Answer:
y= y' 0.67
Explanation:
This is an exercise in refraction of light,
n₁ sin θ₁ = n₂ sin θ₂
where subscript 1 is used for the incident medium and subscript 2 for the refracted medium
sin θ₁ = n2 /n1 sin θ₂
the incident medium is air with refractive index n1 = 1 and the medium where the ray is refracted is water with n = 1.33
let's calculate
sin θ₁ = 1.33 / 1 sin 25
θ₁ = sin⁻¹ (0.56208)
θ₁ = 34.2º
when the ray is refracted we can assume that the adjacent leg (water surface) is the same for the two media
let's use the trigonometry relationship
tan θ₁ = x / y
tan θ₂ = x / y '
tan θ₁ = y’ tan θ₂
y = y ’ tanθ₂ / tan θ₁
to finish exercise you must know the depth of the object
y =y' tan 25/ tan 34.2
y= y' 0.67
The atomic of nitrogen is 7. The number of electrons a neutral atom has is_, and its atomic mass is approximately_amu.
premium
A student releases a marble from the top of a ramp. The marble increases speed
while on the ramp then continues across the floor. The marble travels a total of
150cm in 4.80s.
What was the marble's final speed?
Explanation:
the formula of speed is distance traveled by time it work
31.25 cm/sec
__________________________________________________________
Explanation:We are given:
Distance travelled = 150 cm
Time taken = 4.8 seconds
Final Speed of the Marble:
Speed of the marble = Distance travelled / Time taken
Speed of the marble = 150 / 4.8
Speed of the marble = 31.25 cm/sec
A hockey player skating at 21 m/s comes to a complete stop in 12.0 m.
a) What is its acceleration during this displacement?
b) How long did it take the hockey player to come to a stop?
Answer:
a. 18.375m/s²
b. 1.142s
Explanation:
A. Using v² = u² + 2as
Where;
V = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)
s = distance (m)
In this question, v = 21m/s, u= Om/s, s = 12m
21² = 0² + 2 × a × 12
441 = 0 + 24a
441 = 24a
a = 18.375m/s²
B). Using V = u + at
21 = 0 + 18.375t
21 = 18.375t
t = 21/18.375
t = 1.142s
Consider a compact car that is being driven
at 113 km/h. The acceleration of gravity is 9.8 m/s^2.
From what height would the car have to be
dropped to have the same kinetic energy?
Answer in units of m.
A ball is thrown 24 m/s into the air. How high does it go?
556.4 m
0 m
29.4 m
-556.4 m
Answer:
option c is correct
Explanation:
we know that
2as=vf^2-vi^2
vf=24 m/s
vi= 0 m/s
a=g= 9.8 m/s^2
s=vf^2-vi^2/2a
s=(24)²-(0)²/2*9.8
s=576/19.6
s=29.4 m
therefore option c is correct
the distance from Alex home to his school is 1km and 560 cm. what is this distance?
Answer:
Required Answer:-1meter=100cm1km=1000m[tex]{:}\longrightarrow[/tex][tex]\sf 1km=1000×100cm [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 1km=100000cm [/tex]
Total distance
[tex]{:}\longrightarrow[/tex][tex]\sf 100000+560=1000560cm [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 1000.56m[/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf 1.560km [/tex]
The mass of a paper-clip is 0.50 g and the density of its material is 8.0g/cm'. The total volume of
a number of clips is 20 cm.
How many paper-clips are there?
Answer:
320 paper clips
Explanation:
mass = volume × density = 20cm³ × 8g/cm³ = 160g
mass of 1 paper clip = 0.50g
mass of x paper clips = 160g
x = 160/0.50 = 320
What is the relationship between resistance and current in a circuit with no change in voltage?
A. Current and resistance must be equal in a circuit.
B. A circuit that has more resistance will have smaller current.
C. Current does not depend on resistance in a circuit.
D. A circuit that has more resistance will have a greater current.
Answer:
A
Explanation:
What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?
PLEASE SHOW WORK.
Answer:
The answer is 5.4 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{135}{25} = \frac{27}{5} \\ [/tex]
We have the final answer as
5.4 kgHope this helps you
If two balloons are charged and brought close to each other while hanging you observe that they move away from each other. What would the observation be if the balloons have larger charge?
