Answer:
d = rt
Explanation:
distance = ?
rate = 18 m/s
time = 42 x 60 = 2520 s
distance = 18 x 2520 = 45360 m
Calculate the net force.
The net force acting on the box is 15 N. The force acting on a box is unbalanced because the force on one side is 7 N while the force on other side is 8 N.
What is force?Force is defined as a push or pull exerted on an object as a result of the interaction of the object with another object.
It can also be defined as a factor that can alter how an object moves.
Net force is defined as the force that is the result of all forces concurrently acting on an object.
It can also be defined as all of the forces acting on an object are added together into a vector.
Net force = F1 + F2 + F3
= 7N + 4N + 4N
= 15 N
Thus, the net force acting on the box is 15 N. The force acting on a box is unbalanced because the force on one side is 7 N while the force on other side is 8 N.
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Sound _____. does not need a medium to travel is reflected when it bounces off a shiny surface moves in longitudinal waves is created by electric and magnetic fields
Answer:
reflection
Explanation:
it is a regular reflaction of light
your pet hamster sits on a record player that has constant angular speed . if the hamster moves to a point twice as far from the center, then its linear speed
Exit Slip!
Explain how Earth's tilted
axis and its revolution
around the Sun produces
seasons.
Earth's tilted axis and revolution around the sun causes the seasons.
Different parts of Earth receive the Sun's direct rays, throughout the year. So, it's summer in the Northern Hemisphere when the North Pole tilts toward the Sun. It's winter in the Northern Hemisphere when south Pole tilts toward the Sun.
As we know that the Earth rotates with a constant speed, so that every hour the direct beam will traverse across a single standard meridian. Rotation of Earth 15 degrees is equivalent to unit of one hour . When Earth rotates in such a way that the beam of the sun shifts +1∘ of longitude from East to West and time taken is 4 minutes.
The tilt of the Earth's axis is important because it governs the strength of warming of the sun's energy. The tilt of the surface of the Earth also causes light to be spread across a larger area of land which is termed as called the cosine projection effect.
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Circular Motion
The Trajectory of a mass point is a helix:[tex]r(t)=\left[\begin{array}{ccc}rcos(wt)\\rsin(wt)\\ht\end{array}\right]\\[/tex]
a) Compute the velocity v(t)=r(t) and the acceleration a(t).
b) What is the angel between the velocity and the acceleration?
c) If h=0, what is the radial acceleration and the velocity of the mass point at r=2m and 40rpm
No idea how to answer this question, I have never seen anything like it.
Any help would be awesome :)
Answer and step by step explanation:
First of all, I'm assuming you have had calculus, or this is going to be very awkward. Then, I'm replacing w with the Greek letter omega, it's a pet peeve of mine, sorry.
Shockingly, the derivative of a vector is computed by taking the derivative of each component (by linearity). If you've never heard the word derivative yet, you can think of the x and y components as two harmonic motions out of phase by 90°, and the z component as having constant speed h (harmonic motion is what you see when a mass moves along a circle with constant angular velocity and you look at it from the side).
At this point, let's use the derivative method for question a:
[tex]v(t)=\dot{r}(t)=\left[\begin{array}{ccc}-\omega r\sin(\omega t)\\\omega r\ cos(\omega t)\\h\end{array}\right]; a(t)=\dot v(t)=\ddot{r}(t)=\left[\begin{array}{ccc}-\omega ^2r\cos(\omega t)\\-\omega ^2r\ sin(\omega t)\\0\end{array}\right]\\[/tex]
Point b requires computing angles, which screams dot product to me.
[tex]\vec v(t) \cdot \vec a(t) =\left[\begin{array}{ccc}-\omega r\sin(\omega t)\\\omega r\ cos(\omega t)\\h\end{array}\right]\cdot \left[\begin{array}{ccc}-\omega ^2r\cos(\omega t)\\-\omega ^2r\ sin(\omega t)\\0\end{array}\right] = + \omega^3r^2 sin (\omega t) cos (\omega t) -\omega^3r^2 cos (\omega t) sin (\omega t) = 0 = ||\vec v|| ||\vec a|| cos \theta \implies cos\theta = 0 \implies \theta=\frac \pi2[/tex]
Now, the implied part is granted by the fact that we are assuming neither the velocity nor the acceleration are both zero, so the only option is for the cosine being zero, that makes the two vector orthogonal.
Finally, for point c, let's just take the moduli of both velocity and acceleration [tex]||\vec v|| = \omega r; ||\vec a||=\omega^2 r[/tex] and let's convert the angular velocity in civiliz... err, IS units: [tex]40 rpm = 40\times \frac{2\pi}{60}rad/s = \frac 43 rad/s[/tex]
Let's replace and we get
[tex]v=\frac83= 2.6 m/s\\a= \frac{32}3 = 10.5 m/s^2[/tex]
A solid cylinder (mass 0.274 kg, radius 2.00 cm) rolls without slipping at a speed of 5.00 cm/s. What is its total kinetic energy?
Given data
*The given mass of the solid cylinder is m = 0.274 kg
*The given radius of the cylinder is r = 2.00 cm = 0.02 m
*The given speed is v = 5.00 cm/s = 0.05 m/s
The formula for the total kinetic energy is given as
[tex]\begin{gathered} U_T=U_k+U_R \\ U_T=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \\ =\frac{1}{2}mv^2+\frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 \end{gathered}[/tex]*Here U_K is the translation kinetic energy
*Here U_R is the rotational kinetic energy
*Here 'I' is the moment of inertia of the solid cylinder
Substitute the known values in the above expression as
[tex]\begin{gathered} U_T=\frac{1}{2}(0.274)(0.05)^2+\frac{1}{2}(\frac{1}{2}\times0.274\times(0.02)^2)(\frac{0.05}{0.02})^2 \\ =0.000342+0.000171 \\ =5.13\times10^{-4}\text{ J} \\ =5.13\times10^{-1}\text{ mJ} \end{gathered}[/tex]Hence, the total kinetic energy is U_T = 5.13 × 10^-1 mJ
Imagine you are an alien from a distant galaxy. Your home planet does not have any gravity. You have just landed on Earth and make a few observations about the planet.
What are some ways you might first observe gravity?
How might you test gravitational force?
How would you describe gravity to other aliens back at your home planet?
Use details to support your answer.
The first thing an alien would notice about gravity on earth is how it affects their movement. If their planet is closer to the sun in their solar system, they'd be accustomed to a higher pull of gravity, hence weight. Based on the above, an alien will feel lighter on earth without auto-assistive gravity adjuster technology.
The simplest way to test gravity is by comparing my weight using "homemade" scales on the ship, then checking that against my weight off the ship.
The simplest way to describe Earth's gravity to those the home planet is to use scientific language. The acceleration of gravity at the Earth's surface is approximately 9.8 meters (32 feet) per second every second. This is constant.
What is Gravity?Gravity, often known as gravitation, is a force that exists between all physical things in the universe. The force of gravity seeks to draw any two objects or particles with nonzero mass toward one other. Gravity affects things of all sizes, from subatomic particles to galaxy clusters.
Einstein proposed that the geometry of spacetime is what causes the force we call gravity. A mass (or energy) concentration, such as the Earth or sun, bends space around it in the same way as a boulder bends the flow of a river. As a result, gravity was created.
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Inside the nucleus, two protons are held together by a force which overcomes the repulsion. This force is called
Inside the nucleus, two protons are held together by a strong nuclear force which overcomes the repulsion. This is sometimes referred to as strong interaction.
What is strong nuclear force?The strong nuclear force or the strong interaction is strong enough to overcome the repulsive force between the two positively charged protons and thereby allowing protons and neutrons to stick together even in a small space. Protons are held together by the strong attractive nuclear force that binds together protons and neutrons.
The strong force is one of the four fundamental forces of nature that include gravity, electromagnetism, and the weak nuclear force. The strong force holds protons and neutrons within the nucleus of the atom and thus creating densest environments in nature.
The strong force dies off very fast with distance more than gravity or the electromagnetic force. It's difficult to detect the strong force outside of a nucleus.
