ANSWER:
1177600 joules
STEP-BY-STEP EXPLANATION:
Given:
m = 2300 kg
v = 32 m/s
We can calculate the kinetic energy using the following formula:
[tex]E=\frac{1}{2}m\cdot v^2[/tex]Replacing:
[tex]\begin{gathered} E=\frac{1}{2}\cdot2300\cdot32^2 \\ E=1177600\text{ J} \end{gathered}[/tex]The kinetic energy of the truck is 1177600 joules.
Explain how this sand dune was created using the words weathering, erosion, and deposition.
PLSSS HELPP
By abrasion, the sediment in the wind promotes erosion. The wind scatters sand, sand dunes created. When clay and silt are deposited by the wind. The presence of vegetation ground helps stop wind erosion.
What is an erosion ?Earthen materials were worn away during erosion, a geological process in which they are moved by water or wind. Weathering, a related process that does not involve movement, dissolves and breaks down rock.
What is caused by erosion?The process through which the Earth's surface ages is known as erosion. Natural forces like wind or glacier ice can create erosion. But when it comes to altering the Earth, nothing compares to a slow, constant movement of water, as anyone who has ever seen a picture of a Grand Canyon will attest.
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When you drop a 0.36 kg apple, Earth ex- erts a force on it that accelerates it at 9.8 m/s² toward the earth's surface. According to New- ton's third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98 x 1024 kg, what is the magnitude of the earth's acceleration toward the apple? Answer in units of m/s². Answer in units of m/s^2
The magnitude of the earth's acceleration toward the apple is 5.7x10^-25 m/s^2
Given:
Mass if the apple (m)= 0.36
Mass of the earth (M)=5.98x10^24
Gravitational acceleration (g)= 9.8
we know that the formula is:
Ma=mg
So, according tot the formula
a=(m/M)g
=(0.36/5.98x10^24)9.8
=5.7x10^-25 m/s^2
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if the camera left his hands at 21.9 m/s and took 2.009s to hit the ground, how far down the cliff did it fall?
The camera falls down 25.835 meters under gravity.
While a frame falls, it happens because of gravity, and the force of air will increase with speed. This continues till the force of air equals the weight. Now the accelerates but falls at a constant speed known as the terminal speed.
A motion under gravity refers back to the motion of an item whose vertical movement is laid low with the presence of gravity. The pressure that attracts gadgets downwards is gravity. In fact, gravity works towards the center of the Earth.
S = ut - 1/2at²
= 21.9 × 2.009s - 1/2 × 9.8 × (2.009)²
= 43.997 - 18.162
= 25.835 meter.
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How do I find if the dog is accelerating or not?
We will have the following:
* The dog walks at a constant speed of 1 m/s in a straight line east. [Not accelerating].
* The dog speeds up from 1 m/s to 7 m/s as he runs away from a cat that scared him. [Accelerating].
* The dog walks at a constant speed of 0.5 m/s in a circle around a bowl of food. [Not accelerating].
* The dog stand in one spot on his bed. [Not accelerating].
* The dog slows down to a stop as he approaches the fence at the edge of the yard. [Accelerating].
At room temperature, most metals are
A. Solids
B. Liquids
C. Gases
D. Plasma
Answer:
A. Solids
Explanation:
only Mercury is liquid soo .... solid
Graphing is a way to represent data visually to represent the motion of a rolling ball students craft data on A-line graph which of the following would accurately label the X ax
Such an object lacks a velocity since it lacks a slope. Such an object must have a variable velocity. It's quickening.
Which kind of graph is utilized for motion?Starting off, let's graph a few instances of motion with constant speed. The graph on the right shows three alternative curves, each with a zero initial point. The first thing to note is how straight each graph is. (On a graph, a curve can be any type of line that is drawn.
It is not possible to determine the velocity from a position-time graph's slope when the graph is curved. Only straight lines have slope as a characteristic. Such an object lacks a velocity since it lacks a slope. Words like "the" and "a" are highlighted in this text to emphasize that there isn't a single velocity for these conditions. Such an object must have a variable velocity. It's quickening.
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1) speed of bob?2)Show calculation for the speed of the bob:(write formula first, then insert value with unit, then find the result with unit)
Speed of bob:
[tex]\begin{gathered} \omega=2\pi f \\ _{\text{ }}where\colon \\ f=\frac{1}{T} \\ \omega=\frac{2\pi}{T} \end{gathered}[/tex]so:
[tex]\begin{gathered} \omega=\frac{2\pi}{1.293} \\ \omega=4.83\frac{rad}{s} \end{gathered}[/tex][tex]\begin{gathered} v=\omega\cdot r \\ so\colon \\ v=4.83\cdot0.22 \\ v=1.06\frac{m}{s} \end{gathered}[/tex]
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
The maximum quantity of repetitions that the athlete could complete with their current level of energy was 229 for such a workout.
