Answer: D. 40
Explanation:
Power = Force * Velocity
Force = 600N
Power needs to be converted to Watts:
= 24 * 1,000
= 24,000 W
24,000 = 600 * Velocity
Velocity = 24,000/600
= 40 m/s
A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/s.
a. Calculate the length of the incline
b. calculate the horizontal and vertical components of the velocity at the end of the incline.
If Nell was blocking a defensive lineman for the South Grand Prairie Warriors with a force of 126 Newtons and had a mass of 56.2 kg…what would his acceleration be equal to?
Answer:
2.24 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
m is the mass
f is the force
From the question we have
[tex]a = \frac{126}{56.2} \\ = 2.241992...[/tex]
We have the final answer as
2.24 m/s²Hope this helps you
a wire has a resistance 12 ohm. It is bent in the form of circle. The effective resistance between the points on any diameter will be:. a)12 ohm b)6 ohm c)3ohm d)24ohm
Answer:
a) 12 ohm
that's what I kno
One small speaker is placed 3m to the east of a second speaker, and a listener stands 4m directly south of one of the speakers. That listener finds that if they move in any direction, the sound gets louder. What is the longest possible wavelength of the sound from the speakers
Answer:
The value is [tex]\lambda = 2 \ m[/tex]
Explanation:
From the question we are told that
The distance of the speaker from the second speaker to the east is [tex]d = 3 \ m[/tex]
The distance of the speaker from the listener to the south is [tex]a = 4 \ m[/tex]
Generally given that if the speaker move in any direction, their sound become louder , it then mean that the position of the listener of minimum sound (i.e a position of minima ) ,
Generally the path difference of the sound produce by both speaker at a position of minima is mathematically represented as
[tex]y = \frac{\lambda}{2}[/tex]
Generally considering the orientation of the speakers and applying Pythagoras theorem we see that distance from the second speaker to the listener is mathematically represented as
[tex]b = \sqrt{d^ 2 + a^2 }[/tex]
=> [tex]b = \sqrt{3^ 2 + 4^2 }[/tex]
=> [tex]b = 5[/tex]
Generally the path difference between the two speaker with respect to the listener is
[tex]y = b - a[/tex]
=> [tex]y = 5 - 4[/tex]
=> [tex]y = 1[/tex]
So
[tex]1 = \frac{\lambda}{2}[/tex]
=> [tex]\lambda = 2 \ m[/tex]
A pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.666. What is the weight of the heaviest pallet of bricks that can be supported this way?
A. 400 lb
B. 350 lb
C. 250 lb
D. 266 lb
Answer:
C.250 lb is the answer...A solid is 5 cm tall, 3 cm wide, and 2 cm thick. It has a mass of 129 g. What is its
density?
Answer:
4.3 g/cm³ or 4.3g/cc
Explanation:
Volume(V) = Height × Length × Width
= 5cm × 3cm × 2cm
= 30cm³
Mass(m) = 129gram
So,
Density = m/V
= 129g/30cm³
= 4.3g/cc or 4.3g/cm³
A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The technician measures the oscillating voltage to be a sine wave with a peak voltage of 3.25 V . However, the technician must record the RMS voltage on a report. What value should be reported?
Answer:
RMS voltage is 2.29 V
Explanation:
Given that,
The peak voltge of the voltage = 3.25 V
We need to find the RMS voltage reported by the technician.
We know that the relation between the peak and the RMS voltage is given by the formula as follows :
[tex]V_{rms}=\dfrac{V_o}{\sqrt2}\\\\V_{rms}=\dfrac{3.25}{\sqrt2}\\\\V_{rms}=2.29\ V[/tex]
So, the RMS voltage is 2.29 V.
This is science btw please help me
Answer:
c
Explanation:
A loaded 500 kg sled is traveling on smooth horizontal snow at 5 m/s when it suddenly comes to a rough region. The region is 10 m long and reduces the sled's speed to 3 m/s what average friction force did the rough region exert on the sled?
Answer:
400 N
Explanation:
Change of Kinetic Energy to Friction Wok
∆KE = W
½ x m x (v(5)² - v(3)²) = f x d
½ x 500 x (5² - 3²) = f x 10
250 x (25 - 9) = f x 10
25 x 16 = f
f = 400 N
A seagull flying horizontally over the ocean at a constant speed of 2.85 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 2.45 s after the bird lets go, the fish lands in the ocean.
a. Just before reaching the ocean, what is the horizontal component of the fish's velocity?
b. Just before reaching the ocean, what is the vertical component of the fish's velocity?
c. If the seagull's initial speed were decreased, which of the following regarding the fish's velocity upon reaching the ocean would be true?
1. The horizontal component of the fish's velocity would decrease.
2. The horizontal component of the fish's velocity would increase.
3. The horizontal component of the fish's velocity would stay the same.
4. The vertical component of the fish's velocity would decrease.
5. The vertical component of the fish's velocity would increase.
6. The vertical component of the fish's velocity would stay the same.
Answer:
a) 2.85 m/s
b) -24.01 m/s
c) 1
Explanation:
From the question, we can attest that the motion of the fish that was dropped by the seagul is that of a projectile motion. This motion is made up of two other motions which are, a horizontal uniform motion and a vertical motion, at constant acceleration. From the question, we are asked to find the horizontal motion. And then, the horizontal component of the fish's velocity does not change, therefore its the horizontal component of the fish's velocity is 2.85 m/s
To find the vertical component of the fish's velocity, we use the equation
v(y) = u(y) + gt
Where u(y) is the initial velocity which is zero, and g is the acceleration due to gravity which is -9.8 m/s. We are also told from the question that it took 2.45 s, and that's our time t. Applying this into the equation, we have
v(y) = 0 + -9.8 * 2.45
v(y) = -24.01 m/s
For the 3rd part, the horizontal component of the fish's velocity would decrease
A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg.
Required:
a. If the crate doesn't move, what's the magnitude of the static friction force (in N)?
b. What is the minimum possible value of the coefficient of static friction between the crate and the floor?
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate is not moving then its acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = 0
Fm - Ff = 0.
Fm is the moving force
Ff is the frictional force
Fm = Ff
This means that the moving force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
m is the mass of the crate = 31.2kg
g is the acceleration due to gravity = 9.8m/s²
R = 31.2 × 9.8
R = 305.76N
Recall that;
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-direction with a speed 2v and the second particle is at rest. After the collision, the second particle is moving in the direction 45o below the x-axis and with a speed √2v.
a) Find the velocity of the first particle after the collision. (i.e. find the x-and y-components of the velocity.)
b) Find the total kinetic energy of the two particles before and after the collision.
c) Is the collision elastic or inelastic?
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
An object has mass 4 kg. What is its weight (in newton) on earth?
Answer:
Should be -39.2 N
Explanation:
w=mg
w=4 x -9.8 m/s2
= -39.2 N
A person holds a 4.0 kg block at position A shown above on the left, in contact with an uncompressed vertical spring with a spring constant of 500 N/m. The person gently lowers the block from rest at position A to rest at position B. Which of the following describes the change in the energy of the block-spring-Earth system as a result of the block being lowered?
Answer:
Energy decrease by approximately 1.5J
The change in the energy of the block-spring-Earth system as a result of the block being lowered will be 1.424 J.
Given information;
A person holds a 4.0 kg block at position A in contact with an uncompressed vertical spring with a spring constant, k, of 500 N/m.
The person gently lowers the block from rest at position A to rest at position B. The decrease in height, h, of the object or the compression in spring, x, will be 10 cm.
Due to the decrease in height of the block, the decrease in potential energy will be,
[tex]PE_b=-mgh\\=-4\times9.81\times0.1\\=-3.924\rm\;N[/tex]
Due to compression, the increase in potential energy of the spring will be,
[tex]PE_s=\dfrac{1}{2}kx^2\\=\dfrac{1}{2}500\times(0.1)^2\\=2.5\rm\;J[/tex]
So, the net change in the energy of the system will be,
[tex]E=-3.924+2.5\\=1.424\rm\;J[/tex]
Therefore, the change in the energy of the block-spring-Earth system as a result of the block being lowered will be 1.424 J.
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An object traveling a circular path of radius 5 m at constant speed experiences an acceleration of 3 m/s2. If the radius of its path is increased to 10 m, but its speed remains the same, what is its acceleration?
Answer:
New acceleration = 1.5 m/s²
Explanation:
The acceleration of an object when it moves in a circular path is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
Where
v is speed of an object
r is radius of path
If v remains constant,
[tex]a\propto \dfrac{1}{r}[/tex]
or
[tex]\dfrac{a_1}{a_2}=\dfrac{r_2}{r_1}[/tex]
Put r₁ = 5 m, a₁ = 3m/s², r₂ = 10 m
[tex]a_2=\dfrac{a_1r_1}{r_2}\\\\a_2=\dfrac{3\times 5}{10}\\\\a_2=1.5\ m/s^2[/tex]
Hence, if the radius of its path is increased to 10 m, its acceleration will be 1.5 m/s².
An ideal spring with a constant of 150 N/m is compressed 14 cm. How much energy is stored in the spring?
[tex]\displaystyle U_E = 1.47 \ J[/tex]
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightPhysics
Unit 0
Unit Conversion: 100 cm = 1 m1 N/m (Newton per meter) = 1 J (Joule)Energy
Elastic Potential Energy Formula: [tex]\displaystyle U_E = \frac{1}{2} k \Delta x^2[/tex]
k is spring constantΔx is displacement from equilibriumExplanation:Step 1: Define
k = 150 N/m
Δx = 14 cm
Step 2: Identify Conversions
Unit Conversion 100 cm = 1 m
Step 3: Convert
Δx = 14 cm = 0.14 m
Step 4: Find Energy
Substitute [EPE]: [tex]\displaystyle U_E = \frac{1}{2} (150 \ N/m)(0.14 \ m)^2[/tex]Exponents: [tex]\displaystyle U_E = \frac{1}{2} (150 \ N/m)(0.0196 \ m^2)[/tex]Multiply: [tex]\displaystyle U_E = (75 \ N/m)(0.0196 \ m^2)[/tex]Multiply: [tex]\displaystyle U_E = 1.47 \ N/m[/tex]Redefine: [tex]\displaystyle U_E = 1.47 \ J[/tex]Calculate the escape velocity
the moon's surface given that a man on the moon has 1/6 his weight on earth
Answer:
v = 2.38 × 10³ m/s
Explanation:
Escape velocity, v = √(2gR) where g = acceleration due to gravity on planet and R = radius of planet.
Since it is given that the weight of the man on the moon is 1/6 his weight on earth, and g' = acceleration due to gravity on moon and g = acceleration due to gravity on earth and m = mass of man,
mg' = mg/6
g' = g/6
Since g = 9.8 m/s²,
g'= 9.8 m/s² ÷ 6
g' = 1.63 m/s²
The escape velocity of the moon is thus v = √(2g'R) where R = radius of moon = 1.737 × 10⁶ m.
Substituting these into v, we have
v = √(2g'R)
v = √(2 × 1.63 m/s² × 1.737 × 10⁶ m)
v = √[5.663 × 10⁶ (m/s)²]
v = 2.38 × 10³ m/s
Under conditions of conservation of energy where the initial energy object is only gravitiational potential energy and the final energy is only kinetic, write an equation to calculate the final velocity of the object. Ignore all friction. Show all work
Answer:
The equation to calculate the final velocity of the object.---
Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.
Explanation:
Also just so you know this isn't my answer I got it from here https://brainly.com/question/20016151 from im2slowwwww
A(n) ____ is just an abstract description of what an object will be like if any objects are ever actually instantiated.
Answer:
Class
Explanation:
A class is just an abstract description of what an object will be like if any objects are ever actually instantiated.
Its purpose is to provide an appropriate superclass out of which other classes can confirm take from and even go ahead to share a mutual and particular design. In many other languages, the class name is often used as the name for the template itself, it can also be used as the name for the default constructor of the class too.
Wind gusts create ripples on the ocean that have a wavelength of 2.23 cm and propagate at 4.12 m/s. What is their frequency (in Hz)?
Answer:
The frequency of the wave is 184.75 Hz.
Explanation:
Given;
wavelength of the wind, λ = 2.23 cm = 0.0223 m
speed of the wind wave, v = 4.12 m/s
the frequency of the wave = ?
The frequency of the wave is calculated as;
F = V / λ
Where;
F is the frequency of the wave
Substitute the given values and solve for F
F = (4.12) / (0.0223)
F = 184.75 Hz.
Therefore, the frequency of the wave is 184.75 Hz.
A 1.40 kg block is attached to a spring with spring constant 16.5 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s . What are
Answer:
A = 0.13 m
Explanation:
Given that,
Mass of a block, m = 1.4 kg
Spring constant of the spring, k = 16.5 N/m
While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s or 0.45 m/s
We need to find the amplitude of the subsequent oscillations.
We can use the conservation of energy here. Initially, the kinetic energy of the block is maximum and then it gets converted to the potential energy of the spring.
Mathematically,
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2[/tex]
A is the amplitude of subsequent oscillations.
[tex]A=\sqrt{\dfrac{mv^2}{k}} \\\\A=\sqrt{\dfrac{1.4\times (0.45)^2}{16.5}} \\\\A=0.13\ m[/tex]
So, the amplitude of subsequent oscillations is 0.13 m.
The energy of a photon is ________ proportional to its wavelength.
Answer:
The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.
Explanation:
Plz mark brainliest thanks
How many inner shell electrons are there in Bohr model
Answer:
10 is the correct answer
Explanation:
It is ten because the electrons that encircled the two orbit before the third orbit if the bohr model is 10. In bohr model, the electrons encircled the nucleus of an atom called orbit.
A powered winch is used to pull a sailboat to shore. The winch uses a 900 W motor.
If the motor is used for 30 s, how much work does it do? (Power: P = W/t)
0.03 J
30 J
960 J
27,000 J
Answer: 27,000 J :)
Explanation:
If the winch uses a 900 W motor and the motor is used for 30 s to pull a sailboat then work needed to do this work is 27000 J. hence option D is correct
What is Power?Power is the rate of doing work. Power is also defined as work divided by time. i.e. Power = Work ÷ Time. Its SI unit is Watt denoted by letter W. Watt(W) means J/s or J.s-1. Something makes work in less time, it means it has more power. Work is Force times Displacement. Dimension of Power is [M¹ L² T⁻³]
Given,
Power P = 900W
Time = 30s
Work in joule =?
By using formula,
P = Work ÷ Time
Work = Power × time
W= 900 × 30
W= 27000 J
Hence Work of 27000J is needed to pull a sailboat to shore in 30s.
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If a net horizontal force of 0.8 N is applied to a toy whose mass is 1.2 kg, acceleration is?
Hello!
[tex]\large\boxed{a = \frac{2}{3}m/s^{2}}[/tex]
Use the equation F = m · a to solve. We are given the force (N) and mass (kg), so we can solve for the acceleration by plugging in the given values:
0.8 = 1.2a
0.8 / 1.2 = a
a = 2/3 m/s²
A crane raises a 12,000 N marble sculpture at a constant velocity onto a pedestal 1.5 m above the ground outside an art museum. How much work is done by the crane
For the work, applicate formula:
[tex]\boxed{\boxed{\green{\bf{W = F\times d}}}}[/tex]
According our data:
Replacing:W = 12000 N * 1,5 m
Resolving:W = 18000 J
The work done is 18000 Joules.
what is the role of heat in the formation
of new rocks
Answer: Igneous rocks are created by heat. They start off as magma, which is hot, melted rock deep within a volcano. When magma cools and hardens, igneous rock forms.
Explanation:
Heat or increase in temperature plays a vital role in formation of igneous, metamorphic rocks.
What is the role of heat in the formation of new rocks?On Earth, igneous sedimentary or metamorphic rocks are created. When rocks are heated to their melting point and magma is produced, igneous rocks are created. Rocks that undergo metamorphosis are created when heat and pressure transform the original or parent rock into a whole new rock.
The heat causes a physical weathering process where the rock splits apart into fragments as it expands and contracts. When oxygen or moisture in the air change the chemical makeup of rock minerals, this also contributes to chemical weathering.
Heat and pressure cause an existing rock to change into a new rock, creating metamorphic rocks. When hot magma hits rock, contact metamorphism takes place. Large sections of existing rocks are altered by regional metamorphism as a result of the intense heat and pressure caused by tectonic forces.
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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal force of 98.0 N for 1.40 s.(a) Determine the magnitude of each force acting on the block of ice while you are pulling.Fpull = NFg = NFN = N(b) With what speed is the ice moving after you are finished pulling?m/s
Answer:
a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.
b) The final speed of the block of ice is 9.8 meters per second.
Explanation:
a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.
The weight of the block ([tex]W[/tex]), measured in newtons, is:
[tex]W = m\cdot g[/tex] (1)
Where:
[tex]m[/tex] - Mass of the block of ice, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]m = 14\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the magnitudes of the weight and normal force of the block of ice are, respectively:
[tex]N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]N = W = 137.298\,N[/tex]
And the pull force is:
[tex]F_{pull} = 98\,N[/tex]
The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.
b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:
[tex]m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}[/tex] (2)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the block, measured in meters per second.
[tex]\Sigma F[/tex] - Horizontal net force, measured in newtons.
[tex]\Delta t[/tex] - Impact time, measured in seconds.
Now we clear the final speed in (2):
[tex]v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}[/tex]
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]m = 14\,kg[/tex], [tex]\Sigma F = 98\,N[/tex] and [tex]\Delta t = 1.40\,s[/tex], then final speed of the ice block is:
[tex]v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}[/tex]
[tex]v_{f} = 9.8\,\frac{m}{s}[/tex]
The final speed of the block of ice is 9.8 meters per second.
If a bullet with a mass of 0.005kg is fired from a
gun at a speed of 1000 m/s, what is its
momentum?
what’s the v?
Answer:
5kgm/s
Explanation:
Given parameters:
Mass of bullet = 0.005kg
Speed = 1000m/s
Unknown:
Momentum = ?
Solution:
Momentum is the amount of motion a body possesses.
Mathematically;
Momentum = mass x velocity
Now insert the parameters and solve;
Momentum = 0.005 x 1000 = 5kgm/s
derive an expression for torque experiend by an electric dipole placed in a uniform electric field
Answer:
The torque τ on an electric dipole with dipole moment p in a uniform electric field E is given by τ = p × E where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment.
Explanation: