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2 years ago

Why should you use control while during an experiment?

Physics

1 answer:

Sergio039 [100]2 years ago

8 0

To provide a greater certainty that the observed results are not by chance.

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An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a

Lapatulllka [165]

Answer:

The speed of the electron is $2.55\times 10^3\ m/s$ .

Explanation:

Given that,

The magnitude of electric field, $E=1.32\ kV/m=1.32\times 10^3\ V/m$

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then, the magnitude of electric field and magnetic field is balanced such that :

$evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s$

$v=2.55\times 10^3\ m/s$

So, the speed of the electron is $2.55\times 10^3\ m/s$ . Hence, this is the required solution.

3 0

2 years ago

Which of the following is the best example of a primary circular reaction?

olga55 [171]

Primary Circular Reactions (1-4 months): This substage involves coordinating sensation and new schemas. For example, a child may suck his or her thumb by accident and then later intentionally repeat the action. These actions are repeated because the infant finds them pleasurable.

4 0

2 years ago

What is v^2=0.05-4.9 please i need this asap

Margaret [11]

Answer:

v =2.02

Explanation:

v^2=0.05-4.9

v^2=-4.85

square root both side

v=2.02

^^^^this is a not a perfect square

7 0

2 years ago

If the focal length of a concave mirror is 18cm, find its radius of curvature.

Galina-37 [17]

Given :

The focal length of a concave mirror is 18 cm.

To Find :

The radius of curvature of the concave mirror.

Solution :

We know,

$\text{Focal length}=\dfrac{\text{Radius of curvature}}{2}\\\\F=\dfrac{R}{2}\\\\R = 18\times 2\ cm\\\\R = 36 \ cm$

Therefore, the radius of curvature of concave mirror is 36 cm.

Hence, this is the required solution.

8 0

2 years ago

A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai

Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a) Let us assume the depth be h

As we know that

$Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h$

After solving this,

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

Given that

height of the extra fluid is

$h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}$

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

$Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000$

Pb' = 122 KPa

3 0

2 years ago