Answer:
static force
Explanation:
mark me brainliest
A red pool ball is rolling directly east before it collides with the
white cue ball moving directly north. Due to conservation of
momentum the total momentum of both objects after the
collision would be in which direction?
Answer: North East
Explanation: Trust me, I was just doing this on the Ck-12 and this the answer I choose and It said I'm correct.
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.03 mm apart and position your screen 3.19 m from the slits. Although Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 641 nm . How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe
Answer:
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 2.978 mm
Explanation:
We have the following data:
λ = wavelength of light = 641 nm = 6.41 x 10⁷ m
L = Distance of Screen from slits = 3.19 m
d = slit separation = 1.03 mm = 1.03 x 10⁻³ m
Δx = distance between consecutive bright fringes = fringe spacing = ?
Using formula:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\Delta x = \frac{(6.41\ x\ 10^{-7}\ m)(3.19\ m)}{1.03\ x\ 10^{-3}\ m}[/tex]
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 1.5Δx
distance between central bright fringe and second dark fringe = (1.5)(1.985 mm)
distance between central bright fringe and second dark fringe = 2.978 mm
SIMPLE HARMONIC MOTION 1.0 Objective To study simple harmonic motion by observing the motion of a simple pendulum. 2.0 Simple harmonic motion Oscillatory motion is extremely common in nature. Examples include waves (water, sound, earthquake, etc.) and vibrations produced by musical instruments. If the oscillation is characterized by a constant frequency and amplitude (if the motion reproduces itself in a fixed time period T), then the motion is said to be "harmonic." If the oscillation can be described as a sinusoidal function of time and position, the motion is said to be "simple harmonic." Simple harmonic motion (SHM) occurs when for every applied force or torque, there is a restoring force or torque which is proportional to the displacement of the system from its equilibrium position. 2.0.1 Name two real-world examples of harmonic motion. (Hint: Southern California is known for what sort of natural disasters?)
Answer:
the waves in the sea, leaves of the trees, cables in the bridges, pendulum clock
Explanation:
In nature there are many examples of simple harmonic motion, for example.
* The movement of the waves in the sea is an oscillation movement up and down
* The movement of the leaves of the trees when a wind blows and then stops, but the leaf and branches are oscillating
* The movement of the cables in the bridges, especially in the suspension bridges
* The movement of a pendulum clock
1. A horizontal force of 50 N is applied to push a desk 40 m across a
warehouse floor. How much work is done?
2000 J
1000 J
3000 J
2001 J
Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0
m/s when the cord makes an angle of 30 degrees with the vertical. What
is the speed (V) of the ball when it passes the lowest position?
Answer:
v = 2.57 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. When it is at an angle of 30º
Em₀ = K + U = ½ m v₁² + m g y₁
final point. Lowest position
Em_f = K = ½ m v²
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v₁² + m g y₁ = ½ m v²
Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle
cos 30 = L ’/ L
L ’= L cos 30
y₁ = L -L '
y₁ = L- L cos 30
we substitute
½ m v₁² + m g L (1- cos 30) = ½ m v²
v = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]
let's calculate
v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]
v = 2.57 m / s
Select four of the following that would increase the magnetic field of an electromagnet
Answer:
The correct answers are: A, C, D, E
Explanation:
The magnetic field is a solenoid is given by
B = μ₀ [tex]\frac{N}{L}[/tex] I
where N is the number of turns, I the current and L length of the solenoid.
Using this equation let's examine the different responses to permute increasing the magnetic field
A) True. a thicker wire decreases the resistance and the current can increase the system.
B) False. If there is no voltage source there is no current in the system
C) True. the field is proportional to the number of turns
D) True. the magnetic moments of the core align with the field increasing its value
E) True. When the loops are closer together, more of them can fit per unit length
F) False. If the wire is shorter the number of turns decreases.
The correct answers are: A, C, D, E
A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 6.9 s later. How high is the cliff?
5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.
Answer: 0.8 g/cm
Explanation:
p= m/V
= 4 kg/ 5 liter
= 0.8
Answer:
.8
Explanation:
give them brainliest
A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They use rope with a mass per length of 1.50 kg/m under 6000 N tension. If they are limited to a wave amplitude of 0.500 m, what must be the frequency of waves necessary to transmit power at the average rate of 2.00 kW
Answer:
the required frequency of waves is 2.066 Hz
Explanation:
Given the data in the question;
μ = 1.50 kg/m
T = 6000 N
Amplitude A = 0.500 m
P = 2.00 kW = 2000 W
we know that, the average power transmit through the rope can be expressed as;
p = [tex]\frac{1}{2}[/tex]vμω²A²
p = [tex]\frac{1}{2}[/tex]√(T/μ)μω²A²
so we solve for ω
ω² = 2P / √(T/μ)μA²
we substitute
ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²
ω² = 4000 / 23.71708
ω² = 168.65
(2πf)² = ω²
so
(2πf)² = 168.65
4π²f² = 168.65
f² = 168.65 / 4π²
f² = 4.27195
f = √4.27195
f = 2.066 Hz
Therefore, the required frequency of waves is 2.066 Hz
Which is an example of a noncontract force?
(A) elastic force
(B) Normal force
(C)Applied force
(D) Electric Force
Answer:
electric force
Explanation:
its a contact and noncontact force
Two students Tim and Alane travel to South Dakota. Tim stands on Earth’s surface and enjoys some sunshine. At the same time, Alane descends into a gold mine where neutrinos are detected, Although the photon at the surface and the neutrinos in the mine arrive at the same time, they have had very different histories. Describe the differences.
Answer:
Explanation:
Neutrinos are otherwise called leptons. They are principal particles. A lepton is a rudimentary half-spin molecule that doesn't go through solid reactions. Neutrinos are not usually charged and exceptionally light weighted so they once in a while interface with other matter. Neutrinos are light weighted. Their mass is around 10⁻⁷ kg. A neutrino possesses a small radius, too little to ever be estimated. A little span and very less mass make them imperceptible. Since neutrinos have next to no mass. they travel at almost the speed of light and thus they arrive at the outside of the Sun in only 2 seconds, dissimilar to photons which take convoluted ways to arrive at the Sun's surface in a huge number of years.
The photon and neutrino, both were made in the Sun's center yet on various occasions. The neutrino is only a couple of minutes old though the photon is around 1,000,000 years of age. At the point when the photon was made in the Sun's center. it needed to venture out to the outside of the Sun. in any case, rather because of its hefty mass and cooperation with other matter, it headed out a crisscross way to the surface. Ordinarily, it was repulsed and it was sent back to the middle where it needed to begin once more. It required a large number of years for a photon to arrive at the outside of the Sun.
Nonetheless, when it arrived at the Sun's surface, it required just 8.8 minutes for the photon to arrive at Earth. The neutrino was anyway made only a couple of minutes prior in the Sun's center. Since it has an entirely irrelevant mass, little size, and no charge, it didn't interface with its environmental factors. So it just required 2 seconds for the neutrino to arrive at the Sun's surface. When it arrived at the Sun's surface, it arrived at the earth in about 8.8 minutes. with the photon. So both, photon and neutrino have various histories as the two of them were made at a hole of around 1,000,000 years.
A 2.35-kg rock is released from rest at a height of 21.4 m. Ignore air resistance and determine (a) the kinetic energy at 21.4 m, (b) the gravitational potential energy at 21.4 m, (c) the total mechanical energy at 21.4 m, (d) the kinetic energy at 0 m, (e) the gravitational potential energy at 0 m, and (f) the total mechanical energy at 0 m.
Explanation:
Given that,
The mass of rock, m = 2.35-kg
It was released from rest at a height of 21.4 m.
(a) The kinetic energy is given by : [tex]E_k=\dfrac{1}{2}mv^2[/tex]
As the rock was at rest initially, it means, its kinetic energy is equal to 0.
(b) The gravitational potential energy is given by : [tex]E_p=mgh[/tex]
It can be calculated as :
[tex]E_p=2.35\times 9.8\times 21.4\\\\E_p=492.84\ J[/tex]
(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,
M = 0 J + 492.84 J
M = 492.84 J
Hence, this is the required solution.
Diffraction occurs for all types of waves, including sound waves.
a. True
b. False
Answer:
a. True
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases. A student could verify this statement by measuring the time required for sound to travel a set distance through a solid, a liquid, and a gas.
Mathematically, the speed of a sound is given by the formula:
[tex] Speed = wavelength * frequency [/tex]
Generally, the frequency of a sound wave determines the pitch of the sound that would be heard.
Diffraction occurs for all types of waves, including sound waves.
If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1x104 J of energy each second, how long does it take for the bridge’s oscillations to go from 0.1 m to 0.5 m amplitude?
Answer: Hello, Mark me as Brainliest! :)
If soldiers march across the bridge with a cadence equal to the bridge's natural frequency and impart $$1.00 × 10^4 J$$ of energy each second, how long does it take for the bridge's oscillations to go from 0.100 m to 0.500 m amplitude. $ 5 \times 10^7 \text{J} $ . \\ b) $ 12 \times 10^4 \text{s}$ .
Your Welcome!
Explanation:
As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.
Answer:
102.5N
Explanation:
Given that a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W
The power = 4.12 × 746 = 3073.52 W
Using the formula
Power = force × velocity
3073.52 = force × 30
Force = 3073.52 / 30
Force = 102.5 N
Since most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road, therefore,
the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s is 102.5 N
The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
I don’t understand this
Answer:
true
Explanation:
force or powerbecause he pushes a disk
How can Newton’s laws of motion and the law of gravitation predict the motion of an object, and how can frames of reference be used to describe that motion?
(1) Which appliance is designed to transfer electrical energy to kinetic energy?
D)
A food mbuer
BB kettle
Clamp
D radio
Answer:
bb kettle
Explanation:
it transfres electricsl to kinetic
an inventor makes a clock using a brass rod and a heavy mass as a pendulum.WHAT Happens when the clock get colder?
An inventor makes a clock using a brass rod and a heavy mass as a pendulum. when the clock gets colder then the time clock would gain time
What is thermal expansion?The expansion of any material due to the variation of the temperature is known as thermal expansion. It varied differently for different materials according to their corresponding values of the coefficient of the thermal expansion.
As given in the problem statement that an inventor makes a clock using a brass rod and a heavy mass as a pendulum, when the clock gets colder then the length of the brass decreases due to thermal expansion.
The length of the pendulum gets reduced which further results in the reduction in the time period, as per the formula of the time period for the pendulum
T = 2π√(L/g)
As the length of the brass gets reduced. This means the pendulum of the clock moves faster and the clock would gain time
Thus, if a pendulum made of a heavy mass and a brass rod is used to create a clock by an inventor. The time clock would advance in time as the clock get colder
Learn more about Thermal expansion here
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23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.
Answer:
141.18 ohms
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12
Current (I) = 0.085 A
Resistance (R) =?
The resistance needed can be obtained as follow:
V = IR
12 = 0.085 × R
Divide both side by 0.085
R = 12 / 0.085
R = 141.18 ohms
Therefore, a resistor of resistance 141.18 ohms is needed.
You exert a 138 N push the leftmost of two identical blocks of mass 244 g connected by a spring of stiffness 605 kg/s2. After pushing the block a distance 15 cm, you release it; by this time the rightmost block has moved a distance 5 cm. (a) What is the energy in the oscillations between the blocks
Answer:
the energy in the oscillations between the blocks is 3.025 J
Explanation:
Given the data in the question;
Force f = 138 N
stiffness of spring k = 605 kg/s²
mass of block = 202 g = 0.202 kg
pushing the block a distance 15 cm, the rightmost block has moved a distance 5 cm
i.e
x₁ = 15 cm
x₂ = 5cm
the energy in the oscillations between the blocks will be;
E[tex]_A[/tex] = E[tex]_B[/tex] = [tex]\frac{1}{2}[/tex]k( Δx )²
we substitute
= [tex]\frac{1}{2}[/tex] × k( 15 - 5 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × ( 10 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × 100 × 10⁻⁴
= 3.025 J
Therefore, the energy in the oscillations between the blocks is 3.025 J
Is there a way to see moon and the sun at once?
a pendulum clock having Copper keeps time at 20 degree Celsius it gains 15 second per day if cooled to 0°C celsius calculate the coefficient of linear expansion of copper.
?.............................
a boy throws a ball straight up into the air it reaches the highest point of its flight after 4 seconds how fast was the ball going when it left the boy's hand
Answer:
Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.
Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Erase button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one with 12 V, one with 15 V, and one with 6 V). Don't worry about getting these exact values. You can be off by a few tenths of a volt. Which statement best describes the distribution of the equipotential lines?
1. The equipotential lines are closer together in regions where the electric field is weaker.
2. The equipotential lines are closer together in regions where the electric field is stronger.
3. The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.
Answer:
B or 2
Explanation:
A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius
Answer:
a) σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] , b) σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂, σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Explanation:
a) The very useful concept of charge density is defined by
σ = Q / A
In this case we have a circular disk
The are of a circle is
A = π r²
in this case we have a hole in the center of radius r = b, so
A_net = π r² - π r_ {hollow} ²
A_ {net} = π (a² - b²)
whereby the density is
σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]
b) The density of the other disk is
σ = Q₂ / A₂
σ = [tex]\frac{Q_2}{d^2}[/tex]
c) The total waxed load is requested by the larger circle
Q_ {total} = Q₁ + Q₂
the net charge density, in the whole system is
σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]
the area is
A_{total} = π a²
since the other circle is inside, we are ignoring the space between the two circles
σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
what is the definition of mutual flux?
Answer:
Is where two or more inductors are “linked” so that voltage is induced in one coil proportional to the rate-of-change of current in another
A small dog is trained to jump straight up a distance of 1.1 m. How much kinetic energy does the 7.7 kg dog need to jump this high?
Answer:
83.09 J
Explanation:
The potential energy at the point of the top of the jump is represented by the equation
[tex]mgdeltah[/tex]
when the dog jumps, all the potential energy converts to kinetic energy (1/2mv^2). Plugging in the values:
(7.7)(4.184)(1.1) = 83.0907 J
Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion