The velocity and the acceleration of the ball after it has fallen
for 2.00 s 9.8 m/s.
Calculation:-
S = ut + 1/2 at²
= 0 + 0.5 × 9.8 (2)²
= 4.9 × 4
= 19.6 m
v² = u² + 2aS
v² = 0 +2×9.8 × 19.6
v = √96.04 m
= 9.8 m/s
Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.
Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.
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if the uncertainty in the position of a wave packet representing the state of a quantumsystem particle is equal to its de broglie wavelength, how does the uncertainty in momentum compare with the value of the momentum of the particle?
The uncertainty of momentum compared to the value of the momentum of the particle is Δp ≈ p/2π
In 1924 Louis de Broglie proposed the hypothesis that a moving particle having momentum p could behave as a wave. It is characterized by the presence of a wavelength known as the wavelength of a particle.
The uncertainty in a position is:
ΔxΔp ≈ h
If Δx = λ, the above equity on become as
ΔxΔp ≈ h
λΔp ≈ h
λΔp ≈ h/2π
Δp ≈ h/2πλ
Δp ≈ (h/λ)/2π
Δp ≈ p/2π
With:
p is momentum
h is Planck's constant
λ is the wavelength
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The acceleration due to gravity on Mimas a moon on Saturn, is 0.066m/s^2 . Starting from rest how much time does it take for an object to fall 10m on Mimas?
The equation to obtain the vertical distance cobered by the object is,
[tex]h=ut+\frac{1}{2}gt^2[/tex]The acceleration due to gravity on saturn is 0.066 m/s2.
Plug in the known values,
[tex]\begin{gathered} 10\text{ m=(0 m/s)t+}\frac{1}{2}(0.066m/s^2)t^2 \\ t^2=\frac{2(10\text{ m)}}{0.066m/s^2} \\ t=\sqrt[]{303.03s^2} \\ \approx17.4\text{ s} \end{gathered}[/tex]Thus, the time taken by object to fall is 17.4 s.
A rocket moves straight upward, starting from rest with an acceleration of 129. 4 m/s2. It runs out of fuel at the end of 4. 00 s and continues to coast upward, reaching maximum height before falling back to earth. Find the velocity the instant before the rocket crashes on the ground.
Velocity of rocket instant before the rocket as it crashes is 135.81 m/s.
What is Velocity?The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity. It is the speed at which distance changes.It is the displacement change rate.Velocity can be defined as the rate at which something moves in a specific direction as the speed of a car driving north on a highway or the pace at which a rocket takes off.Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.The vector quantity velocity (v), denoted by the equation v = s/t, quantifies displacement (or change in position, s), over change in time (t).To learn more about Velocity from the given link :
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Aircraft carriers use catapults to launch jets from their deck. One such catapult accelerates a 18,900 kg aircraft from rest to 61 m/s in 3.3 s. What is the magnitude of the force (in N) required to do this?
Given that the mass of catapult, m = 18900 kg
The initial velocity is
[tex]v_o=\text{ 0 m/s}[/tex]As the body is at rest initially.
The final speed is
[tex]v_f=\text{ 61 m/s}[/tex]The time taken is t = 3.3 s
We have to find the force.
According to Newton's second law, the formula to calculate force is
[tex]\begin{gathered} F=\text{mass}\times acceleration \\ =m\times\frac{(v_f-v_o)}{t} \end{gathered}[/tex]Substituting the values, the force will be
[tex]\begin{gathered} F=18900\times\frac{61-0}{3.3} \\ =349363.63\text{ N} \end{gathered}[/tex]Thus, the force is 349363.63 N
What should be the nature of fuse wire?
Answer: a fuse wire is a wire of high resistance and low melting point.
A rock is placed into a graduated cylinder and the volume of water increases from 50 ml to 60 ml. If the rock has a mass of 25 grams, what is its density?
The density of the rock, having a mass of 25 grams which increases the volume of water from 50 mL to 60 mL is 2.5 g/mL
How do I determine the density of the rock?To obtain the density of the rock, we must know the volume of the rock. The volume of the rock can be obatined as follow:
Volume of water = 50 mL Volume of water + rock = 60 mL Volume of rock =?Volume of rock = (Volume of water + rock) - volume of water
Volume of rock = 60 - 50
Volume of rock = 10 mL
Now, we shall determine the density of the rock. This is illustrated below:
Volume of rock = 10 mL Mass of rock = 25 grams Density of rock = ?Density = mass / volume
Density of rock = 25 / 10
Density of rock = 2.5 g/mL
Thus, we can say that the density of the rock is 2.5 g/mL
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Energy can be transferred between objects or systems; momentum cannot be transferred and is only gained or lost from an object or system.
In a collision when two or more objects collide, the momentum will be transferred between the objects involved in the collision. But the total momentum of a system will always be conserved. Thus the system will never lose or gain momentum.
Thus the given statement is false.
A wheel rolls without slipping. Which is the correct velocity vector for point p on the wheel?.
The velocity vector for the point P will be the resultant velocity. So, option (C) is correct.
What do we mean by plane of motion?Plane motion is the movement of an item when all of its pieces move in a straight line along a set plane. Two different types of plane motion are listed below:
1. Pure rotational motion: In this motion, the rigid body revolves around a fixed axis that is parallel to a fixed plane. In other words, with respect to an inertial frame of reference, the axis is stationary and does not move or change direction.
2. The overall plane motion: In this case, the motion can be characterized as a combination of pure translational motion in a plane that is fixed and pure rotational motion along an axis perpendicular to the plane.
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A rock is thrown upward with a velocity of 27 meters per second from the top of a 35 meter high cliff on the way back down. When will the Rick be 5 meters from ground level? Round your answer to two decimal places
Given data
*The given veloicty of the rock is u = 27 m/s
*The given height of the cliff is h = 35 m
*The given height of the Rick from the ground level is s = 5 m
*The value of the acceleration doue to gravity is g = 9.8 m/s^2
The formukla for the time taken by the Rick at 5 meters from the ground level is given as
[tex]s=h+ut+\frac{1}{2}at^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} 5=(35)+(27)(t)+\frac{1}{2}(9.8)t^2 \\ 4.9t^2-27t-30=0 \\ t=6.45\text{ s} \end{gathered}[/tex]Hence, the time taken by the Ric at 5 meters above the ground lel is t = 6.45 s
Answer:
6.45 s
Explanation:
df = final position = 5 m a = accel of gravity = -9.81 m/s^2
do = original position = 35 m vo = original velocity = 27m/s
df = do + vot + 1/2 at^2
5 = 35 + 27t - (1/2)9.81t^2
-4.905 t^2+ 27t +30 = 0
Use quadratic formula a = -4.905 b = 27 c = 30
to find t = 6.45 s
(Ignore the negative value found with the Quadratic Formula)
I already did part a and the magnitude for all the parts. I need help finding the direction.
b)
Given that,
A=10 N at 0°
B=20 N at 180°
As we can see these two vectors are in opposite directions.
Thus the resultant vector will be in the direction of the vector with the higher magnitude.
In vector addition we can write,
[tex]C=10\cos 0^0+20\cos 180^0[/tex]Where C is the resultant vector.
Thus magnitude of R will be equal to,
[tex]R=-10\text{ N}[/tex]The negative sign indicates that the direction of the C is along the negative x-axis.
Thus the magnitude of C is 10 N and the direction of the C is 180°
a)
Given,
A= 10 N at 0°
B=20 N at 0°
The the vector C is
[tex]C=10\cos 0^0+20\cos 0^0=30\text{ N}[/tex]The direction of R will be along the direction of these two vectors, as they are in the same direction.
Thus the magnitude of the vector C will be 30 N and the direction of the vector C is 0°
c)
Given,
A=10 N at 180°
B=20 N at 180°
Thus the vector C is
[tex]C=10\cos 180^0+20\cos 180^0=-30\text{ N}[/tex]The direction of the vector C will be along the direction of the negative x-axis, as these two vectors are along the negative x-axis.
