2.25 m/s² of acceleration is required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds.
To find the answer, we need to know about the acceleration.
What is acceleration?Acceleration is given as the ratio of velocity to time.Mathematically, acceleration= velocity/time.What is the acceleration required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds?Here change in velocity of the car is 51-26= 25 mi/h. As 1 mi/h = 0.45 m/s. So 25mi/h = 11.25 m/s.Acceleration= (11.25m/s)/5s = 2.25 m/s².Thus, we can conclude that the constant acceleration is 2.25 m/s².
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What is weight??????????????????
Answer:
Weight is a body's relative mass, or the quantity of matter contained by it.
Explanation:
Hope this helps
Select odd one out and also give the reason:
a) Gravitational force , magnetic force , electrostatic force
Gravitational force is not related to other force because it is related to gravity.
How magnetic force and electrostatic force are related?Electrostatic force and magnetic force are link to each other because both have charges. Electrostatic force is type of force that is present between two electrically charged particles. They can either be a repulsive or attractive force.
So we can conclude that Gravitational force is not related to other force because it is related to gravity.
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Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.
The launch velocity of the projectile is 13.28 m/s.
What is projectile motion?
The motion of an object thrown in the air under the force of gravity is known as projectile motion.
Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,
s=u*t+(1/2)at^2.
where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.
For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.
-19.3 = (0)*t+(1/2)*(-9.8)*t^2
19.3 = (1/2)*(9.8)*t^2
t=1.98 s
Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,
X=vt
where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.
For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.
26.3=v*(1.98)
v=13.28 m/s
The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.
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You travel in a circle, whose circumstances is 8 kilometers, at an average speed 8 kilometers/hour. If you stop at the same point you started from what is your average velocity
Answer:
The Average Velocity is 2.22m/s.
Explanation:
Average Velocity = displacement(m) ÷ time(s)
8km = 8000m
8km/hr = 2.22m/s
Finding Time
Time = distance÷speed
Time = 8000m ÷ 2.22m/s = 3603.6s = 3604s
Finding Average Velocity
Average Velocity = displacement ÷ time
Average Velocity = 8000m ÷ 3604s
Average Velocity = 2.22m/s
Therefore the Average Velocity is 2.22m/s.
which two forms of energy do burning substances produce
Answer:thermal (heat) energy and radiant (light) energy.
Explanation:
Answer:
Answer: We use the chemical energy in fuels by burning them and transforming them into other types of energy: thermal energy, as when we burn fuel for heat; and kinetic energy, as when we burn gasoline to power our car's motion
Explanation:
hope it will help you
. A microwave oven operates at 2.4GHz with an intensity inside the oven of 2500 W/m^2. What are the amplitudes of the oscillating electric and magnetic fields
Answer: A microwave oven operates at 2.4GHz with an intensity inside the oven of 2500 W/m^2. Then, the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]
Explanation: To find the correct answer, we have to know more about the EM Waves.
What are EM waves?Electromagnetic waves can be defined as the waves, that can interact with the matter and are created as a result of vibrations of electric and magnetic fields.Maxwell theoretically predicted the existence of electromagnetic waves.A time varying magnetic field is a source of electric field.Similarly, an oscillating electric field gives rise to an oscillating magnetic field.The amplitudes of electric and magnetic fields can relate as,[tex]E_0=cB_0[/tex]
The expression for intensity of electromagnetic waves in terms of amplitude of magnetic field can be written as,[tex]I=\frac{1}{2u_0} cB_0^{2}[/tex]
How to solve the problem?From the above expression of intensity, we can find the expression for amplitude of magnetic field as,[tex]B_0=\sqrt{\frac{2u_0I}{c} }=\sqrt{\frac{2*4*3.14*10^{-7}*2500 }{3*10^{8} } } =4.57*10^{-6} T[/tex]
Thus, from the amplitude of magnetic field, we get, the amplitude of electric field as,[tex]E_0=cB_0=3*10^{8} *4.57*10^{-6} =1371 V/m[/tex]
Thus, from the above calculations, we get the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]
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25 points please help PLEASE
Journal prompt to be answered in 2 fully developed paragraphs
Prompt: What are some products (or programs) that you could purchase to help your performance in your current physical activity? How would the product (or program help)? Do you really think it is effective? Use specific examples from your experience.
