimately raising the temperature of both of these objects.

5 0

3 years ago

Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b

fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² = 376 J

After coming to a flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

8 0

3 years ago

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

Readme [11.4K]

Answer:

KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= $100\times \dfrac{2\pi}{60}$

=10.47\ rad/s

now, finding the position of center of mass of the system

r₁ + r₂ = 0.4 m.....(1)

equating momentum about center of mass

m₁r₁ = m₂ r₂

0.3 x r₁ = 0.6 r₂

r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

r₂ = 0.4/3

r₁ = 0.8/3

now, calculation of rotational energy

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0

4 years ago