Projectile Motion for an Object Launched at an Angle
When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity.The analysis of the motion involves dealing with the two motions independently.
Explanation:
and Thais
Projectile Motion for an Object Launched at an Angle
When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity.The analysis of the motion involves dealing with the two motions independently.
Projectile Motion for an Object Launched at an Angle
When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity.The analysis of the motion involves dealing with the two motions independently.
A body initially at rest is accelerated at the rate of 0.2m/s for 5 seconds under contant force of 50 N. Calculate the workdone on the body
Answer:
the correct answer is 125 joule
Obtain a copy of the Student Guide for this lab. Your teacher may provide a copy, or you can click the link to access it. Be sure to read the entire Student Guide for this lab.
Did you read the Student Guide carefully?
yes
no
I cannot provide the information for the student guide as I do not have access to the specific lab or the guide. So, I cannot confirm whether I have read the student guide or not. Please provide me with the necessary information or a specific lab for me to provide accurate and helpful information. Instead of this I can give a brief about the student guide for a physics lab.
What is the Student Guide for the physics lab?A student guide for a physics lab is a document that provides students with information about the expectations, procedures, and safety guidelines for the lab. The guide usually includes the following information:
1. Introduction: A brief overview of the lab, including its purpose and objectives.
2. Materials and equipment: A list of the materials and equipment that students will use during the lab, as well as instructions on how to use them.
3. Procedure: A step-by-step guide on how to perform the experiment, including any calculations or data analysis that may be required.
4. Safety guidelines: A list of safety guidelines that students should follow during the lab, including proper handling of materials, use of protective equipment, and emergency procedures.
5. Pre-lab assignments: Assignments that students need to complete before the lab, such as reading the lab manual or reviewing relevant concepts.
6. Post-lab assignments: Assignments that students need to complete after the lab, such as analyzing data, writing lab reports, or answering questions.
Therefore, A student guide is an essential tool that helps students prepare for the lab, understand the concepts and procedures involved, and stay safe while conducting the experiment. It is important for students to read the entire guide carefully to ensure they have a clear understanding of the lab expectations and procedures.
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8. When wind dies down or stops blowing
happens.
Answer:
Deposition
Explanation:
Deposition occurs when water slows or ceases moving, the wind dies or stops blowing, or glaciers melt. The deposited material can also be used to construct new landforms. Waves, for example, can dump sediment in places offshore, where it might accumulate to form sand dunes.When the wind calms down or vegetation stops or slows the breeze, the sediment particles begin to fall. Water is another factor that may erode, move, or deposit sediment. Flowing water is a key erosive agent. Water transports dirt and rock fragments as it moves. Warm, wet air will not travel if wind systems are not present. Water will still evaporate, but it will not move, therefore everywhere else than a major body of water will dry up. Lakes may be fine since evaporating water will flow back into them, and the sea will be fine, but everywhere else will become extremely dry very rapidly. Wind is constantly blowing somewhere on the world at any given time. Winds are usually quiet near the middle of a high pressure system. Wind is the passage of air from a high pressure location to a low pressure area.... So essentially air is always moving. Weathering and erosion are caused by wind. Weathering is caused by wind blowing debris against cliffs and huge rocks. This wears down the rock, reducing it to sand and dust. Sand and dust are also eroded by wind. 2. Rocks are tough and durable, but they don't last forever. Weathering and erosion are processes that occur as a result of forces such as wind and water breaking down rocks. Weathering is the process through which rocks deteriorate. Weathering is caused by a variety of factors, including climate change.
Deposition occurs when wind dies down or stops blowing happens.
What is deposition?Deposition refers to the depositing of sediment carried by wind, flowing water, the sea, or ice. Sediment can be carried in the form of pebbles, sand, and mud, or as salts dissolved in water. Organic activity may later deposit salts.
The fractured rock fragments or soil particles are accumulated in an environment where the energy of the flowing river is reduced. It settles the particles, which compact and form sedimentary rocks over time.
Deposition can occur when a river enters a shallow area of water or when the volume of water decreases, such as after a flood or during a drought. Deposition is common at the mouth of a river near the end of its journey.