Thus the magnitude of the vector C is 30 N and the direction of the vector C is 180°
d)
Given,
A= 10 N at 0°
B=20 N at 90°
The x-component of the vector C will be,
[tex]C_x=10\cos 0^0+20\cos 90^0=10\text{ }\hat{\text{i}}[/tex]The y- component of C will be,
[tex]C_y=10\sin 0^0+20\sin 90^0=20\text{ }\hat{j}[/tex]The magnitude of vector C is given by,
[tex]\begin{gathered} C=\sqrt[]{C^2_x+C^2_y}_{} \\ =\sqrt[]{10^2+20^2} \\ =22.36\text{ N} \end{gathered}[/tex]The direction of the vector C will be,
[tex]\begin{gathered} \theta=\tan ^{-1}\frac{C_y}{C_x} \\ =\tan ^{-1}\frac{20}{10} \\ =63.43^0 \end{gathered}[/tex]Thus the magnitude of C will be 22.36 N and the direction of C will be 63.43°
e)
Given that
A= 10 N at 90°
B=20 N at 0°
The x-component of the vector C will be
[tex]C_x=10\cos 90^0+20\cos 0^0=20\text{ }\hat{\text{i}}[/tex]The y-component of the vector C will be
[tex]C_y=10\sin 90^0+20\sin 0^0=10\text{ }\hat{j}[/tex]The magnitude of the vector C will be
[tex]\begin{gathered} C=\sqrt[]{C^2_x+C^2_y}_{} \\ =\sqrt[]{20^2+10^2} \\ =22.36\text{ N} \end{gathered}[/tex]The direction of vector C will be,
[tex]\begin{gathered} \theta=\tan ^{-1}\frac{C_y}{C_x} \\ =\tan ^{-1}\frac{10}{20} \\ =26.56^0 \end{gathered}[/tex]Thus the magnitude of the vector C will be 22.36 N and the direction of the vector C will be 26.56°
can some help me find the slope of the line / graph?
The given points are
(0.47, 0.4905)
(0.52, 0.6867)
(0.58, 0.8829)
(0.64, 1.0791)
The formula for calculating slope is expressed as
slope = (y2 - y1)/(x2 - x1)
where
y2 and y1 are final and initial coordinates of y
x2 and x1 are final and initial coordinates of x
Picking these points from the given points, we have
when x1 = 0.52, y1 = 0.6867
when x2 = 0.58, y2 = 0.8829
slope = (0.8829 - 0.6867)/(0.58 - 0.52) = 0.1962/0.06
slope = 3.27
We would plot the corresponding x and y values on the line. The graph is shown below
Which one of the following is true for a chemical reaction?
a) The number of protons and/or neutrons in the atoms char
b) The electron clouds of atoms change.
C) Matter can be converted into energy.
D) The number of atoms can change.
Main answer- C (matter can be converted into energy)
supporting answer-Nuclear fission is a reaction in which particles or atoms split into two smaller and lighter nuclei, which can then be converted into energy.
final answer -hence final answer is C
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what happens to the brightness of light emitted by a light bulb when the current flowing through it increases?
A wide range of applied voltages will allow it to function; as the voltage rises, the current rises as well, making the light bulb brighter.
When the current rises, what happens to the bulb's brightness?The quantity of electrons passing through the circuit each second determines the current that runs across it. When employing batteries, raising voltage also causes the circuit's current to grow. When the brightness of the bulb rises, the current is seen to rise.
What takes place when there is current?When electric current runs, electrons pass through a conductor. All substances restrict the flow of electric current to some degree. Resistance is the name given to this quality.
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A wheel of diameter 26 cm turns at 1500 rpm. How far will a point on the outer rimmove in 2.0 s?1) 3.1 m12) 41 m3) 90 m4) 180 m
Given,
The diameter of the wheel, d=26 cm=0.26 m
Therefore the radius of the wheel, r=d/2=0.13m
Time period, t=2.0 s
Frequency of the revolution, f=1500 rpm
1 relovution=2π radian.
Therefore the angular velocity is calculated as,
[tex]\omega=\frac{1500\times2\pi}{60}=157.1\text{ rad/s}[/tex]The angular velocity and angular displacement are related as,
[tex]\omega=\frac{\theta}{t}[/tex]On rearranging the above equation and substituting the known values,
[tex]\theta=\omega t=157.1\times2.0=314.2\text{ rad}[/tex]We can calculate how far the rim has moved by using the following relationship,
[tex]s=r\theta[/tex]Where 's' is the distance of the movement of the rim,
On substituting the known values in the above equation,
[tex]s=0.13\times314.2=40.8\text{ m}\approx41\text{ m}[/tex]Therefore the correct answer is option 2, 41 m
What is the wavelength of the wave in the string
The standing wave in the figure has 4 nodes and 3 antinodes.
The wavelength of a standing wave is given by,
[tex]\lambda=\frac{2}{n}L[/tex]Where n=number of nodes-1
From the figure n=3
Therefore the wavelength is,
[tex]\lambda=\frac{2}{3}L[/tex]Therefore the wavelength of the stationary wave produced in the given string of length L is (2/3)L.
A 6.25-gram bullet travelling at 365 m/s strikes and enters a 4.50-kg crate. The crate slides 0.15 m along a wood floor until it comes to rest.1. Find the friction force between the crate and the floor.
In order to determine the friction force, proceed as follow:
Take into account that total momentum of the system must conserve, then, you have:
m*v = (m + M)v'
where,
m: mass of the bullet = 6.25g = 0.00625kg
v: initial speed of the bullet = 365m/s
M: mass of the crate = 4.50kg
v': speed of both crate and bullet after the impact = ?
Solve the equation above for v', replace tha values of the other parameters and simplify:
[tex]\begin{gathered} v^{\prime}=\frac{mv}{m+M} \\ v^{\prime}=\frac{(0.00625kg)(365\frac{m}{s})}{0.00625\operatorname{kg}+4.50\operatorname{kg}} \\ v^{\prime}\approx0.506\frac{m}{s} \end{gathered}[/tex]Now, consider that the work done by the friction force is given by:
W = Fr*d
where,
Fr: friction force = ?
d: distance = 0.15m
Furthermore, the work done is equal to:
W = 1/2*(m+M)v'^2 that is, the change in kinetic energy is equal to the work
Then, you can equal the previous expressions for W, solve for Fr, replace and simplify:
[tex]\begin{gathered} F_rd=\frac{1}{2}(m+M)v^{\prime}^2 \\ F_r=\frac{1}{2}\frac{(m+M)v^{\prime2}}{d} \\ F_r=\frac{1}{2}\frac{(0.00625kg+4.50kg)(0.506\frac{m}{s})^2}{0.15m} \\ F_r\approx3.846N \end{gathered}[/tex]Now, take into account that the friction force can be written as follow:
Fr = μN = μ(m+M)g
where,
μ: coefficient of kinetic friction between crate and floor
g: gravitational acceleration constant = 9.8m/s^2
Solve the equatio above for μ, replace the values of the other parameters and simplify:
[tex]\begin{gathered} \mu=\frac{F_r}{(m+M)g} \\ \mu=\frac{3.846N}{(0.00625kg+4.50kg)(9.8\frac{m}{s^2})} \\ \mu\approx0.087 \end{gathered}[/tex]Hence, the coefficient of kinetic friction is approximately 0.087
a collection of hydrogen atoms in the ground state is illuminated with ultraviolet light of wavelength 75.0 nm. find the kinetic energy of the emitted electrons.
The kinetic energy of the emitted electrons is 11.24 eV.
What is kinetic energy?The following is the physics definition of kinetic energy:
The amount of effort that a moving object may accomplish is measured by its kinetic energy.
