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Anna [14]
3 years ago
7

When doing scientific research ,the sources used should be ?

Physics
2 answers:
quester [9]3 years ago
6 0

Answer: Reliable and trusted

Sliva [168]3 years ago
4 0

The sources used in the scientific research purpose should be reliable.

<u>Explanation: </u>

In doing the scientific research, the sources must not be of popular kind but should be representative kind which could represent the material and method adopted in the research very well. The sources are never termed as new or old sources, as the source is not bounded by time but space.

The source applicability is defined by the parameter used in the specified location and not time. Therefore, the source must be reliable which could give definite information and trustworthy.

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The area under acceleration time garph represents?​
Keith_Richards [23]

Answer:

Change in Velocity because

at = v

Explanation:

Remeber area is length times Width. In this case, the area under a accleraton vs time graph is Accleration Times Time. Which is velocity

8 0
2 years ago
Read 2 more answers
The equatorial diameter of venus is 7,523 miles. if a mile equals 1.609 km, what is venus's diameter in kilometers?
polet [3.4K]

The diameter of venus in km is 12104.507Km.

<h3 /><h3>What is Unit conversion?</h3>

By definition, unit conversion refers to the division or multiplication operation used to convert measurements of the same quantity between various units. The act of converting something from one form to another in mathematics, such as from inches to millimetres or from litres to gallons, is known as conversion.

the diameter of venus = 7,523 miles

1 mile = 1.609 km

so,

diameter of venus = 7523 × 1.609 Km

= 12104.507Km

to learn more about unit conversion go to - brainly.com/question/13016491

#SPJ4

6 0
1 year ago
A proton moving at 8.00 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-1
gladu [14]

Answer:

19.1 deg

Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ

Sinθ = 0.327

θ = 19.1 deg

4 0
3 years ago
Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
8 0
3 years ago
A 3. 5-a current is maintained in a simple circuit with a total resistance of 1500 ω. What net charge passes through any point i
melamori03 [73]

Electric Current:

Electric current is the flow of charge through a given circuit per unit time. Electric current is one of the components needed to calculate the electric power that a device needs to operate and do work. Electric current is measured in amperes (A), which is equal to:

            1A = 1 C/ s

Recall that the coulomb (C) is the unit for charge while the second (s) is the unit for time

Given: I = 3.5

A is the current

Δt =30 s is the time interval

                     A =ΔQ/ΔT

Net charge = 100C

Electricity is produced when an electric current runs through a circuit.

How does electric current work?

A current of electricity is a steady flow of electrons. When electrons move from one place to another, round a circuit, they carry electrical energy from place to place like marching ants carrying leaves. Instead of carrying leaves, electrons carry a tiny amount of electric charge.

Learn more about Electric current :

brainly.com/question/27003377

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7 0
1 year ago
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