Thus, the answer is deposition.
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HELP QUICK
What is the correct definition of the word clan?
a.
a cluster of multistory structures made out of adobe
b.
a large family, or group of families, related through common ancestory
c.
a portable dome or arch shaped home made out of wooden sticks and covered in animal skin, mud, or grass
d.
religious leader who was believed to have the power to heal as well as predict and control future events
Answer:
The answer is B.
Explanation:
A Clan is usually just one family, but can be multiple families that are related. They were usually very close knit. Sometimes a clan is also a group of people with a common interest.
Why would the lamp turn off eventually after this circuit has been switched on for a long time?
I'll mark you as brainliest for this
An 9.20 kg ball, hanging from the ceiling by a light wire 130 cm long, is struck in an elastic collision by a 2.80 kg ball moving horizontally at 3.50 m/s just before the collision. Find the tension in the wire just after collision.
The tension in the wire is caused by the elastic collision between the balls and the magnitude of this tension is determined as 18.81 N.
Speed of the balls after the collisionThe speed of the balls after the collision is determined by applying the principle of conservation of linear momentum as shown below.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
9.2(0) + 2.8(3.5) = 9.2v₁ + 2.8(v₂)
9.8 = 9.2v₁ + 2.8v₂ -----(1)
One - directional linear velocityu₁ + v₁ = u₂ + v₂
0 + v₁ = 3.5 + v₂
v₂ = v₁ - 3.5 ---(2)
Substitute (2) into (1)
9.8 = 9.2v₁ + 2.8(v₁ - 3.5)
9.8 = 9.2v₁ + 2.8v₁ - 9.8
19.6 = 12v₁
v₁ = 19.6/12
v₁ = 1.63 m/s
Tension on the wirev = √T/μ
v² = T/μ
T = v²μ
where;
μ is mass per unit length of the wire = 9.2/1.3 = 7.08 kg/m
T = (1.63²) x 7.08
T = 18.81 N
Thus, the tension in the wire just after collision is 18.81 N.
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A teenager is standing at the rim of a large horizontal uniform wooden disk that can rotate freely about a vertical axis at its center. The mass of the disk (in kg) is M and its radius (in m) is R. The mass of the teenager (in kg) is m. The disk and teenager are initially at rest. The teenager then throws a large rock that has a mass (in kg) of m_rock. As it leaves the thrower's hands, the rock is traveling horizontally with speed v (in m/s) relative to the earth in a direction tangent to the rim of the disk. The teenager remains at rest relative to the disk and so rotates with it after throwing the rock. In terms of M, R, m, m_rock and v, what is the angular speed of the disk
Answer:
[tex]\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}[/tex]
Explanation:
[tex](inertia.of.the.disk) Id =\frac{M R^{2}}{2}\\(inertia.of.the.solid) Is = m R^{2}\\\\Itotal = Id + Is = \left(\frac{M}{2}+m\right) R^{2}\\\\\\L=\vec{\gamma} \times \vec{p}={mvr} \quad \& \quad L=I \omega\\\\Mrock \timesv \times R=\left(\frac{M}{2}+m\right) R^{2} \times \omega\\\\\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}\\[/tex]
The required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].
Given data:
The mass of disk is, M.
The radius of disk is, R.
The mass of teenager is, m.
The mass of rock is, [tex]m_{rock}[/tex].
The speed of rock is, v.
Here we need to apply the concept of angular momentum to obtain the value of angular speed of disk. The expression for the angular momentum is given as,
[tex]L = I_{total} \times \omega[/tex] .......................................(1)
Here, [tex]\omega[/tex] is the angular speed and [tex]I_{total}[/tex] is the total moment of inertia of system.
And its value is,
[tex]I_{total} = I_{disk}+I_{solid}\\\\I_{total} = \dfrac{MR^{2}}{2}+mR^{2}[/tex] ..........................................(2)
And the angular momentum is also expressed as,
[tex]L = p \times R\\L =( m_{rock}v) \times R[/tex] ....................................................(3)
Then, using the equation (1), (2) and (3) we have,
[tex]( m_{rock} \times v) \times R = (\dfrac{MR^{2}}{2}+mR^{2}) \times \omega\\\\m_{rock} \times v = (\dfrac{MR}{2}+mR) \times \omega\\\\\\\omega = \dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex]
Thus, we can conclude that the required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].
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5. A car travels at a constant speed of 20 m/s around a horizontal circular track of radius 80 m. The tires roll without slipping. We can conclude that the coefficient of static friction between the tires and track:
Answer:
.5
Explanation:
First we define the variables
v=20
r=80
Next we construct our formula, since our tires do not slip, our centripetal force is equal to our friction
ma=mg*u
We calculate our centripetal force through finding our acceleration (v^2/r) and assuming our mass is a variable m (we cancel it out later).
mv^2/r = mg*u
All we have left is substitution and simplification
400m/80=mg*u
5=g*u
G is our gravitational acceleration (I will use 10, use 9.8 instead if you prefer)
5=10*u
u=.5
which equation can be used to solve for acceleration?
Answer:
the answer is the first one.
Explanation:
one way to find out is if you look at the other equations they are mixed matched. For an example a=d/t should be s= d/t.
Answer:
a
Explanation:
i did it
A train that travels 100.0 kilometers in 4.0 hours is traveling at what speed?
Answer:
25 kilometers an hour.
Explanation: Since there is 100 kilometers, you dived it by four hours and you get the kilometers per hour.
Answer:
25 kmph
Explanation:
100 divided by 4 is 25
hope it helps
After an airplane takes off, it travels 10.4 km west, 8.7 km north, and 2.1 km up. How far is it from the take off point?
The displacement of the airplane from the take off point after travelling the given distance is 15 km.
Displacement of the airplaneThe displacement of the airplane is the shortest distance between the initial position of the airplane and the final position of the airplane. The displacement of the of the airplane is calculated as follows;
total upward distance = 8.7 km + 2.1 km = 10.8 kmtotal horizontal distance = 10.4 kmR² = Rx² + Ry²
R² = 10.4² + 10.8²
R² = 224.8
R = √224.8
R = 15 km
Thus, the displacement of the airplane from the take off point is 15 km.
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Which trait is found in early embryos of both birds and humans but then disappears from each develpment
A 57 kg person in a rollercoaster moving through the bottom of a curved track of radius 42.7 m feels a normal force of 995 N. How fast is the car moving?
Answer:
Use Fc centripetal force as positive and W the weight as negative
N = m v^2 / R + m g
v^2 = (N - m g) R / m
v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2
v = 18.1 m/s
Note: N - m g is the net force producing the centripetal force
An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)
Answer:
Acceleration = 8.27 cm/s²
Explanation:
We are given;
initial velocity; v_i = 10.5 cm/s
Initial position; x_i = 2.72 cm
Time; t = 2.30 s
final position; x_f = 5.00 cm
To find the acceleration, we will make use of the formula;
x_f - x_i = (v_i * t) - (½at²)
Plugging in the relevant values, we have;
5 - 2.72 = (10.5 × 2.3) - (½ × a × 2.3²)
2.28 = 24.15 - 2.645a
24.15 - 2.28 = 2.645a
2.645a = 21.87
a = 21.87/2.645
a = 8.27 cm/s²
A skier of mass 58.0 kg starts from rest at the top of a ski slope of height 70.0 m . Part APart complete If frictional forces do −1.04×104 J of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be g = 9.80 m/s2 . 31.8 m/s Previous Answers Correct Part BPart complete Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.250. If the patch is of width 68.0 m and the average force of air resistance on the skier is 150 N , how fast is she going after crossing the patch? 18.1 m/s Previous Answers Answer Requested Part C After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.30 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?
Answer:
A) v = 31.8 m/s
B) v_f = 18.1 m/s
C) F_avg = 4130.7 N
Explanation:
A) From law of conservation of energy, we know that;
mgh = W_f + ½mv²
v is the speed at which she is going at the bottom of the slope.
Thus, making v the subject, we have;
v = √[2gh - (2W_f)/m)]
We are given;
h = 70 m
m = 58 kg
W_f = 1.04 × 10⁴ J = 10400 J
Thus;
v = √[(2 × 9.8 × 70) - (2 × 10400)/58)]
v = 31.8 m/s
B) Total force will be given by the formula;
F_t = F_k + F_r
Where;
F_k is force of kinetic friction = mg•μ_k
μ_k = 0.25
So, F_k = 58 × 9.8 × 0.25
F_k = 142.1 N
We are given force of air resistance(F_r) as 150 N
Thus;
F_t = 142.1 + 150
F_t = 292.1 N
Final velocity is gotten from the formula;
v_i² - v_f² = 2F_t•L/m
Thus;
v_f = √[v_i² - (2F_t•L/m)]
Where, v_i = 31.8 m/s, F_t = 292.1, m = 58 kg, L = 68 m
Thus;
v_f = √[31.8² - (2 × 292.1 × 68/58)]
v_f = 18.1 m/s
C) the average force exerted on her by the snowdrift as it stops her is given by the formula;
F_avg = m•v_f²/2l
F_avg = 58 × 18.1²/(2 × 2.3)
F_avg = 4130.7 N
A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?
Answer:
[tex]v=697.2km/h[/tex]
Explanation:
Hello.
In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:
[tex]V=\frac{d}{t}[/tex]
The distance is clearly 1743 km and the time is:
[tex]t=2h+30min*\frac{1h}{60min} =2.5h[/tex]
Thus, the velocity turns out:
[tex]v=\frac{1743km}{2.5h}\\ \\v=697.2km/h[/tex]
Which is a typical velocity for a plane to allow it be stable when flying.
Best regards.
Which bWhich piece of furniture will have the most inertia and give the furniture movers the most difficulty in moving?
a 50kg couch
a 5kg dining chair
a 75kg dining table
an 10kg end tableest describes friction?
Answer:
The answer is C: a 75kg dining table
Explanation:
Took it on the Unit Test: 2020
A scientist is building a 3-D model of a small and fast meat-eating dinosaur. How will the dinosaur look in the model?
A fast meat eating dinosaur must have strong claws for catching prey as well as strong teeth for eating meat.
What is a dinosaur?Dinosaurs are ancient reptiles that occurred in different forms and shapes and are currently considered a part of the evolutionary sequence.
A carnivorous dinosaur must have strong claws for catching prey as well as strong teeth for eating meat.
The image of a carnivorous dinosaur is shown in the image attached to this answer.
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A car with a mass of 950 kg is tied to a crane by a strong cable and being lowered at a rate of 1.2 m/s2. Draw a free-body diagram showing all the forces acting on the car. Include the values of each force. Show all calculations.
Answer:
Data that we know:
Mass of the car = 950kg.
Gravitational acceleration = -9.8m/s^2.
Acceleration of the car = -1.2 m/s^2.
(the negative signs are because that accelerations point downwards)
Now, the total acceleration of the car is:
-1.2m/s^2 = -9.8m/s^2 + Ac.
Where Ac is the acceleration caused by the cable, that is opposite to the acceleration of the car:
Ac = 9.8m/s^2 - 1.2m/s^2 = 8.6m/s^2.
Now we have the accelerations, remember that by second's newton law we have:
F = m*a
Then the force of the cable is:
Fc = 8.6m/s^2*950kg = 8,170N.
The gravitational force is:
Fg = -9.8m/s^2*950kg = -9,310N
The net force is:
Ft = 8,170N - 9,310N = -1,140N.
Below is the free body-diagram.
What is the standard unit of measure for pressure?
Answer:
Pascal
Explanation:
The standard SI unit for pressure measurement is the Pascal (Pa) which is equivalent to one Newton per square meter (N/m2) or the KiloPascal (kPa) where 1 kPa = 1000 Pa.
A dragster crosses the finish line with a velocity of 140m/s . Assuming the vehicle maintained a constant acceleration from start to finish, what was its average velocity for the race?
Answer:
Answer is attached
The average velocity for the race is 70 m/s if the dragster crosses the finish line with a velocity of 140 m/s.
What is linear acceleration?It is defined as the rate of change in linear velocity with respect to time. It is also known as linear acceleration.
It is given that:
A dragster crosses the finish line with a velocity of 140m/s.
The vehicle maintained a constant acceleration from start to finish.
As we know,
The average velocity can be given as:
V(avrg) = (v1 + v2)/2
v1 = 140 m/s
v2 = 0 m/s
v(avrg) is the average of the velocity.
V(avrg) = (140 + 0)/2
V(avrg) = 140/2
V(avrg) = 70 m/s
Thus, the average velocity for the race is 70 m/s if the dragster crosses the finish line with a velocity of 140 m/s.
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[tex]hello \: [/tex]
[tex]good \: evening[/tex]
[tex]what \: is \: acceleration[/tex]
Have a great day.
Answer:
acceleration, rate at which velocity changes with time, in terms of both speed and direction. A point or an object moving in a straight line is accelerated if it speeds up or slows down. Motion on a circle is accelerated even if the speed is constant, because the direction is continually changing. For all other kinds of motion, both effects contribute to the acceleration.
Because acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is also a vector quantity. Acceleration is defined as the change in the velocity vector in a time interval, divided by the time interval. Instantaneous acceleration (at a precise moment and location) is given by the limit of the ratio of the change in velocity during a given time interval to the time interval as the time interval goes to zero (see analysis: Instantaneous rates of change). For example, if velocity is expressed in metres per second, acceleration will be expressed in metres per second per second.
Explanation:
An object whose weight is 10kg is placed on smooth plane inclined at 30° to the horizontal. find the acceleration of the object as it moves down the plane
please i need help
Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
5. 500 kg gold with a volume of 0.026 m
A. 0.193 kg/m3
B. 13 kg/m3
C. 19230 kg/m3
4
Answer:
19230 kg/m^3
Explanation:
Divide its mass by its volume to get its density:
500 kg/0.026 m = 19230 kg/m^3
Also, please provide the question next time since other's didn't know what they should solve for in this problem-- I only knew because I had this same problem.
Answer:
19230 kg/m3
Explanation:
I took the test, that is the answer
Describe cytochrome c and how it can be used to provide evidence of evolutionary relationships. Then, explain how a scientist might determine which two species in a group of five species are more closely related using evidence from cytochrome c. You may use the evidence in the graph to support your answer.
Write 4-6 sentences:
Answer:
the c is in the cycle lab where they were created and then it cycles then it reoves it to the right.
Explanation:
One scientific investigation may result in
A.
further investigations.
B.
the generation of new ideas or phenomena to study.
C.
the development of new methods or procedures for investigation.
D.
all of these
A 1,120 kg car is traveling with a speed of 40 m’s .find it.s energy
Answer:
896000 Joule.
Explanation:
Solution,
Given,
Mass (m) = 1,120 kg.
Velocity (v) = 40 m/s.
Now,
Kinetic Energy = 1/2mv²
= 1/2 × 1,120 × 40²
= 896000 Joule.
Help as soon as possible
PLS
Answer:
B-A-C
Explanation:
The idea is to evaluate the slopes of the three graphs. Notice that the three graphs are a representation of position (x) as a function of time (t).
Then, the slope of those lines are giving information of the change in position over the change in time (rise over the run). For the graph with positive slope the velocity is positive, for those with negative slope (line going down) the velocity is negative. But we are asked to compare speeds, which are the magnitude of each velocity (slope) so it is important to determine the graph with the largest slope (graph C), and that with the smallest (graph B)
Then the order is: B-A-C (third option answer option)
A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a doorframe and a weight with mass 7.00 kg is hung from the other end. The final length of the spring is 44.5 cm. (a) Find its spring constant (in N/m).
When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
a train accelerates at a -1.5 m/s2 for 10 seconds. if the train had an initial speed of 32 m/s, what is its new speed?
Answer:17m/s
Explanation:
In this question we use first law of motion which is
vf=vi+at
By putting values
vf=32+10×-1.5
vf=17m/s
Answer:
yeah its 17m/s
Explanation: