Answer:
Kinetic energy
Explanation:
The swing will have potential energy at all points before the lowest, where it will all be converted to kinetic.
What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?
42 degrees
48 degrees
57 degrees
61 degrees
Answer:
61 degrees, I just did the test.
Explanation:
Answer: 61 degrees
Explanation:
I just did the question and got it right
A particle is fallingdown into a medium with an initial velocity of 30m/s. If the acceleration of the particle is =(−4t)/m/s^2 , where t is in seconds, determine the distance traveled before the particle stops.
Answer:
The distance traveled by the particle before it stops is 41.06 m.
Explanation:
We can find the distance traveled by the particle using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex] v_{f}[/tex]: is the final velocity = 0 (when it stops)
[tex] v_{0}[/tex]: is the initial velocity = 30 m/s
a: is the acceleration = -4t m/s²
t: is the time
d: is the distance
First, we need to calculate the time:
[tex] v_{f} = v_{0} + at [/tex]
[tex] 0 = 30 m/s + (-4t m/s^{2})t [/tex]
[tex]0 = 30 m/s - 4t^{2} m/s^{3}[/tex]
[tex]t = 2.74 s[/tex]
Now, the acceleration is:
[tex] a = -4t = -10.96 m/s^{2} [/tex]
Hence, the distance is:
[tex] d = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{-(30 m/s)^{2}}{2*(-10.96 m/s^{2})} = 41.06 m [/tex]
Therefore, the distance traveled by the particle before it stops is 41.06 m.
I hope it helps you!
A stone sphere of radius 7.00 m rests in a flat field. Relative to the ground, what is the gravitational potential energy of a 90.0-kg person sitting on the very top of the sphere?
a. 7.76 x 104 J
b. 1.23 x 104 J
c. 3.88 x 104 J
Answer:
(B) 1.23 x 10⁴ J
Explanation:
Given;
radius of the sphere, r = 7.0 m
diameter of the sphere, d = 2r = 14.0 m
mass of the person sitting on the sphere, m = 90.0 kg
The gravitational potential energy of the person is given by;
P.E = mgh
where;
g is acceleration due to gravity = 9.8 m/s²
h is the height above the ground level = d = 14.0 m
P.E = mgh
P.E = (90)(9.8)(14)
P.E = 12348 J
P.E = 1.2348 x 10⁴ J
Therefore, the gravitational potential energy of the person is 1.2348 x 10⁴ J
Average formula is...?
Answer:
total sum of all numbers/ number of items in the set
Explanation:
big ideas
key questions and terms notes
Answer:
what i and confused
Explanation:
Answer:
nerd
Explanation:
I H4+3 U
Calculate the height from which a body is released rest if its velocity just before hitting the ground is 30ms-1
Answer:
45.9m
Explanation:
Given parameters:
Final velocity = 30m/s
Initial velocity = 0m/s
Unknown:
Height of fall = ?
Solution:
The motion equation to solve this problem is given below;
V² = U² + 2gH
V² = 0 + (2 x 9.8 x H)
30² = 19.6H
H = [tex]\frac{900}{19.6}[/tex] = 45.9m
Suppose we wish to use a 8.0 m iron bar to lift a heavy object by using it as a lever. If we place the pivot point at a distance of 1.0 m from the end of the bar that is in contact with the load and we can exert a downward force of 562 N on the other end of the bar, find the maximum load that this person is able to lift (pry) using this arrangement (neglect the mass of the bar in this problem).
Answer:
W = 3934 N , m = 401.43 kg
Explanation:
This problem can be solved using the rotational equilibrium relation, where we place the reference system at the pivot point and assume that the counterclockwise turns are positive.
∑τ = 0
F 7 - W 1 = 0
W = F 7/1
W = 562 7
W = 3934 N
W = mg
m = W / g
m = 3934 / 9.8
m = 401.43 kg
Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30 meter per seconds?
Explanation:
u = 0m/s
v = 30m/s
a = g = 10m/s²
s = H = ?
using the formula,
v² = u² + 2as
30² = 0² + 2(10)(H)
900 = 20H
H = 900/20
H = 45m
Answer:
the answer is
45m - height
how many significant figures are in 46 m?
Is a light bulb that is on potential or kinetic?
Answer:
pretty sure its kinetic
Explanation:
Answer:
Kinetic
Explanation:
The stored chemical potential energy of a battery converts to electrical kinetic energy to transport electricity to a light bulb, which radiates thermal kinetic energy.
A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N
Answer:
The correct option is D
Explanation:
From the question we are told that
The maximum electric field strength is [tex]E = 1.0 *10^{11} \ V/m[/tex]
The area is [tex]A = 5.0 \ mm^2 = 5.0 *10^{-6} \ m^2[/tex]
Generally the force the laser applies is mathematically represented as
[tex]F = \epsilon_o * E ^2 * A[/tex]
Here [tex]\epsilon_o = 8.85*10^{-12} C/(V \cdot m)[/tex]
[tex]F = 8.85*10^{-12} * (1.0 *10^{11}) ^2 * 5.00*10^{-6 }[/tex]
=> [tex]F = 4.43 *10^{5} \ N[/tex]
The chemists'_
is another name for the periodic table.
Answer here
Answer: Calendar
Explanation: I'm pretty sure it is calendar.
Some nitrogen at 80.0 psi gauge pressure occupies 13.0 ft^3. Find its volume, in ft^3, at 50.0 psi gauge pressure.
Answer:
20.8 ft³
Explanation:
The following data were obtained from the question:
Initial pressure (P1) = 80 psi
Initial volume (V1) = 13 ft³
Final pressure (P2) = 50 psi
Final volume (V2) =?
The new volume of the gas can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
80 × 13 = 50 × V2
1040 = 50 × V2
Divide both side by 50
V2 = 1040 / 50
V2 = 20.8 ft³
Thus, the volume of the gas at a pressure of 50 psi is 20.8 ft³
Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.400 mmmm wide. The diffraction pattern is observed on a screen 2.90 mm away. Define the width of a bright fringe as the distance between the minima on either side.
Required:
a. What is the width of the central bright fringe?
b. What is the width of the first bright fringe on either side of the central one?
Answer:
a
[tex]y_1 = 0.004589 \ m[/tex]
b
[tex]y_2 =0.009179 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the red light is [tex]\lambda _r = 633 \ nm = 633 *10^{-9} \ m[/tex]
The width of the slit is [tex]d = 0.40 mm = 0.40 *10^{-3} \ m[/tex]
The distance of the screen from the point of diffraction is [tex]D = 2.9 \ m[/tex]
Generally the width of the central bright fringe is mathematically represented as
[tex]y_1 = \frac{\lambda * D}{d}[/tex]
=> [tex]y_1 = \frac{633 *10^{-9} * 2.90 }{0.40 *10^{-3}}[/tex]
=> [tex]y_1 = 0.004589 \ m[/tex]
Generally the width of the first bright fringe on either side of the central one is mathematically represented as
[tex]y_2 = 2 * y_1[/tex]
=> [tex]y_2 = 2 * 0.004589[/tex]
=> [tex]y_2 =0.009179 \ m[/tex]
The capacitors are reconnected in series, and the combination is again connected to the battery. From the same choices, choose the one that is true.
a) The potential difference across each capacitor is the same, and the equivalent capacitance is greater than any of the capacitors in the group.
b) The capacitor with the smallest capacitance carries the largest charge.
c) The potential differences across the capacitors are the same only if the capacitances are the same.
d) All capacitors have the same charge, and the equivalent capacitance is greater than the capacitance of any of the capacitors in the group.
e) The capacitor with the largest capacitance carries the smallest charge.
Answer:
This question appear incomplete
Explanation:
This question appear incomplete. However, when capacitors are connected in series, the total capacitance is usually less than the capacitance of individual capacitor. This is because the total capacitance can be calculated as 1 ÷ 1/C₁ + 1/C₂ +1/C₃...
The formula above does not refer to the charge; this is because capacitors connected in series have the same charge (regardless of the capacitance). Also, the largest potential difference occur in the capacitor with the smallest/lowest capacitance.
A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. If the merry-go-round makes 4.9 rev/min, what is the velocity of the child in m/s?
a.
9.2 m/s
b.
4.6 m/s
c.
0.74 m/s
d.
1.75 m/s
Answer:
b. 4.6 m/sExplanation:
the formula for calculating the linear velocity is expressed as;
v = wr
w is the angular velocity
r is the radius
r = d/2 = 18/2
r = 9 m
Given that 1rev/min = 0.10472rad/s
4.9 rev/min = x
x = 4.9 * 0.10472
x = 0.513128 rad/s
Substitute into the given formula;
v = 9 * 0.513128
V = 4.61m/s
Hence the velocity of the child in m/s is 4.6m/s
What is the index of refraction of a material for which the wavelength of light is 0.671 times its value in a vacuum
Explanation:
We need to find the index of refraction of a material for which the wavelength of light is 0.671 times its value in a vacuum.