A. There is no further effect. The repulsion magnitude is always the same.
B. There is no further effect. The magnitude of the force is determined by the charge signs not their magnitude
C. The balloons will lift in addition to separating as now they can start to overcome gravitational forces
D. The balloons will separate further as the repulsion magnitude increases
E. The balloons will come closer as the charges create larger polarization forces
F. The balloons will spin around each other since the electric force can produce rotational motion
Answer:
D. The balloons will separate further as the repulsion magnitude increases.
A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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A child and sled with a combined mass of 54.8
kg slide down a frictionless hill that is 11.5 m
high at an angle of 36 degrees
from horizontal.
The acceleration of gravity is 9.81 m/s^2.
If the sled starts from rest, what is its speed
at the bottom of the hill?
Answer in units of m/s.
Answer:
15.02 m/s.
Explanation:
Given that the height of the hill, h= 11.5 m.
Combined mass, m= 54.8 kg
The initial velocity of the combined mass, u=0
Acceleration due to gravity, [tex]g = 9.81 m/s^2[/tex].
Angle of the path the horizontal, [tex]\theta = 36[/tex] degree.
Let A be the initial position and B be the final position of the sled as shown in the figure.
The path is frictionless so the drag force =0
The gravitational force acting on the combined mass in the downward direction, [tex]F= mg\cdots(i)[/tex]
The component of force acting in the direction of motion = [tex]F\sin \theta.[/tex]
Let [tex]a[/tex] be the acceleration of the combined mass, m, So,
[tex]F\sin \theta= ma[/tex]
[tex]\Rightarrow mg \sin \theta= ma[/tex] [ from equation (i)]
[tex]\Rightarrow a = g \sin \theta \cdots(ii).[/tex]
Let v be the final velocity of the combined mass.
Now, by using the equation of motion,
[tex]v^2=u^2+2as\\\\\Rightarrow v^2=0^2+2as\\\\ \Rightarrow v^2=2as\cdots(iii)[/tex]
Here, s is the displacement in the direction of motion,
So, s= AB
Now, in the right-angled triangle ABO,
[tex]\sin\theta = OA/AB= h/AB\\\\\Rightarrow AB = h/ \sin\theta\\\\\Rightarrow s = h/ \sin\theta\cdots(iv)[/tex]
Now, from equations (ii), (iii) and (iv), we have
[tex]v^2= 2\times g \sin \theta \times \frac {h}{\sin\theta}\\\\\Rightarrow v^2= 2gh\\\\\Rightarrow v= \sqrt{2gh}[/tex]
By using the given values, we have
[tex]v= \sqrt{2\times 9.81\times 11.5}=\sqrt {225.63}\\\\\Rightarrow v = 15.02 m/s[/tex]
Hence, the speed of the combined mass at the bottom = 15.02 m/s.
The speed of the sled at the bottom of the hill is [tex]15.02m/s[/tex]
The speed of sled is calculated by using Newton's law of motion,
[tex]v^{2} =u^{2} +2gh[/tex]
where u is initial velocity, v is final velocity , g is acceleration due to gravity and h is height.
Given that, [tex]u = 0, g = 9.81m/s^{2}[/tex] and [tex]h = 11.5 m[/tex]
Substitute values in above equation.
[tex]v^{2}=0^{2}+2*9.81*11.5\\\\v^{2}=225.63\\\\v=\sqrt{225.63}=15.02m/s[/tex]
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answer plz answer plzzz I am a little confused with full time
PLEASE HELP PLS ITS DUE IN 10 MINUTES!!
Suppose you’re standing at the top of a 100m high building. How much potential energy would you lose if you walked down the stairs by 20m?
You are now standing at a height of 80m. How much potential energy would you lose if you went down by 20m?
Suppose now you decide to go down by 20m more (you’re standing at 60m high). How much potential energy would you need to get to 40? Can you draw any conclusions about potential energy from this observation?
Answer:
I AM SO SORRY
but here is the answer
Explanation:
20 j
a car with a mass of 1200 kg travels a distance of 150 M as it moves from one stoplight to the next at its fastest the car travels at 22 m per second what is its kinetic energy at this point
Answer:
ke = 1/2mv^2
= 1/2 * 1200 * 22^2
= 600 * 484
= 290400 J
Hope it helped u,
pls mark this as brainliest
and put thanks
^_^
Answer:
The kinetic energy at this point 290,400
Climate is the day-to day condition of an area including temperature, pressure, and precipitation. Weather is the usual pattern of temperature, pressure, and precipitation pf an area over time. True or False
Answer:
False
Explanation:
Weather is the day-to day condition of an area including temperature, pressure, and precipitation. Climate is the usual pattern of temperature, pressure, and precipitation pf an area over time.