Kinetic energy is a scalar quantity that can only be fully explained by its magnitude.
Given parameters:
Wavelength of ultraviolet light : λ = 75.0 nm.
Energy of the ultraviolet light: E = hc/λ = 1242/75.0 eV= 22.84 eV
Potential energy of a electron of hydrogen atoms in the ground state is -13.6 eV.
This energy of light supplies required potential energy to the emitted electron and kinetic energy of electron.
So, the kinetic energy of the emitted electrons = 22.84 eV - 13.6 eV = 11.24 eV.
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Which of the following accurately describes a property of an insulator?
a. An insulator will inhibit the flow of neutrons,
b. An insulator will inhibit the flow of electrons.
c. An insulator will allow the flow of neutrons.
d. An insulator will allow the flow of electrons.
An organ pipe closed at one end and open at the other has a length of 1.8 m.
Given,
The length of the pipe, L=1.8 m
Speed of the sound, v=340 m/s
To form the standing wave with the longest wavelength, the number of nodes of the thus formed standing wave should be equal to 1.
The wavelength of the standing wave created in a closed-end pipe is given by,
[tex]\lambda=\frac{4}{1}L[/tex]On substituting the known values,
[tex]\begin{gathered} \lambda=\frac{4}{1}\times1.8 \\ =7.2\text{ m} \end{gathered}[/tex]Thus the longest possible wavelength is 7.2 m
The frequency of this standing wave is given by,
[tex]f=\frac{v}{\lambda}[/tex]On substituting the known values,
[tex]\begin{gathered} f=\frac{340}{7.2} \\ =47.22\text{ Hz} \end{gathered}[/tex]Thus the frequency of this standing wave is 47.22 Hz.
What was the tension in the cable when the craft was being lowered to the seafloor?.
Assuming that the "craft" refers to an early submersible craft, the tension in the cable, when it was being lowered to the seafloor, would be the tension when it was stationary minus the drag force.
Say, for example, the tension in the cable when an early submersible craft was 7000 N. When the craft was either lowered to or raised from the seafloor at a steady raid, the motion in through the water it experienced added a 1500 N drag force.
When it's being lowered, the tension in the cable would be the difference between the two forces = 7000 N - 1500 N = 5500 N.
When it's being raised, the tension in the cable would be the sum of the two forces = 7000 N + 1500 N = 8500 N.
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momentum problems practice class
The answers to the momentum questions with given velocity and mass attached as an image are as follows:
70,000kgm/s35,000kgm/s2m/s5kg40,000kgm/s28.57m/sWhat is momentum in physics?Momentum is the tendency of a body to maintain its inertial motion. It is the product of a body's mass and velocity, or the vector sum of the products of its masses and velocities.
Momentum = mass × velocity
According to the questions asked in the attached image, a tractor-trailer truck has a velocity of 35m/s. The momentum at different masses can be calculated as follows:
Truck of 2000kg, momentum = 2000 × 35 = 70,000kgm/s. Car of 1000kg, momentum = 1000 × 35 = 35,000kgm/sVelocity of a bowling ball: 16kgm/s ÷ 8kg = 2m/smass of beach ball: 0.25kgm/s ÷ 0.5m/s = 5kgmomentum of a truck: 4000kg × 10m/s = 40,000kgm/svelocity of a car: 40,000kgm/s ÷ 1,400kg = 28.57m/sThe 4kilogram ball rolling along the same path at speed of 1m/s will take more force to stop in 10s than the 8kg ball traveling at a speed of 0.2m/s because the former has more momentum.Learn more about momentum at: https://brainly.com/question/24030570
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Astronomers have measured light from stars in galaxies that
are billions of years old. Which of the following must be true
about energy for this to be possible?
When stars emit energy in the form of light it travels to Earth faster
than other places in the Universe.
Energy in the form of light can only travel toward objects.
When energy in the form of light is redshifted it is no longer
measurable.
Energy in the form of light travels across space and cannot be
destroyed
The statement that is true about energy is that "energy in the form of light travels across space and cannot be destroyed" Option D.
