1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
marta [7]
2 years ago
15

What are some sources of error in a refraction lab using pins and a glass block?

Physics
1 answer:
ser-zykov [4K]2 years ago
6 0

DONT WORRY

ANSWER KO YAN NG MADALI

PERO BRAINLIEST MUNA

MUAAAAAH

TENCHUUU

You might be interested in
why is the earth more attracted to the sun gravitational pull than the gravitational pull of the moon
Ierofanga [76]
First, let's take a look at the equation for the force of gravity between two objects:

F = (GMm)/r², where, 

G = gravitational constant = 6.67 x 10⁻¹¹
M = mass of one object
m = mass of the other object
r = distance between the two objects

From this equation, we can see that the force of gravity is directly proportional to the mass of the two objects and inversely proportional to the distance between them. We can then say that the Earth is <span>more attracted to the sun than the moon because of the massive mass of the Sun (1.9891 x 10</span>³⁰)<span> compared to moon (7.3577 x 10</span>²²<span>). Although, the moon is nearer to the Earth, it has little effect to bring down the gravitational pull of the Sun. </span>
5 0
3 years ago
What is the name of the process where rocks are being broken down into small grains of sediment or soil?
elena-s [515]

Answer:

<em>a</em><em>.</em><em> </em><em>w</em><em>e</em><em>a</em><em>t</em><em>h</em><em>e</em><em>r</em><em>i</em><em>n</em><em>g</em><em>.</em>

Explanation:

<em>w</em><em>e</em><em>a</em><em>t</em><em>h</em><em>e</em><em>r</em><em>i</em><em>n</em><em>g</em><em> </em><em>i</em><em>s</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>p</em><em>r</em><em>o</em><em>c</em><em>e</em><em>s</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em>b</em><em>r</em><em>e</em><em>a</em><em>k</em><em>i</em><em>n</em><em>g</em><em> </em><em>d</em><em>o</em><em>w</em><em>n</em><em> </em><em>r</em><em>o</em><em>c</em><em>k</em><em>s</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>m</em><em>i</em><em>n</em><em>e</em><em>r</em><em>a</em><em>l</em><em>s</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>s</em><em>m</em><em>a</em><em>l</em><em>l</em><em>e</em><em>r</em><em> </em><em>p</em><em>i</em><em>e</em><em>c</em><em>e</em><em>s</em><em> </em><em>b</em><em>y</em><em> </em><em>w</em><em>a</em><em>t</em><em>e</em><em>r</em><em>,</em><em> </em><em>w</em><em>i</em><em>n</em><em>d</em><em>,</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>i</em><em>c</em><em>e</em><em>.</em>

4 0
3 years ago
A sound travels down a hallway that is 115 m long. Then it echoes and
yan [13]

Answer:

0.66 s

Explanation:

∆x = v∆t → 2 × 115 = 350 ∆t → ∆t = 230/350 = 0.66 s

7 0
2 years ago
I need answers and solvings to these questions​
den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) 15.7 rad/s^2

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f is the final angular velocity

\omega_i is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

\omega_i = 0 (since it starts from rest)

\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s is the final angular velocity

t = 2 s

Substituting, we find

\alpha = \frac{31.4-0}{2}=15.7 rad/s^2

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

a=-\omega^2 x

where

a is the acceleration

x is the displacement

\omega is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

a=-100 x

which implies that

\omega^2 = 100

And so

\omega = \sqrt{100}=10 rad/s

Also, the angular frequency is related to the frequency f by

f=\frac{\omega}{2\pi}

Therefore, the frequency of this simple harmonic oscillator is

f=\frac{10}{2\pi}=1.6 Hz

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

mg=kx

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5

The angular frequency of the spring is given by

\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz

And therefore, its period is

T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the hose, with r_1 = 5 mm being the radius of the hose

v_1 = 4 m/s is the speed of the petrol in the hose

A_2 = \pi r_2^2 is the cross-sectional area of the nozzle, with r_2 being the radius of the nozzle

v_2 = 16 m/s is the speed in the nozzle

Solving for r_2, we find the radius of the nozzle:

\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm

So, the diameter of the nozzle will be

d_2 = 2r_2 = 2(2.5)=5.0 mm

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

\frac{F_1}{A_1}=\frac{F_2}{A_2}

where

F_1 is the force that must be applied to the small piston

A_1 = \pi r_1^2 is the area of the first piston, with r_1= 2 cm being its radius

F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N is the force applied on the bigger piston (the weight of the car)

A_2 = \pi r_2^2 is the area of the bigger piston, with r_2= 15 cm being its radius

Solving for F_1, we find

F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0
3 years ago
Energy from solar radiation may be ________ or taken in by a surface or an object.
tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

7 0
3 years ago
Other questions:
  • Which statement is true about industrial-organizational psychologists? They apply psychological principles to the workplace. The
    7·2 answers
  • As a student is performing a double slit experiment to determine the wavelength of a light source, she realizes that the nodal l
    7·1 answer
  • if the speed of sound through the air is rounder 80 meters per second what will be the change in temperature of the air​
    10·1 answer
  • The molecules of a substance are moving rapidly in a container. The molecules do not interact with each other. If the container
    15·2 answers
  • Which of the following is a method used to prevent soil erosion?
    6·1 answer
  • When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same
    12·2 answers
  • You drop a 10 kg bowling ball off a 2 story building, what is the force generated by
    8·1 answer
  • Join not for bâd purpose​
    8·2 answers
  • An airplane travels 1200 km in 90 minutes. What is the average speed in m/s for this trip?
    11·1 answer
  • A t-rated switch may be used to (its/a) ______ current capacity when controlling an incandescent lighting load.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!