The wavelength in any medium is given by :
[tex]\lambda_m=\dfrac{\lambda}{n}[/tex]
Where, [tex]\lambda[/tex] is wavelength in vacuum and n is refractive index of the medium.
[tex]n=\dfrac{\lambda}{\lambda_m}\\\\n=\dfrac{\lambda}{0.671\lambda}\\\\=\dfrac{1}{0.671}\\\\=1.49[/tex]
So, the medium can be PMMA (acrylic, plexiglas, lucite, perspex) which have a refractive index of 1.49.
The index of refraction of the material for which the wavelength of light is 0.671 times, the value of it in a vacuum is 1.49.
What is the index of refraction of a material?The index of refraction of a material is the ratio of wavelength of the wave in vacuum to the wavelength of the wave in the given medium.
The index of refraction is inversely proportional to the wavelength. It can be find out using the following formula.
[tex]n=\dfrac{\lambda}{\lambda_m}[/tex]
Here, (λ) is the wavelength in vacuum and (λm) is the wavelength in a medium.
The index of refraction of a material for which the wavelength of light is 0.671 times its value in a vacuum (0.671×λ).
Plug in the values in the above formula,
[tex]n=\dfrac{\lambda}{0.671\times\lambda}\\n=1.49[/tex]
Thus, the index of refraction of the material for which the wavelength of light is 0.671 times, the value of it in a vacuum is 1.49.
Learn more about the index of refraction here;
https://brainly.com/question/10729741
Why is science important for society? (2 points)
a
It helps make decisions based on scientific reasoning.
b
It proves that all questions can be answered using science.
c
It emphasizes the importance of human choice in decision making.
d
It shows that all social problems can be solved using scientific knowledge.
Answer:
i think its d
Explanation:
i took this quiz and got 100 percent
Scientists monitoring and predicting future volcanic eruptions can givr people living near Mt/ St. Helens prior warning so that they can take action. What additional measures could help reduce the negative impacts of a future eruption?
Answer:
first thing first they really have to concern about this,
and second one is people should leave those hazardous places
Answer:
-Emergency evacuation roadways could be designed.
-An Evacuation plan could be organized.
-Warning Sirens could be designed.
Explanation:
Those 3 are the answer.
WILL MARK BRAINLIEST -- The Earth’s velocity is shown by a red vector. Which best describes the Earth’s velocity?
a. It is constant in magnitude and direction
b. It is constant in magnitude but not constant in direction
c. It is not constant in magnitude but is constant in direction
d. It is not constant in magnitude or direction
Question #2
Like liquids, are gases made of particles?
Yes
No
Maybe
Answer:
Yes
Explanation:
Answer:
YesExplanation:
Remember that all solids, liquids, and gases are made up of atoms, molecules, and / or ions. ( Also remember that the answer "maybe" most likely won't be the answer in most quizzes. )
Which answers your question, gases are made up of particles.
Hope this helps! <3
A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour her average speed is____ block/hour
Answer:
2 m/s
Explanation:
The total time = 1 hour
The vertical displacement = 1 - 1
Vertical displacement = 0
Horizontal displacement = 4 - 2
Horizontal displacement = 2
Total displacement = sqrt (2^2 - 0^2)
Displacement - 2
Average velocity is displacement/time
= 2x1
= 2 m/s
The average velocity is 2 metres per second.
A car is filled up with 20 gallons of gas. The car uses .25 gallons per minute. How much time will the car travel ?
A mass M is attached to a spring with spring constant k. When this system is set in motion with amplitude A, it has a period T. What is the period if the mass is doubled to 2M
Answer:
T' = o.707T
Explanation:
[tex]T = 1/2\pi \sqrt{k/m} \\\\m'=2m then\\T' = 1/2\pi \sqrt{k/m'}\\ = 1/2\pi \sqrt{k/2m}\\T' = \frac{(1/2\pi \sqrt{k/m}\\)}{1.4} T' = \frac{T}{1.4}\\T' = 0.707T[/tex]
during a basketball game what are the most points you can score from shooting and getting fouled
Answer:
If a player is fouled while shooting a three-point shot and makes it anyway, he is awarded one free throw. Thus, he could score four points on the play. Inbounds- If fouled while not shooting, the ball is given to the team the foul was committed upon.
A block is at rest. The coefficients of static and kinetic friction are s = 0.81 and k = 0.69, respectively. The acceleration of gravity is 9.8 m/s^2.
Required:
a. What is the largest angle which the incline can have so that the mass does not slide down the incline?
b. What is the acceleration of the block down the incline if the angle of the incline is 44°?
