which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload

Why is ammeter connected in series in an electric circuit?

topjm [15]

In order to measure the size of the electrical current flowing in the circuit,
the current must pass through the meter.

8 0

4 years ago

A new band sensation is playing a concert and recording it for a live album to be released this summer. The band asks the sound

Amanda [17]

The sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

<h3>Amplitude of sound wave</h3>

The amplitude of a sound wave is the maximum vertical displacement of the sound wave.

The sound mixer will need to increase the amplitude of the sound wave produced by the singer.

The increase in the amplitude of the sound wave produced by the lower tune singer will result in increased loudness of the sound.

Thus, the sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

Learn more about sound waves here: brainly.com/question/1199084

8 0

2 years ago

A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-> of magnitude 4.8

n200080 [17]

Answer

4.8 N

If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.

The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.

6 0

3 years ago

What are 3 different forces that act on objects on the earth ΩvΩ

Stolb23 [73]

A force is a push or pull acting upon an object as a result of its interaction with another object. There are a variety of types of forces. a variety of force types were placed into two broad category headings on the basis of whether the force resulted from the contact or non-contact of the two interacting objects.

Contact Forces

Action-at-a-Distance Forces

Frictional Force
Gravitational Force
Tensional Force
Electrical Force
Normal Force
Magnetic Force
Air Resistance Force
Applied Force
Spring Force

These are types of individual forces

Applied Force
Gravitational Force
Normal Force
Frictional Force
Air Resistance Force
Tensional Force
Spring Force

6 0

3 years ago

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

3 years ago