What is energy?We know that in physics, energy is generally defined as the ability to do work. There are several kinds of energy that are in existence and we know that light energy is only one of the kinds of energy.
Energy in the form of light is produced by stars and these stars have been in existence for several billion years and they are still able to produce light that can be visible to the astronomers that study the outer space.
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after being struck by the bullet, the disk rotates at 4.00 rad/srad/s . what is the horizontal velocity of the bullet just before it strikes the disk?
The component of horizontal velocity of bullet is 8m/s for 4 radians per second.
Given - Rotation rate = 4 rad per sec.
This means 4 radians for one seconds . i.e. one seconds is equivalent to 60min.
To resolve the velocity following equation can be employed-
[tex]\beta =vt+\frac{\alpha t^2}{2}\\ \\\beta = rotation\\\\v= velocity\\\\t= time\\\\\alpha =acceleration[/tex]
Now substituting all values in this equation
4= 0+α/2
α=4 x 2 ⇒8m/s²
since the rotation is for one seconds so velocity = acceleration.
[tex]acceleration=\frac{dv}{dt}[/tex]
dt =1
so α=v=8m/s
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A 40. 0-kg child stands at one end of a 40. 0-kg boat that is 4. 00 m in length. The boat is initially 3. 00 m from the pier. The child notices a turtle on a rock near the far end of the boat and proceeds to walk to that end to catch the turtle. Neglecting friction between the boat and the water, where is the child relative to the pier when he reaches the far end of the boat?.
When the boy reaches the far end of the boat, his position in relation to the pier is 61.5 m away.
Explain about the friction?The force produced when two surfaces slide against and touch one another is referred to as frictional force. There are a few things that affect the frictional force: The surface texture of these forces and the amount of force pressing them together have the biggest impact.
The imperfections on the two surfaces in contact, friction is produced. Small irregularities can be found on even the smoothest surfaces, and when two surfaces interact with one another, they cause friction.
Two surfaces interacting produces the common force known as friction. This interaction raises the heat energy of the two surfaces as they slide against one another (the temperature goes up).
Boat mass M = 40 kg
The child weighs 40kg.
3 meters separate the boat's near end from the pier.
Boat length is 4 meters.
= 40 *(3/2) + 40 *3/ 40+ 40
= 60 +120/ 80
=61.5 meter
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you hold a bucket in one hand. in the bucket is a 540 g rock. you swing the bucket so the rock moves in a vertical circle 2.6 m in diameter. part a what is the minimum speed the rock must have at the top of the circle if it is to always stay in contact with the bottom of the bucket? express your answer to two significant figures and include the appropriate units.
The minimum speed at the bucket so the rock always stays in contact with the bottom of the bucket is 3.6 m/s.
When the bucket is at the top of the circle, two forces are working on the bucket. The force of gravity and the rope tension. If the rock must have at stay in contact with the bottom of the bucket the rope tension equal to zero (T = 0). All force in the vertical circle is equal as centripetal force.
∑F = Fc
T + w = Fc
0 + w = Fc
w = Fc
m × g = m × v²/R
g = v²/R
v² = gR
[tex]v = \sqrt{gR}[/tex]
where
Acceleration due to gravity = 9.8 m/s²Mass of the object = m (kg)R = radius of a circle (m)T = rope tension (N)w = weight of the object (N)v = velocity (m/s)Fc = centripetal force (N)From the problem, given
Mass of the rock = m = 540 grams = 0.54 kgDiameter circle = D = 2.6 mRadius of circle = R[tex]v = \sqrt{gR}[/tex]
[tex]v = \sqrt{9.8 \times 1.3}[/tex]
[tex]v = \sqrt{12.74}[/tex]
v = 3.6 m/s
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What change could be made in the setup in diagram 1 to increase the amount of force necessary to move the block of wood?
Answer:
Add mass or weight to the block, roughen the surface of the table or the bottom of the block, increase the surface area of the wood, etc.
When a block of steel at 90 degrees celsius is placed in a bucket of water at 30 degrees celsius, what happens?.
When a block of steel at 90 degrees Celsius is placed in a pail of water at 30 degrees Celsius, there is a differential in temperature.
What is the formula for temperature change?Where T is the final temperature and Ti is the initial temperature, gives the change in temperature. Every substance has a unique specific heat that is measured in either K units depending on how T is expressed.
What causes changes in temperature?The balance between energy entering and leaving the planet's system determines the temperature of Earth. The Earth warms as incoming solar energy is absorbed by the Earth system. Earth does not warm when solar energy is reflected back into space.
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I need help with question 3 you don’t have to explain it I just need the answers thank you.
Weight of the bag of sugar on the moon = 3.34 N
Weight of the bag of sugar on Venus = 18.09 N
Mass of the bag of sugar on the Earth = 2.04 kg
Mass of the bag of sugar on the moon = 2.04 kg
Mass of the bag of sugar on venus = 2.04 kg
Explanation:The weight of the object on the earth
[tex]\begin{gathered} W_E=4.50\text{ lb} \\ 1\text{ lb = }4.45\text{ N} \\ W_E=4.50\text{ }\times4.45 \\ W_E=20.03N \end{gathered}[/tex]The acceleration due to gravity on the earth:
[tex]g_E=9.81m/s^2[/tex][tex]\begin{gathered} W_E=m_Eg_E_{} \\ 20.03=m_E\times9.81 \\ m_E=\frac{20.03}{9.81} \\ m_E=2.04\operatorname{kg} \end{gathered}[/tex]The mass of the bag of sugar on the earth = 2.04 kg
Note that the mass of the sugar is the same everywhere
Acceleration due to gravity on the moon is one-sixth of acceleration due to gravity on the earth
[tex]\begin{gathered} g_m=\frac{1}{6}g_E \\ g_m=\frac{1}{6}(9.81) \\ g_m=1.635 \end{gathered}[/tex]The weight of the bag of sugar on the moon is:
[tex]\begin{gathered} W_m_{}=m_mg_m_{} \\ W_m=2.04\times1.635 \\ W_m=3.34N \end{gathered}[/tex]The acceleration due to gravity on the Venus is 0.904 times that of the earth
[tex]\begin{gathered} g_v=0.904g_E \\ g_v=0.904(9.81) \\ g_v=8.87m/s^2 \end{gathered}[/tex]The weight of the bag of sugar on the venus is:
[tex]\begin{gathered} W_v=m_vg_v \\ W_v=2.04(8.87) \\ W_v=18.09\text{ N} \end{gathered}[/tex]Mass of the bag of sugar on the Earth = 2.04 kg
Mass of the bag of sugar on the moon = 2.04 kg
Mass of the bag of sugar on venus = 2.04 kg
in terms of the torque needed to rotate your leg as you run, would it be better to have a long calf and short thigh, or vice versa?
Running torque is the amount of torque required by a device to maintain the constant angular velocity of a rotating part while in operation. This value provides important operating information for rotation machinery and must be determined through testing.
What is running torque ?Running torque is also known as "friction torque." It is the torque required to turn the nut or bolt while the nut is "loose." A locking nut's locking feature (typically a plastic insert or a distorted bore section of the threads that directly squeezes together on them) requires some effort to overcome.
Since torque value is an attempt to measure the stretch in a bolt as the rotary sliding wedge of the threads is applied to pull the two ends apart, any rotational resistance forces that are added, not directly applying the clamping force to the part, will skew the final result.
If you torque to 100 inlbs and it takes 30 inlbs of torque just to turn the nut because it's a lock nut, you'll only have applied 70 inlbs of "stretch" to the bolt when the wrench trips at 100 inlbs. So, if you measure 30 inlb while spinning the nut down before it clamps down, add it to the 100 inlb value, turn the wrench to 130, and you'll have the correct 100 inlb application to the bolt or stud.
To learn more about running torque refer to :https://brainly.com/question/20691242
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Please help me, I am following along diligently. V(t) = t^2 -9t+18, with distance, s measured in meters, left or right of 0, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t=0 sec is 1 meter right of 0, that is s(0)= 1Part I: Average velocity over the interval 0 to 8 secondsPart II: The instantaneous velocity and speed at time 5 secsPart III: The time intervals when the particle is moving rightPart IV: The time intervals when the particle is going faster, and slowing downPart V: Total distance the particle has traveled between 0 and 8 seconds
Given that the velocity at any time t is
[tex]v(t)=t^2-9t+18[/tex]Also, the time interval is from t = 0 to t = 8 seconds
The position at time t = 0 s is s(0) = 1 m towards right of zero.
The initial time is t = 0 s, so the initial velocity will be
[tex]\begin{gathered} v_i(t=0)=0^2-9\times0+18\text{ } \\ v_i(0)\text{ = 18 m/s} \end{gathered}[/tex]The final time is t = 8 s, so the final velocity will be
[tex]\begin{gathered} v_f(t=8)=8^2-9\times8+18 \\ v_f(8)\text{ = 64-72+18} \\ =\text{ 10 m/s} \end{gathered}[/tex]The average velocity will be
[tex]\begin{gathered} v_{av}=\frac{v_i+v_f}{2} \\ =\frac{18+10}{2} \\ =14\text{ m/s} \end{gathered}[/tex]Thus, the average velocity is 14 m/s.
Part II:
The instantaneous velocity at time t =5 s will be
[tex]\begin{gathered} v(t=5)=5^2-9\times5+18 \\ =25-45+18 \\ =-2\text{ m/s} \end{gathered}[/tex]The instantaneous speed is the magnitude of instantaneous velocity.
Thus, the instantaneous speed will be 2 m/s.
Part III:
The particle will move towards the right when v(t) > 0
The time intervals will be
[tex]\begin{gathered} t^2-9t+18>0 \\ t^2-6t-3t+18>0 \\ t(t-6)-3(t-6)>0 \\ (t-6)(t-3)>0 \\ t-6>0\text{ or t>6} \\ t-3>0\text{ ot t>3} \end{gathered}[/tex]Thus, time intervals are t > 3 and t > 6 when the particle is moving towards the right.
Part IV :
The particle will move faster if the acceleration, a(t) > 0
The particle will slow down if the acceleration, a(t) < 0
So, first, we need to find the acceleration, it can be calculated as
[tex]\begin{gathered} a(t)=\text{ }\frac{d(v(t))}{dt} \\ =\frac{d(t^2-9t+18)}{dt} \\ =2t-9 \end{gathered}[/tex]For the particle moving faster,
[tex]\begin{gathered} a(t)>0 \\ 2t-9>0 \\ 2t-9+9>9+0 \\ 2t>9 \\ \frac{2t}{2}>\frac{9}{2} \\ t>\frac{9}{2} \\ t>4.5\text{ s} \end{gathered}[/tex]For particle slowing down,
[tex]\begin{gathered} a(t)<0 \\ 2t-9<0 \\ 2t-9+9<9+0_{} \\ 2t<9 \\ \frac{2t}{2}<\frac{9}{2} \\ t<4.5\text{ s} \end{gathered}[/tex]The total distance can be calculated as
[tex]\begin{gathered} s(t)=\int ^8_0v(t)dt \\ =\text{ }\int ^8_0(t^2-9t+18)\mathrm{d}t \\ =\lbrack\frac{t^3}{3}\rbrack^8_0-9\lbrack\frac{t^2}{2}\rbrack^8_0+18\lbrack t^{}\rbrack^8_0 \\ =\frac{1}{3}\lbrack512-0\rbrack-9\lbrack64-0\rbrack+18\lbrack8-0\rbrack \\ =\text{ 170.67-576+144} \\ =-261.33\text{ m} \end{gathered}[/tex]Here, the negative symbol indicates it is towards the left from zero.