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Part 1 Assume that as a wave travels from one particular medium to another, its speed decreases. What happens to the wave's frequency?A. Its frequency increases.B. Its frequency decreases.C. Its frequency remains the same.D. It is impossible to determine without more data.Part 2Which best explains the correct answer to Part 1?A. A wave's frequency always remains the same. B. A wave's frequency is inversely proportional to its speed.C. Given that the wave's wavelength remains the same, if the wave's speed decreases, then the frequency must decrease.D. Given that a wave's speed is equal to the product of its wavelength and frequency, the wavelength must be known to determine the frequency.
Part 1
C. Its frequency remains the same.
Part 2
A. A wave's frequency always remains the same.
Explanation:The relationship between wavelength, frequency, and speed is given as:
[tex]v=\lambda f[/tex]When the frequency of a wave increases, its wavelength also increases.
The speed of a wave does not change as its frequency changes, and vice-versa.
Therefore, as the speed of the wave decreases, its frequency remains the same
A wave's frequency always remains the same regardless of the speed and wavelength
9.A stock car is moving at 25.0 m/s when the driverapplies the brakes. If it stops in 3.00 s, what is itsaverage acceleration?
The average acceleration a of an object that changes its speed from v_0 to v_f during a time interval t is:
[tex]a=\frac{v_f-v_0}{t}[/tex]Since the car is initially moving at 25.0m/s and it stops in 3.00 seconds, then its final speed is 0m/s.
Replace v_f=0, v_0=25.0m/s and t=3.00s to find the average acceleration of the car:
[tex]a=\frac{0\frac{m}{s}-25.0\frac{m}{s}}{3.00s}=-8.333\ldots\frac{m}{s^2}[/tex]Therefore, the average acceleration of the car when it stops from 25.0m/s in 3.00 seconds, is -8.33m/s^2.
What is the electric field between the plates of a capacitor that has a charge of 10.75 microC and voltage difference between the plates of 97.87 Volts if the plates are separated by 2.11 mm?
The eletric field on a capacitor is given by the following formula:
[tex]E=\frac{Q}{A\epsilon_0}=\frac{V}{d}[/tex]Where Q is the difference of charge between plates, A is the area, and e0 is a constant. It can also be written as the second formula, where V is the voltage, and d is the distance between plates.
When we use the second formula, with the values from the exercise, we get:
[tex]E=\frac{97.87}{2.11*10^{-3}}=46383.886[\frac{V}{m}][/tex]Thus, our final answer is E=46383.886 V/m
A 0.35-kg tennis racquet moving to the right at 20 m/s hits a 0.06-kg tennis ball that is moving to the left at 30 m/s. After the collision, the racquet continues to the right, but at a reduced speed of 10 m/s. What is the ball's velocity after the collision?
From conservation of momentum we have that:
[tex]0.35(20)-0.06(30)=0.35(10)+0.06v[/tex]Solving for v we have that:
[tex]\begin{gathered} 0.35(20)-0.06(30)=0.35(10)+0.06v \\ 7-1.8=3.5+0.06v \\ 0.06v=7-1.8-3.5 \\ 0.06v=1.7 \\ v=\frac{1.7}{0.06} \\ v=28.33 \end{gathered}[/tex]Therefore the velocity of the ball after the collision is 28.33 m/s
A spaceship of mass m has its engines switched off and is moving in a circular orbit at
height R above the surface of a planet of mass M and radius R.
a) Derive an expression for total mechanical energy E of the orbiting spaceship, in terms of G, m,
M and R.
b) Derive an expression for the minimum speed V the spaceship would need to escape from this
orbit into deep space, in terms of system parameters. (The engines can’t fire for the whole trip;
they can only give the spaceship one boost so it obtains this velocity. Ignore all other celestial
objects.)
The total mechanical energy of a spaceship will be E = - GMm/2(R+h)
which is moving in circular orbit at height R above the surface of a planet of mass M and radius R . Here, m is the mass of spaceship and G is gravitational force.
Mechanical energy, also known as kinetic energy or potential energy, is the energy that an object possesses when it is in motion or the energy that an object stores due to its location. Renewable energy is also fueled by mechanical energy. In order to efficiently produce electricity or convert energy, many sources of renewable energy depend on mechanical energy.The force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them, according to Newton's universal law of gravitation.Therefore, when the engine of spaceship of mass (m) is switch offed and moving in circular orbit of a planet at height R above the surface of a planet of mass M and radius R then the mechanical energy is given by
E = - GMm/2(R+h)
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For friction on an incline, what direction does Fp point?
The force of friction on an inclined plane, will act upward to oppose the downward motion of the object on the inclined plane.
What is force of friction?
The force of friction is the force that opposes the motion of an object.
Since the force of friction opposes the motion of an object, it acts upward along the plane for an object moving along an inclined plane.
As the object moves downwards, the force of friction will act upwards to oppose the downward motion of the object.
Mathematically, the magnitude of force of friction on an object moving along inclined plane is given as;
Ff = μmg
where;
μ is the coefficient of frictionm is the mass of the objectg is acceleration due to gravityThus, the force of friction on an inclined plane, will act upward to oppose the downward motion of the object on the inclined plane.
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The complete question is below:
For friction on an incline, what direction does force of friction (Fp) point?
In the experiment, the pressure of the gas is 1.2 x10^5 Pa at a temperature of 25.0°C.
When the cylinder is heated, the pressure reaches 2.1x10 Pa. Calculate the
temperature of the gas (in "C) at this pressure.
The temperature of the gas (in °C) at this pressure is 248.5 °C
Temperature is a bodily quantity that expresses the hotness of count or radiation. There are 3 kinds of temperature scales. Temperature is the degree of hotness or coldness of an item.
Temperature is a degree of the common kinetic energy of the debris in an object. whilst the temperature increases, the motion of those particles also increases. Temperature is measured with a thermometer or calorimeter.
Given;
P₁ = 1.2 x10⁵ Pa = 1.18430792
T₁ = 25.0°C = 298 K
P₂ = 2.1x10 Pa = 0.000207253886
T₂ = ?
using ideal gas equaion:-
PV = nRT
P₁/T₁ = P₂/T₂
T₂ = P₂T₁ /P₁
= 2.1x10⁵ x 298 / 1.2 x10⁵
= 521.5 K
= 248.5 °C
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WHAT IS CHROMATOGRAPHY ?
Answer:
Chromatography is a process for separating components of a mixture. To get the process started, the mixture is dissolved in a substance called the mobile phase, which carries it through a second substance called the stationary phase.
Explanation:
hope it will be helpful
Answer:
Chromatography is a process for separating components of a mixture.
Explanation:
To get the process started, the mixture is dissolved in a substance called the mobile phase, which carries it through a second substance called the stationary phase.
Have a good day!
A projectile is fired horizontally at an angle of 70° with the speed of 0.11km/s. Find the range and height. The final answer should be in meters.
In the given question projectile is fired horizontally at an angle of 70° with the speed of 0.11km/s ,(=110m/s).
to find the range and height
height=544.66m
time= 10.54 second
range=793.63m
vy=0
h=v^2sin^2θ/ 2g
v=0.11km/s =110m/s
θ=70°
range=v^2sin2θ/ g
horizontal velocity,
vx= 110cos70=37.622 m/s
vertical velocity,
vy=110sin70=103.36 m/s
using the equation of motion
s=ut+1/2 at^2
range, s=0, t=total time taken
range= 110^2 sin 2*70/9.8
range=793.63m
t= vsinθ/ g
=110 sin 70/9.8
time= 10.54 second
s=(110sin 70)*10.54 +1/2(-9.8)* (10.54^2)
s=1089-544.34
height=544.66m
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The object represented by line E does not stop. Is this true or false?
The given graph is about velocity and time, the dependent and independent variable, respectively. As we can observe, line E crosses the zero level to the negative zone, which means the object didn't stop but changed its direction.
Therefore, the statement is true.A object has a mass of 4 kg and is accelerating at 3 m/s2.The net force acting on the object is [] N.
In order to calculate the net force acting on the object, we can use the second law of Newton:
[tex]\sum ^{}_{}F=m\cdot a[/tex]So, for a mass of 4 kg and an acceleration of 3 m/s², we have:
[tex]\begin{gathered} F=4\cdot3 \\ F=12\text{ N} \end{gathered}[/tex]Therefore the net force is 12 Newtons.
PLS help due today number in the picture are 15 and 0
As the horizontal component of the velocity is constant throughout the projectile, the time of the flight of the projectile is independent of the horizontal velocity, therefore the correct answer is option E .
What is a projectile motion ?It is the motion of any object or body when it is ejected off the surface of the earth and follows any curving path while being affected by the gravitational pull of the planet .
The total time of the flight of the projectile = 2uSinθ / g
Thus , the time of the flight of the projectile is independent of the horizontal velocity, therefore the correct answer is option E.
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Giant electric eels can deliver a shock of 590 V with up to 1.0 A of current for a brief time. A snorkeler in salt water has a body resistance of about 770 Ω. A current of about 500 mA can cause heart fibrillation and death if it lasts too long.
A)What is the maximum power max a giant electric eel can deliver to its prey?
B)If the snorkeler is struck by the eel, what current will pass through her body?
C)Is this current large enough to be dangerous?
D)What power received does the snorkeler receive from the eel?
Answer:
Explanation:
Given:
U = 590 V
I = 1.0 A
R₁ = 770 Ω
I* = 500 mA = 0.5 A
_________________
P - ?
A)
Power:
P = U·I = 590·1.0 = 590 W
B)
Сurrent:
I₁ = U / R ₁ = 590 / 770 = 0.770 A = 770 mA
C)
This current is life-threatening!!!
I₁ > I* (700 mA > 500 mA)
D)
Energy received:
E = I₁²·R₁ = 0.770²· 770 ≈ 460 W
Set the cannon to have an initial speed of 15 m/s. For which situation do you think the cannon ball will go the highest: if it is set at a 25-degree angle, or if it is set at a 35-degree angle?
Question 3 options:
25 degrees
35 degrees
The cannon ball will travel the highest distance when the angle of projection is 35 degrees.
What is the maximum height of a projectile?The maximum height reached by a projectile is calculated using the following formula.
H = u²sin²θ/2g
where;
u is the initial velocity of the projectile θ is the angle of pojectiong is acceleration due to gravitywhen the angle of projection is 25 degrees;
H = (15² (sin 25)²) / (2 x 9.8)
H = 2.05 m
when the angle of projection is 35 degrees;
H = (15² (sin 35)²) / (2 x 9.8)
H = 3.78 m
Thus, the cannon ball will travel the highest distance when the angle of projection is 35 degrees.
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Question 8 of 10 A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment? A. The brand of salt used B. The time it takes for the water to boil C. The kind of bottles used D. The amount of salt added to the water SUBMIT it's B
Answer:
Explanation:
A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment? A. The time it takes for the water to boil B. The amount of salt added to the water C. The kind of bottles used D. The brand of salt used
If the displacement, velocity and acceleration at an instant of a particledescribing S.H.M are respectively 7.5m, 7.5m/s, 7.5m/s?. Calculate themaximum velocity of the particle.
We are given the displacement, velocity, and acceleration of a particle that describes Simple Harmonic Motion. We are asked to determine the maximum velocity of the particle. To do that we can use the following equation for the maximum velocity of a particle describing SHM:
[tex]v_{máx}=-A\omega[/tex]Where:
[tex]\begin{gathered} A=\text{ amplitude} \\ \omega=\text{ angular frequency} \end{gathered}[/tex]To determine the values of amplitude and angular frequency we can use the expressions for displacement, velocity, and acceleration. The expression for displacement is:
[tex]x=A\cos (\omega t)[/tex]The expression for the velocity is:
[tex]v=-A\omega\sin (\omega t)[/tex]And the expression for acceleration is:
[tex]a=-A\omega^2\cos (\omega t)[/tex]Now, from the expression for the displacement we can solve for the amplitude, like this:
[tex]\frac{x}{\cos(\omega t)}=A[/tex]Now we can replace this in the expression got eh acceleration:
[tex]a=-\frac{x}{\cos(\omega t)}\omega^2\cos (\omega t)[/tex]Simplifying:
[tex]a=-x\omega^2[/tex]Now we can solve for the angular frequency:
[tex]-\frac{a}{x}=\omega^2[/tex]Taking square root to both sides:
[tex]\sqrt[]{-\frac{a}{x}}=\omega[/tex]replacing the values:
[tex]\sqrt[]{-\frac{-7.5\frac{m}{s^2}}{7.5m}}=\omega[/tex]Solving the operations:
[tex]1s^{-1}=\omega[/tex]Now we divide the formula for the displacement and the formula for the velocity, we get:
[tex]\frac{v}{x}=\frac{-A\omega\sin (\omega t)}{A\cos (\omega t)}[/tex]Simplifying we get:
[tex]\frac{v}{x}=-\omega\tan (\omega t)[/tex]Replacing the known values:
[tex]\frac{-7.5\frac{m}{s}}{7.5m}=-(1s^{-1})\tan (t)[/tex]Simplifying:
[tex]-1=-\tan (t)[/tex]Solving for "t":
[tex]t=\arctan (1)=0.78[/tex]Now we can replace these values in the formula for displacement to get the value of the amplitude:
[tex]x=A\cos (\omega t)[/tex]Solving for the amplitude:
[tex]\frac{x}{\cos (\omega t)}=A[/tex]Replacing the known values:
[tex]\frac{7.5m}{\cos (0.78)}=A[/tex]Solving the operations:
[tex]10.6m=A[/tex]Now we replace the values in the formula for the maximum velocity:
[tex]v_{\text{max}}=A\omega[/tex]Replacing:
[tex]\begin{gathered} v_{\max }=(10.6m)(1s^{-1}) \\ v_{\max }=10.6\text{ m/s} \end{gathered}[/tex]Therefore, the maximum velocity is 10.6 meters per second.
how long must a 50 N force act on a 400 kg mass to raise its speed from 10 m/s to 12 m/s
Answer:
see below
Explanation:
F=ma
50 = 400 a
A = 50/400 = .125 m/s^2
Accel = change in v / change in t
.125 = (12-10) / t Shows t = 16 seconds
1) What is the horizontal force on block A due to block B?2) What is the net horizontal force on block B?3) What is the horizontal force on block B due to block C?
Given data:
Mass of each block;
[tex]m=12\text{ kg}[/tex]Acceleration;
[tex]a=1.2\text{ m/s}^2[/tex]The free-body diagram for A,
The free-body equation for A is given as,
[tex]F-N_1=ma\ldots(1)[/tex]The free-body diagram for B is given as,
The free-body equation for B is given as,
[tex]\begin{gathered} F_B=N_1-N_2 \\ ma=N_1-N_2\ldots(2) \end{gathered}[/tex]The free-body equation for C is given as,
The free-body equation for C is given as,
[tex]\begin{gathered} F_c=ma \\ N_2=ma\ldots(3) \end{gathered}[/tex]Equating equation (2) and (3),
[tex]\begin{gathered} N_1-N_2=N_2_{} \\ N_1=2N_2 \\ N_1=2ma \end{gathered}[/tex]Part (1),
The horizontal force on block A due to block B is given as,
[tex]\begin{gathered} F_{AB}=N_1 \\ =2ma \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} F_{AB}=2\times(12\text{ kg})\times(1.2\text{ m/s}^2) \\ =28.8\text{ N} \end{gathered}[/tex]Therefore, the net horizontal force on block A due to block B is 28.8 N.
Part (2)
The net horizontal force on block B is given as,
[tex]\begin{gathered} F_B=N_1-N_2 \\ =2ma-ma \\ =ma \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} F_B=(12\text{ kg})\times(1.2\text{ m/s}^2) \\ =14.4\text{ N} \end{gathered}[/tex]Therefore, the net horizontal force on block B is 14.4 N.
Part (3)
The horizontal force on block B due to block C is given as,
[tex]\begin{gathered} F_{BC}=N_2 \\ =ma \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} F_{BC}=(12\text{ kg})\times(1.2\text{ m/s}^2) \\ =14.4\text{ N} \end{gathered}[/tex]Therefore, the horizontal force on block B due to block C is 14.4 N.
A 2.55kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0360m . The spring has force constant 870N/m . The coefficient of kinetic friction between the floor and the block is 0.45 . The block and spring are released from rest and the block slides along the floor.
What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0160 m .)
The speed of the block when it has moved a distance of 0.0200 m from its initial position is 0.29 m/s.
The mass of the block is 2.55 Kg. Coefficient of kinetic friction is 0.45. Spring constant is 870N/m. The final compression of the spring is 0.0160 meters. The block is moved by distance of 0.0200 meters.
By considering the block-spring system, we can use Work energy theorem here,
According to work energy theorem,
Work done by conservative forces on the body + Work done by none conservative forces = Total change in kinetic energy of the body.
Work done by Conservative forces = 1/2Kx².
Where,
K is spring constant,
x is the compression in spring,
Work done by non-conservative forces = umgd
Where,
u is the coefficient of kinetic friction,
m is mass of the block,
d is the distance by which the block moves,
Here, we will take spring force as positive because the spring force and displacement both are in same direction,
Putting all the values,
1/2Kx² - umgd = 1/2mV² - 1/2mU²
1/2(870)(0.016)² - 0.45(2.55)(9.8)(0.02) = 1/2(2.55)V²
0.11136 - 0.22491= (1.275)V²
0.11355/1.275 = V²
V = 0.29 m/s.
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Set the cannon to have an initial speed of 25 m/s at an angle of 30-degrees. Find how long it takes to hit the ground. Will the time it takes to get to the highest point be less than ½ the time, more than ½ the time, or ½ the time?
Question 7 options:
1/2 the time
more than 1/2 the time
less than 1/2 the time
The time taken by the projectile to reach the highest point would be one half of the total time of the projectile , therefore the correct answer is option D .
What is a projectile motion ?It is the motion of any object or body when it is ejected off the surface of the earth and follows any curving path while being affected by the gravitational pull of the planet .
As given in the problem Set the cannon to have an initial speed of 25 m / s at an angle of 30 degrees .
Thus, the right response is option D because it would take the projectile half as long to reach the highest position.
To learn more about projectile motion here,
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Element x is a non metal in which position of the periodic table can element x be found
Answer:
To the right
Explanation:
Metals are to the left, nonmetals are to the right. There's a border I drew that separates nonmetals and metals
A steel railroad track has a length of 24 mwhen the temperature is 8°C.What is the increase in the length of therail on a hot day when the temperature is36 °C? The linear expansion coefficient ofsteel is 11 x 10-6(°C)-1.Answer in units of m.Answer in units of mSecond part:- Suppose the end of rail are rigidly clamped eight Celsius degrees to prevent expansion calculate the thermal stress in the rail if it’s temperature is raised to 36 Celsius degrees Young’s module for Steel is 20×10 to the power of 10 and N/m to the power of two answer in units of N/m to the power of two.
Given:
• Length = 24 m
,• Temperature, T1 = 8°C
,• Expansion coefficient = 11 x 10⁻⁶(°C)⁻¹
Let's solve for the following:
• (a). The increase in length of the rail when the temperature is 36 °C
To find the increase in length, apply the formula:
[tex]L=L_o*\alpha *(T_2-T_1)[/tex]Where:
Lo = 24 m
α = 11 x 10⁻⁶(°C)⁻¹
T2 = 36 °C
T1 = 8°C
Thus, we have:
[tex]\begin{gathered} dL=24*11\times10^{-6}*(36-8) \\ \\ dL=24*11\times10^{-6}*(28) \\ \\ dL=0.0074\text{ m} \end{gathered}[/tex]The increase in length is 0.0074 m.
• (b). Let's calculate the thermal stress.
To find the thermal stress, we have the formula:
[tex]\text{ thermal stress = }Y\frac{dL}{L}[/tex]Where:
Y is the young modulus = 20 x 10¹⁰ N/m
dL is the change in length = 0.0074 m
L is the length = 24 m
Input values in the formula and solve:
[tex]\begin{gathered} \text{ thermal stress = 20}\times10^{10}*\frac{0.0074}{24} \\ \\ \text{ thermal stress = 6.17}\times10^7\text{ N/m}^2 \end{gathered}[/tex]The thermal stress is 6.17 x 10⁷ N/m².
ANSWER:
• (a). 0.0074 m.
• (b). 6.17 x 10⁷ N/m²