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Allushta [10]
3 years ago
12

If the actual mass of a marble is 3.53 g, and you measured it to be 3.22 g, is your measurement accurate?

Physics
1 answer:
adelina 88 [10]3 years ago
4 0
No because it’s too much different between the real value and measured value
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What is the frequency of blue light that has a wavelength of 446 nm?
Maslowich
When I went through with the math, the answer I came upon was: 
<span>6.67 X 10^14 </span>

<span>Here is how I did it: First of all we need to know the equation. </span>

<span>c=nu X lamda </span>
<span>(speed of light) = (frequency)(wavelength) </span>
<span>(3.0 X 10^8 m/s) = (frequency)(450nm) </span>

<span>We want the answer in meters so we need to convert 450nm to meters. </span>
<span>450nm= 4.5 X 10^ -7 m </span>
<span>(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) </span>

<span>Divide the speed of light by the wavelength. </span>
<span>(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- </span>

<span>Answer: 6.67 X 10^14 s- hope this helps</span>
7 0
4 years ago
Read 2 more answers
0.10-kilogram model rocket’s engine is designed to deliver an impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75
UkoKoshka [18]

Answer:

8.0 N

Explanation:

Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).

Mathematically, Fore is expressed as

F = ma ........................... equation 1

Where F = force, m = mass, a = acceleration.

and

I = mΔv

Δv = I/m ............................ Equation 2

Where I = impulse, m = mass, Δv = change in velocity

Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.

Substituting into equation 2

Δv = 6.0/0.1

Δv = 60 m/s.

But

a = Δv/t

where t = time = 0.75 seconds.

a = 60/0.75

a = 80 m/s²

Substitute the values of a and m into equation 1.

F = 0.1(80)

F = 8.0 N.

Thus the average force produced = 8.0 N

6 0
3 years ago
3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of
Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

5 0
2 years ago
This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14
Harrizon [31]

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

3 0
3 years ago
Describe the sun. Include its structure (layers), relative size, location, and effects of its magnetic field. Where does the sun
Salsk061 [2.6K]

Answer: Nuclear fusion.

Explanation: The sun is a medium-sized star, its radius is  695.510 km and its mass is equivalent to that obtained by bringing together about 110 planets equal to Earth (6371 km is its radius).

It has six layers: The core, the radioactive zone, the convective zone, the photosphere, the chromosphere and the corona.

Magnetic field disruptions near active regions can generate strong explosions in the sun such as sun flashes and coronal mass ejections. The degree of complexity of the sun´s magnetic field increases and decreases with the course of each sunspot cycle.  

Sir Arthur Eddington was the first to evaluate all the data and dared to conjecture that nuclear fusion, the process that creates heavy elements from the fusion of lighter ones, could be responsible for the great production of the sun´s energy; this process make the sun´s energy was taken for the earth and the planet get back to the sun recycled energy. The sun has a very large and complex magnetic field; the average magnetic field of the sun is approximately 1 Gauss, almost twice as strong as the average magnetic field of the Earth´s surface (approximately 0.5 Gauss). Because the surface of the sun is more than 12.000 times larger than the Earth, the overall influence of the sun´s magnetic field is immensely large.

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3 years ago
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