It is true that the climate is the day-to day condition of an area including temperature, pressure, and precipitation. Weather is the usual pattern of temperature, pressure, and precipitation pf an area over time.
What is the relation between climate and weather?The weather is the current state of the atmosphere for a particular place and at a definite or short period of time. The atmospheric conditions which are considered are temperature, cloudiness, dryness, humidity, rain, sunshine and wind.
The climate is the atmospheric conditions of a particular place over a long time period.
It is the condition of the Earth and the atmosphere which tells us about the extent at either it is hot or cold, wet or dry, or it can be calm or stormy type. Basically the weather changes and happened at the least level of the atmosphere and the layer is known to be the troposphere that is found just below the stratosphere. Troposphere is the layer present in the most lower level forming the Earth's atmosphere. In this layer of has 75% mass out of atmospheric mass and 99% of the total mass of water vapor.
So, the given statement is true.
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What is the difference between renewable energy non-renewable energy? Use in your own words.
answer plese lang po
Answer:
hi
Explanation:
how to convert gumbauan baboy to kung burger p
The radius of the earth’s orbit is r = 1.5 × 1011 m and its orbital period is T = 365.3 days. What is the radial acceleration (in m/s2) of the earth toward the sun, assuming a circular orbit?
A. 5.91 × 10−3 m/s2
B. 3.23 × 10−3 m/s2
C. 7.23 × 10−4 m/s2
D. 5.91 × 10−2 m/s2
The radial acceleration of a body a distance R from the center of a circular path and with period T has magnitude a such that
a = 4 π ² R / T ²
The given period is
T = 365.3 days ≈ 3.156 x 10⁷ s
(i.e. multiply the number of days by 24 hours/day and 3600 seconds/hour)
So the Earth's radial acceleration is
a = 4 π ² (1.5 x 10¹¹ m) / (3.156 x 10⁷ s)²
a ≈ 0.0059 m/s²
a piping system consists of 100 ft of 2-inch pipe, a sudden expansion to 3-inch pipe, and then 50 ft of 3-inch pipe. Water is flowing at 100 gal/min through the system. What is the pressure difference from one end of the pipe to the other
Answer:
16+15+19= ??
Am just messign with u lol
Explanation:
anwser s 19 inches
i
experiment to show that light travels in a straight line
A kid runs and slides down a slip-n-slide. Once the kid hits the slide they have 200N of friction force acting on them, and they decelerate at 2.5m/s. What is the mass of the kid?
A cathode ray tube is made of glass with a small amount of some kind of gas in it. It has metal electrodes at each end to pick up an electric current. The electrodes are named "positive” and "negative." What are one of the main uses of this device ? to speed up the flow of current flowing through a wire to examine a beam of charged particles to magnetize different atoms within a space to change the strength of a proton within an atom
One of the main uses of this device is to examine a beam of charged particles
The cathode ray tube is a device used to determine the charge flowing in a gas. When an electric field is set up with the help of metal electrodes, the cathode ray tends to bend towards the positive electrode.
Since the cathode ray bends towards the electrodes, it implies that it has a charge and the electrodes present help us determine the charge of the beam of charged particles and thus examine the beam of charged particles.
So, one of the main uses of this device is to examine a beam of charged particles.
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Answer:
B
Explanation:
Two blocks with different masses are dropped, hitting the ground with the same velocity. Which of the following is true?
They have same change in velocity but different changes in kinetic energy
The lighter object started at a smaller height.
The heavier object started at a smaller height
They started at the same height
They have same change in kinetic energy but different changes in velocity
Answer: • They have same change in velocity but different changes in kinetic energy
•They started at the same height.
Explanation:
First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.
Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.
Therefore, based in the explanation, the correct options are:
• They have same change in velocity but different changes in kinetic energy
•They started at the same height.
A 0.20 kg mass (m1) hangs vertically from a spring and an elongation of the spring of 9.50 cm (r1) is recorded. With a mass (m2) of 1.00 kg hanging on the spring, a second elongation (r2) of 12.00 cm is recorded. Calculate the spring constant k in Newtons per meter (N/m). (Note: The equilibrium position is not zero.)