Energy efficiency is the ratio of input power to energy output. As just a result, a cookie contains 54.0 kcal in calories. An athlete utilizes the 54.0 kcal in just this cookie for input energy.
The following formula could be employed to measure efficiency:
Energy intake minus power generated is efficiency.
Efficiency = output energy / input energy
After rearrangement above equation.
output energy = efficiency × input energy
Put the given data in above equation.
Output energy = 0.25 × 54 kcal = 13.5 kcal.
Again for power capacity, the lifting exercise was already carried out n times. The source of total output energy is mgh within the form of electromagnetic potential.
So, Energy per repetition = mgh
Where, m will represent mass g will represent gravity h is height.
By entering the data provided above.
Energy per repetition = 103 kg × 9.8 m/[tex]s^{2}[/tex] × (2.45 × 0.1m)
Energy per repetition= 247.303 J
Energy per repetition = 0.059 kcal.
So, quantity of repetitions = overall output energy / energy per repetition
quantity of repetitions = 13.5 kcal / 0.059 kcal
quantity of repetitions will be equal to 229 repetitions.
Therefore, the quantity of repetition would be be 229.
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7 The distance-time graph for a motorcyclist riding off from rest is shown in Figure 1.2.18.
a Describe the motion.
b Calculate how far the motorbike moves in 30 seconds.
c Calculate the speed.
The distance travelled by the motorbike is 600 m and the speed of the motorbike is 20 m/s.
What is the motion of the motorbike?The motion of the motorbike can be described as linear type of motion since the distance travelled by the motorbike increases with increase in the time of motion.
There is directly linear relationship between the distance travelled by the motorbike and the time of motion of the motorbike.
The distance travelled by the motorbike is calculated as follows;
from the given graph, at time 30 seconds, the corresponding the distance of the motorbike on the vertical axis is 600 m.
distance = 600 m
The speed of the motorbike is calculated as follows;
v = d/t
v = 600 m / 30 s
v = 20 m/s
Thus, the motion of the motorbike is a linear type of motion, the distance travelled by the motorbike increases with increase in the time of motion and the speed of the motorbike depends on the distance and time of motion.
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What is the equation for the X axis of this fbd? Remember the sign of convention discussed in the module
F - f = ma
Explanation:The sum of forces on the x-axis = ma (since it is not stated that the forces are in equilibrium)
[tex]\sum fx=ma[/tex]By resolving the Free Body
[tex]F+(-f)=ma[/tex]Simplifying the equation above, we have:
F - f = ma
A cell of a constant e. m.f is connected in series with a parallel arrangement of 8ohms coil and a uniform wire of length (Lcm). The total external resistance in the circuit is found to be 6ohms. calculate the resistance of the wire and its length. Resistivity of the wire material = 6.0×10^-5 ohm/meter, cross-sectional area of the wire = 6.0×10^-3cm².
The resistance of the wire will be 24ohm and the length of the wire will be 24×10⁴cm.
As given in the question,
Rₜₒₜₐₗ = 6ohm
Resistivity = 6×10⁻⁵ohm/m = 6×10⁻⁵/100ohm/cm
Area(A)= 6×10⁻³cm²
Let R = resistance of the wire, therefore;
Rₜₒₜₐₗ = R×8 / R+8
6 = R×8 / R+8
6R + 48 = 8R
48 = 2R
R = 24ohm
R = 24ohm
we know that resistance (R)=rho×l/A
Therefore,
rho×l/A = 24
6×10⁻⁵/100 × l/6×10⁻³ = 24
10⁻⁴ × l = 24
l = 24×10⁴cm
So, the resistance of the wire will be R=24ohm and the length of the wire will be 24×10⁴cm.
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An Earth satellite moves in circular orbit 602 km above the earths surface with a period of 96.53 min. What are the speed and the magnitude of the centripetal acceleration of the satellite.
The speed and the magnitude of the centripetal acceleration of the satellite are 7.568 km/s and 8.205 x 10^-3 m / s^2
Here this problem we are dealing with the speed and centripetal acceleration where the speed of the satellite is the speed required to attain adjustment between gravity's pull on the satellite and the inertia of the satellite's motion whereas centripetal acceleration is referred to as the acceleration of a body navigating a circular way.Since we are given the orbital radius which is 602 km and the period which is 96.53 min.
The radius of the satellite's orbit, r = earth radius + orbit radius
=>r =6378 km + 602 km = 6980 km
Since the formula for the orbital circumference is :
2πR
=>2 • 3.14 •6980 = 43,834.4 km
Now divide the circumference by the time period we get the orbital speed which is,
43,834.4 km / (96.53 min • 60) = 7.568 km/s
the formula we are referring for calculating the centripetal acceleration is:
a = v²/ r ,
=> a = (7.568 ^2) / 6980 = 8.205 x 10^-3 m / s^2
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What would happen if an unbalanced force was acting on an object? Would it move at a constant speed? Explain why or why not.
According to Newton's Second Law of Motion, if a net force acts on an object, the object accelerates acording to the equation:
[tex]\Sigma F=ma[/tex]Where m is the mass of the object.
Then, if an unbalanced force is acting on an object, the net force is different from 0, which means that the object will accelerate in the direction of the net force.
If an object has a uniform acceleration, it won't have a constant velocity, but it may have a constant speed anyway if the force is always perpendicular to the velocity, in that case the object would be moving in a circular trajectory.
In other words, an unbalanced force will cause an acceleration, this can either change the speed of an object or the direction of its velocity (or both).
Then, it might be possible for an object to have a constant speed under the influence of an unbalanced force if that object follows a circular trajectory.
What is the range of motion(far,medium,or short) in these state of matters:
Neon:
Argon:
Oxygen:
Water:
thank you!
Which is true of the force pair of Newton's third law?a.)The two forces are in the same direction.b.)The two forces never produce an acceleration. c.)The two forces act on different objects.
Given:
Force pair of Newton's third law.
Required:
To find the true statement for the force pair of Newton's third law.
Explanation:
According to Newton's third law
This can be understood by taking an example.
Suppose a person hits the wall with his hand, the person hitting the wall is the action.
This action results in pain in the person's hand which is the reaction.
Here, the hand exerted force on the wall and the wall exerted an equal force in opposite direction.
The conclusion can be drawn as
Action and reaction act on two different objects.
The forces act in opposite direction.
Final Answer: The two forces act on different objects.
During which season does the north pole experience constant sunlight?
A. spring
B. winter
C. summer
D. autumn
During the summer season, the north pole gets constant sunlight.
How does the north pole experience constant sunlight?In the summer season, the north pole experience constant sunlight. Because in the summer the shadows are short and the sun remains higher in the sky. In the summertime, the sun remains above the horizon at the north pole which circles the pole every day.
In the summer the sun's rays reach the north pole while in winter season the north pole is tilted away from the incoming sunshine. Mostly the regions of the north pole remain cold because the sunlight does not directly fall on the north pole. At the north pole, there are only two types of seasons, one is summer and the other is winter season. The annual average temperature of the winter season at the north pole is -40 degrees Celsius while in summer season the average temperature is 32 F.
So we can conclude that during the months of the summer season the north pole gets maximum sunlight.
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a 10 kg block of ice slides a ramp 20 m long. inclined at 10 degrees to the horizontal, if the ramp frictionless what is the acceleration of the ice
The acceleration of the ice is obtained as 0.97 m/s².
What is the acceleration?We have to recall that the acceleration has to do with the rate at which the velocity is change per unit time. Now, we know that from the parameters that have been given in the question that is before us here;
μmgcos(θ) = ma
μgcos(θ) = a
Hence;
μ = coefficient of friction
g = acceleration due to gravity
θ = the angle of inclination of the plane
We can now find the acceleration by the use of the formula as shown above hence we have;
a = (0.1)(9.8) cos(10)
a = 0.97 m/s²
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A= 1/2 W (R+S) for W
The change of the subject in the formula is given by; W = 2A/(R + S)
How do we make a subject in formula?When we have a formula, each of the components of the formula are the parameters of the formula. In this case, we have a formula that says A= 1/2 W (R+S). The parameters that are in this formula are A, W, R and S.
According to the question, we have been asked to make W the subject of the formula.
The first step is to multiply both sides by 2
2A = W(R + S)
And secondly, we divide both sides by the sum of R and S and we have;
W = 2A/(R + S)
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Please help me
how law of motion applied in basketball explain
Explanation:
According to the first law of motion, the basketball is always moving in one direction, unless acted on by another force.
7. Determine the resistance of a lamp that draws 3 amperes when connected to a120-volt supply
A lamp draws 3 amperes of current and is connected to a 120 volt supply.
We are asked to find the resistance of the lamp.
Recall the Ohm's law is given by
[tex]V=IR[/tex]Where V is the voltage, I is current, and R is the resistance.
Re-arrange the equation for resistance
[tex]R=\frac{V}{I}[/tex]Substitute the given values
[tex]\begin{gathered} R=\frac{120}{3} \\ R=40\; \Omega \end{gathered}[/tex]Therefore, the resistance of the lamp is 40 ohms.
I don't know what I'm doing wrong I just can't seem to get the answers right. Help please.
The current and potential difference in the lumped element model of electrical circuits are dealt with by Kirchhoff's circuit laws, which are two equalities. German scientist Gustav Kirchhoff published the first account of them in 1845. This was done before James Clerk Maxwell and generalized Georg Ohm's work.
Because the sum of the currents entering and leaving the junction is equal, Kirchhoff's first law is based on the conservation of charge. The algebraic sum of potential drops in a closed circuit must equal zero, according to Kirchhoff's second law. Therefore, it is based on energy conservation.
A stream of charged particles—such as electrons or ions—moving through a conductor for electricity or into empty space is known as an electric current. It is determined by measuring the net rate of electric charge flow through a surface or into a control volume.
The amount of resistance in an electrical circuit represents the resistance to current flow. The Greek letter omega (Ω), which stands for resistance, represents ohms.
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In (Figure 1) a 6.00-m-long, uniform beam is hanging from a point 1.00 m to the right of its center. The beam weighs 155 N and makes an angle of 30.0∘ with the vertical. At the right-hand end of the beam a 100.0 N weight is hung; an unknown weight w hangs at the left end.If the system is in equilibrium, what is w ? You can ignore the thickness of the beam.
Given::
Length = 6.00 m
Weight of beam = 155 N
Angle = 30.0 degrees
Weight added at the right-end = 100 N
If the system is in equilibrium, let's find the weight, w, at the left-end.
Since the system is in equilibrium, the net torque of the system will be zero.
Now, we have the equation:
[tex](100)(2.00sin30)-w(4.00sin30)-(155)(1.0sin30)=0[/tex]Where w is the weight at the left end.
Let's solve for w.
We have:
[tex]\begin{gathered} (100)(1)-w(2)-(155)(0.5)=0 \\ \\ 100-2w-77.5=0 \\ \\ 100-77.5-2w=0 \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} 22.5-2w=0 \\ \\ 2w=22.5 \end{gathered}[/tex]Divide both sides by 2:
[tex]\begin{gathered} \frac{2w}{2}=\frac{22.5}{2} \\ \\ w=11.25\text{ N} \end{gathered}[/tex]Therefore, the weight hung at the left end is 11.25 N
ANSWER:
11.25 N
The double-sided mirror would produce virtual image to which objects? A)1,2 and 4 B)1,2 and 3 C)3 and 5 D)4 and 5
Objects 3 and 5.
Virtual images are always formed by convex mirrors and are formed by concave mirrors when the object is placed in front of f.
A railroad locomotive is at rest with its whistle shrieking and then it starts moving toward you. Describe what happens to the frequency and wavelength of the sound that hits your ear when:A) It is approaching you.B) It pulls away from you.What happens to the amplitude in both of these scenarios?
a) Frequency increases , the time between waves decreases , wavelenghts decrease
b) Frequency decreases , wavelenght increases.
The doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. AS the source approaches the listener , the sound waves get closer together, increasing their frequency and decreasing their wavelenghts.
Amplitude remains the same.
Jimmy launches his rocket straight up with an initial velocity of 25 m/s.a. What is the velocity as the rocket reaches maximum height?b. How long will it take the rocket to reach its maximum height?
Given that the initial velocity is u = 25 m/s.
We have to find the final velocity and time at its maximum height.
First, we need to find the maximum height.
[tex]h\text{ =}\frac{u}{2g}[/tex]Here, g = 9.81 m/s^2 is the acceleration due to gravity.
Substituting the values, the height will be
[tex]\begin{gathered} h=\text{ }\frac{25}{2\times9.81} \\ =1.274\text{ m} \end{gathered}[/tex](a) Let the final velocity be v.
The formula to calculate final velocity is
[tex]\begin{gathered} v^2=u^2-2gh^{} \\ v=\sqrt[]{u^2-2gh^{}} \end{gathered}[/tex]Substituting the values, the final velocity will be
[tex]\begin{gathered} v=\sqrt[]{\lbrack(25)^2-(2\times9.81\times1.274)\rbrack} \\ =\sqrt[]{600.01} \\ =24.49\text{ m/s} \end{gathered}[/tex](b) The time taken can be calculated by the formula,
[tex]\begin{gathered} t=\frac{v-u}{-g} \\ =\frac{24.49-25}{-9.81} \\ =0.051\text{ s} \end{gathered}[/tex]Thus, the final velocity is 24.49 m/s and time is 0.051 s
The rocket’s velocity just before it hits the ground is the same magnitude as the initial velocity. Use the appropriate kinematics equation to show that this is true.
WILL MARK BRAINLIEST LOTS OF POINTS!
The rocket’s velocity just before it hits the ground is the same magnitude as the initial velocity because change in length in y direction will zero.
if u and v are the initial and final velocities of the rocket then from kinematic equations it may be written as
[tex]s=ut+\frac{1}{2} at^{2}\\0=ut+\frac{1}{2} at^{2}\\\frac{1}{2} at^{2} =-u t\\at=-2u[/tex]
since
v=u + at
v=u - 2u
v = - u
Hence the rocket hit the ground with same velocity as it projected.
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A ball is thrown downwards from a height of 25m with an initial velocity of 10 m/s.
Find the velocity of the ball just before it hits the ground.
•V₁=10mls
h=25m
The velocity of the ball just before it hits the ground is 24.29m/s.
What is acceleration due to gravity?
The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s2 is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on earth's surface is 9.8 m/s2.
Given:
Initial velocity u = 10m/s
height h = 25m
we know that the acceleration due to gravity g is : 9.8m/s
Therefore using the formula,
v² - u² = 2gs where s is the height in this case and v is the final velocity.
v² - 10² = 2 x 9.8 x 25
=> v² = 490 + 100
=> v = √590
=> v = 24.29m/s
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3. A parallel-plate capacitor has a capacitance of 1.35 pF. If a 12.0 V battery is connected to this capacitor, how much electrical potential energy would it store?
Answer:
9.72 x 10^(-11)
Explanation:
The energy stored in a capacitor can be calculated as:
[tex]E=\frac{CV^2}{2}[/tex]Where C is the capacitance and V is the Voltage. So, replacing C by 1.35pF or 1.35 x 10^(-12) F, and V by 12.0 V, we get:
[tex]E=\frac{1.35\times10^{-12}F(12.0V)^2}{2}=9.72\times10^{-11}J[/tex]Therefore, the capacitot would store 9.72 x 10^(-11) J of energy.
The electrical potential energy that would be stored in the capacitor is 9.72 x 10^(-11) J. It is the energy which is present in the object when it is at rest.
What is Potential energy?Potential energy is the energy which is stored inside an object which is at rest.
For the parallel plate capacitors, capacitance is dependent upon its geometry, which is given by the formula:
C=ϵ⋅Ad or C = ϵ ⋅ A d ,
where, C is the value of the capacitance,
A is the area of each plate,
d is the distance between the plates, and
ϵ is the permittivity of the material between the plates of the parallel capacitor.
The energy stored in a capacitor can be calculated as:
E = CV²/2
where, C is the capacitance and V is the Voltage.
So, by replacing C with 1.35pF or 1.35 × 10⁻¹²F, and V by 12.0V, we get:
E = 1.35 × 10⁻¹² × (12)²/2
E = 9.72 × 10⁻¹¹ J
Therefore, the capacitor would store 9.72 × 10⁻¹¹ J of energy.
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Which measurements or observations are needed to calculate density
Answer:
mass and volume measurements are required to calculate density.
Can anyone tell me the final displacement of the puck
The final displacement of the puck is 4.14i + 14.14j
What is displacement?Displacement is the change in position of an object.
How to find the final displacement of the puck?Since for the first displacement of the puck from the goal is 20 m at 45°, we have that its displacement vector is d = (20cos 45°)i + (20sin45°)j.
= (20 × 0.7071)i + (20 × 0.7071)j.
= 14.14i + 14.14j
Also, for the second displacement is 10 m to the left. Its displacement vector is d' = -10i
So, we see that the total displacement is D = d + d'
So, adding the displacements, we have D = d + d'
= 14.14i + 14.14j + (-10i)
= 14.14i - 10i + 14.14j
= 4.14i + 14.14j
So, we see that the final displacement of the puck is 4.14i + 14.14j
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