Answer:
Explanation:
a ) The angle required = angle of repose = θ
Tanθ = .81
θ = 39⁰
b ) when angle of incline θ = 44
Net force on the block = mg sinθ - μ mg cosθ where μ is coefficient of kinetic friction
acceleration = gsinθ - μ g cosθ
= 9.8 ( sin44 - μ cos44 )
= 9.8 ( .695 - .69 x .72 )
= 9.8 ( .695 - .497 )
= 1.94 m /s²
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force of the earth were to act on it (ie., neglect the forces from the sun and other solar system objects), the spacecraft would eventually crash into the earth The mass of the earth is Me and its radius is Re. Neglect air resistance throughout this problem, since the spacecraft is primarily moving through the near vacuum of space
Find the speed s of the spacecraft when it crashes into the earth Express the speed in terms of M, Re, and the universal gravitational constant G.
Answer:
Speed of the spacecraft right before the collision: [tex]\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}[/tex].
Assumption: the earth is exactly spherical with a uniform density.
Explanation:
This question could be solved using the conservation of energy.
The mechanical energy of this spacecraft is the sum of:
the kinetic energy of this spacecraft, andthe (gravitational) potential energy of this spacecraft.Let [tex]m[/tex] denote the mass of this spacecraft. At a distance of [tex]R[/tex] from the center of the earth (with mass [tex]M_\text{e}[/tex]), the gravitational potential energy ([tex]\mathrm{GPE}[/tex]) of this spacecraft would be:
[tex]\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}[/tex].
Initially, [tex]R[/tex] (the denominator of this fraction) is infinitely large. Therefore, the initial value of [tex]\mathrm{GPE}[/tex] will be infinitely close to zero.
On the other hand, the question states that the initial kinetic energy ([tex]\rm KE[/tex]) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.
Right before the collision, the spacecraft would be very close to the surface of the earth. The distance [tex]R[/tex] between the spacecraft and the center of the earth would be approximately equal to [tex]R_\text{e}[/tex], the radius of the earth.
The [tex]\mathrm{GPE}[/tex] of the spacecraft at that moment would be:
[tex]\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}[/tex].
Subtract this value from zero to find the loss in the [tex]\rm GPE[/tex] of this spacecraft:
[tex]\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}[/tex]
Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the [tex]\rm GPE[/tex] of this spacecraft would be equal to the size of the gain in its [tex]\rm KE[/tex].
Therefore, right before collision, the [tex]\rm KE[/tex] of this spacecraft would be:
[tex]\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}[/tex].
On the other hand, let [tex]v[/tex] denote the speed of this spacecraft. The following equation that relates [tex]v\![/tex] and [tex]m[/tex] to [tex]\rm KE[/tex]:
[tex]\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2[/tex].
Rearrange this equation to find an equation for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}[/tex].
It is already found that right before the collision, [tex]\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}[/tex]. Make use of this equation to find [tex]v[/tex] at that moment:
[tex]\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}[/tex].
Is a man kicking ball potential or kinetic?
The mass of Earth is 5.972×1024 kg and its orbital radius is an average of 1.496×1011 m . Calculate its linear momentum.
A rock of mass 0.340 kg is spun horizontally at the end of a wire that has a diameter of 1.00 mm. If the wire gets stretched by 2.00 mm when the rock moves at a speed of 19.0 m/s, what is the Young's modulus of the wire?
Answer:
Y = 78.13 x 10⁹ Pa = 78.13 GPa
Explanation:
First we will find the centripetal force acting on the wire as follows:
F = mv²/r
where,
F = Force = ?
m = mass of rock = 0.34 kg
v = speed = 19 m/s
r = length of wire
Therefore,
F = (0.34)(19)²/r
F = 122.74/r
now, we find cross-sectional area of wire:
A = πd²/4
where,
A = Area = ?
d = diameter of wire = 1 mm = 0.001 m
Therefore,
A = π(0.001)²/4
A = 7.85 x 10⁻⁷ m²
Now, we calculate the stress on wire:
Stress = F/A
Stress = (122.74/r)/(7.85 x 10⁻⁷)
Stress = 1.56 x 10⁸/r
Now, we calculate strain:
Strain = Δr/r
where,
Δr = stretch in length = 2 mm = 0.002 m
Therefore,
Strain = 0.002/r
now, for Young's modulus (Y):
Y = Stress/Strain
Y = (1.56 x 10⁸/r)/(0.002/r)
Y = 78.13 x 10⁹ Pa = 78.13 GPa