Answer:
k=320N/m
Explanation:
Step one:
given data
Let the initial/equilibrum position be x
mass m1= 0.2kg
F1= 0.2*10= 2N
elongation e= 9.5cm= 0.095m
mass m2=1kg
F2=1*10= 10N
elongation e= 12cm= 0.12m
Step two:
From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F
F=ke
2=k(0.095-a)
2=0.095k-ka----------1
10=k(0.12-a)
10=0.12k-ka----------2
solving equation 1 and 2 simultaneously
10=0.12k-ka----------2
- 2=0.095k-ka----------1
8=0.025k-0
divide both side by 0.025
k=8/0.025
k=320N/m
A beam contains 4.9 × 108 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 4.6 × 105 m/s. What is (a) the magnitude of the current density and (b) direction of the current density (c) what additional quantity do you need to calculate the total current i in this ion beam?
Answer:
[tex]72.12\ \text{A/m}^2[/tex]
south
cross sectional area of the beam
Explanation:
v = Velocity of ions = [tex]4.6\times 10^5\ \text{m/s}[/tex]
Number of ions per [tex]\text{cm}^3[/tex] = [tex]4.9\times 10^8[/tex]
Charge density would be the product of number of ions per [tex]cm^3[/tex] and the charge of electrons multiplied by 2 as they are doubly charged.
[tex]\rho_q=4.9\times 10^8\times 10^6\times 2\times 1.6\times 10^{-19}\\\Rightarrow \rho_q=0.0001568\ \text{C/m}^3[/tex]
Current density is given by
[tex]J=\rho_qv\\\Rightarrow J=0.0001568\times 4.6\times 10^5\\\Rightarrow J=72.12\ \text{A/m}^2[/tex]
The current density is [tex]72.12\ \text{A/m}^2[/tex]
The direction of the current density is opposite to the movement of the charged particle. The particles are moving north so the direction of current density will be to the south.
Current is given by
[tex]I=JA[/tex]
where A is the cross sectional area of the beam .
So the cross sectional area of the beam is required to determine the total current in this ion beam.
Sandra pays $11 for 2.75 pounds of cheese. What is the cost per pound?
$0.40
x
$0.44
$4.00
$4.40
Hi
Answer:
$4.00
Explanation:
The cost per pound is $4.00, $11 divided by 2.75 is 4.
Answer:
Th cost per pound is $4.00.
Explanation:
11/2.75= 4
What is magnet made of
Answer:
metals like iron or nickel
Explanation:
When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) what is the effective force constant of the springs? in N/m (b) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?
Answer:
Explanation:
F=kx
x=F/k
F=2000 kg
x=100 cm=9*10^-3
effective spring constant=k=F/x
k=2000/9*10^-3=2.2*10^-5
now frequency
f=1/2π√k/m
f=1/2*3.14√2.2*10^-5/310
f=1/6.28√7.097*10^-8
f=1/6.28*2.7*10^-4
f=0.16*2.7*10^-4
f=4.32*10^-5
The effective spring constant of the springs is 33755.55 N/m.
The frequency of the car's vibration is 2.07 Hz.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
Weight of the four people: F = 310 × 9.80 N = 3038 Newton.
The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.
Hence, the effective spring constant of the springs: k= force/compression
= 3038 N/0.90 × 10⁻² m
= 33755.55 N/m.
The frequency of the car's vibration is: f = 1/2π√(k/m)
=1/2π√(33755.55/2000)
= 2.07 Hz.
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A car slows from 37.3 m/s to 8.8 m/s with a constant acceleration of -3.3 m/s2. How long in seconds does it require?
Answer:
The time it takes the car to stop is 8.64 s.
Explanation:
Given;
initial velocity of the car, u = 37.3 m/s
final velocity of the car, v = 8.8 m/s
constant acceleration of the car, a = -3.3 m/s²
The time it takes the car to stop is given by;
[tex]t = \frac{v-u}{a}[/tex]
Substitute the givens;
[tex]t = \frac{v-u}{a}\\\\t = \frac{8.8-37.3}{-3.3}\\\\t = 8.64 \ s[/tex]
Therefore, the time it takes the car to stop is 8.64 s.