Answer:
você tem um camping de em Rio do Rio de Janeiro ou em lojas em SP e SP e RJ e região metropolitana de Porto Velho de Porto Velho Rio
Long distance running is the physical activity I engage in now, and there are a number of items and programs that can greatly improve my performance.
A GPS powered watch is one such item which I think is quite useful. I can monitor my speed, distance, heart rate and other important information while running with this device. This data lets me change my pace, track my effort, and make sure I stay within my desired heart rate range. The watch also provides post-run analysis, which I can use to identify my weaknesses and modify my training appropriately. In my opinion, GPS running watches have changed the game by allowing me to run harder and push my physical limits.
A strength training program designed for runners has also been very helpful in enhancing my performance. The program's primary emphasis is on functional workouts that target the major running muscle groups, such as the core, glutes and legs.
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A rocket takes off from Earth's surface, accelerating straight up at 31.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 90.4 kg, including her space suit.
The normal force is 2820.48N in the negative y direction.
According to Newton's second law of motion,
Force = mass × acceleration
F = m×A
Note that rocket takes off from Earth's surface, accelerating straight up at 31.2 m/s² .
The rocket accelerates upwards, hence the acceleration will be negative because it defies gravity's law (it keeps going into space without coming down)
Acceleration of the rocket = -31.2m/s²
Mass of the astronaut = 90.4kg
Normal force acting on the astronaut = -31.2 × 90.4kg
= -2820.48N
Therefore, the normal force is 2820.48N in the negative y direction.
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A box weighing 660 N is sliding across a cement floor and the coefficient of sliding
friction between the box and the floor is 0.15. If the force pushing the box is 500 N, what
is the acceleration of the box?
O 6 m/s²
O 2 m/s²
O 10 m/s²
pts
O 9 m/s²
The acceleration of the box is 6 m/s²; option A.
What is the acceleration of the box?The acceleration of the box is determined as follows:
Frictional force = 660 * 0.15 = 99 N
Net force = 500 - 99 = 401 N
Force = mass * accelerationMass of the box = 660/9.8 = 67.3 kg
Acceleration = net force/mass
Acceleration = 401/67.3
Acceleration = 6 m/s²
In conclusion, the acceleration of the box is determined from the net force and the mass of the box.
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With a mass of 111.1 kg, is moving with a velocity of 3.000 m/s to the right. Mass is initially at rest and has a mass of 222.2 kg. After the elastic collision, bounces off with a velocity of 1.000 m/s to the left. Calculate the velocity of after the collision.
Let us assume the positive \hat{x} direction is towards the right.
During any collision, the total momentum is conserved.
That means, the total momentum before the collision is the same as the total momentum after the collision.
p before = P after
Given,
Mass 1, m_1 = 111.1kg
Mass 2, m_2 = 222.2kg
Initial Velocity of Mass 1, 1,1 3.00.î m/s
Initial Velocity of Mass 2, \overrightarrow v_{2,i} =0
Velocity of Mass 1 after collision, \overrightarrow v_{1,f} =-1.000\hat{x}\hspace{0.1cm}m/s
The total Momentum of the system before the collision is,
\overrightarrow p_{before} =m_1\overrightarrow v_{1,i} + m_2\overrightarrow v_{2,i}
=> \overrightarrow p_{before} =(111.1)(3.00\hat{x}) + 0
=> \overrightarrow p_{before} =333.3\hspace{0.1cm}\hat{x}\hspace{0.1cm}kgm/s
The total Momentum of the system after the collision is,
\overrightarrow p_{after} =m_1\overrightarrow v_{1,f} + m_2\overrightarrow v_{2,f}
=> \overrightarrow p_{after} =(111.1)(-1.000\hat{x}) + (222.2)\overrightarrow v_{2,f}
=> \overrightarrow p_{after} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}
We know the total momentum before the collision is the same as the total momentum after the collision.
p before = P after
=>333.3\hat{x} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}
=>(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f} = 333.3\hat{x}
=> (222.2)\overrightarrow v_{2,f} = 333.3\hat{x} + 111.1\hat{x}
=> (222.2)\overrightarrow v_{2,f} = 444.4\hat{x}
=> \overrightarrow v_{2,f} = \frac{444.4}{222.2}\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s
=> \overrightarrow v_{2,f} = 2.00\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s
Therefore the velocity of m_2 after collision is v_{2,f} = 2.00\hspace{0.1cm}m/s going to the right
Please check the attached file for a brief answer.
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The constant G in Newton's equation _______. View Available Hint(s)for Part A The constant G in Newton's equation _______. was measured by Newton shows gravity to be a relatively huge force produces equilibrium makes the units of measurement consistent
The constant G in Newton's equation makes the units of measurement consistent was measured by Newton shows gravity to be a relatively huge force produces equilibrium makes the units of measurement consistent.
Newton's law:The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact.
Newton's First lawNewton's Second lawNewton's Third lawNewton's first law, an object will not change its motion unless a force acts on it.
Newton's second law, the force on an object is equal to its mass times its acceleration.
Newton's third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
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In a test run, a rocket-powered car is driving down a test track at a constant speed when the rockets are fired. it then accelerates at 4 m/s2 for 8.0 s covering a distance of 224 m. at what speed was the car travelling when the rockets were fired?
With the use of the second equation of motion formula, the speed the car travelled when the rockets were fired is 12 m/s.
Speed in a Linear MotionLinear motion depicts movement in a straight line. While speed is the distance travelled per time taken.
Given that In a test run, a rocket-powered car is driving down a test track at a constant speed when the rockets are fired. it then accelerates at 4 m/s2 for 8.0 s covering a distance of 224 m.
The given parameters are;
Accelerates a = 4 m/s2Time t = 8sDistance s = 224m
The speed the car travelled when the rockets were fired can be calculated by using the below formula
s = ut + 1/2a[tex]t^{2}[/tex]
Substitute all the parameters into the formula
224 = 8u + 1/2 x 4 x [tex]8^{2}[/tex]
224 =8u + 2 x 64
224 = 8u + 128
224 - 128 = 8u
8u = 96
u = 96/8
u = 12 m/s
Therefore, the speed the car travelled when the rockets were fired is 12 m/s.
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An object weighing 2.7n while in air and 1.2n when completely immersed in water.find the relative density of the object
Answer:
1.8 = relative density (there are no units for relative density)
Explanation:
It displaces water equal to it's volume and gets buoyancy equal to that amount of water
2.7 - 1.2 = 1.5 N of buoyancy
density of water = 1 gm /cc
1.5 N = m (9.81)
m of water displaced = .1529 kg
152.9 cc of water will produce this buoyancy....this is the volume of the object
find mass of object 2.7 = m (9.81) shows m = .2752 kg = 272.5 gm
density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc
Relative to water (which is 1 gm / cc) the relative density is 1.8
====> it is 1.8 times denser than water and will sink when in water....
Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth
if dropped from 10.0 m on Earth and on the Moon. The gravitational strength on
the Moon is 1.6 N/kg.
The time needed for the hammer to reach the surface of the Earth is 3.54 s.
Time of motion of the hammer
The time of motion is calculated as follows;
t = √(2h/g)
where;
h is height of fallg is acceleration due to gravityt = √(2 x 10 / 1.6)
t = 3.54 s
Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.
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A 450-kg sports car accelerates from rest to 100 m/s in 20.0 s. What magnitude force does a 90.0 kg passenger experience during the acceleration
Magnitude of force experienced by the 90Kg passenger is 450N.
To find the answer, we need to know about the pseudo force.
What's the pseudo force experienced by a passenger traveling in a vehicle?When a passenger inside a vehicle, it experiences pseudo force in the opposite direction of motion of the vehicle.The pseudo force= mass of the passenger × acceleration of the vehicleWhat's the acceleration of the vehicle when it achieves a speed of 100 m/s from rest in 20s?Acceleration= Change in velocity/timeHere, change in velocity= 100m/s and time = 20sSo, acceleration= 100/20= 5m/s²What's the pseudo force experienced by the 90Kg passenger?Pseudo force= 90×5=450N.
Thus, we can conclude that the force experienced by the passenger is 450N.
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Verification of laws of reflection of sound (Record Work)
The Angle of Incidence is equal to the Angle of Reflection.
What is the Verification of laws of reflection of sound?Take a wooden drawing board and fix a white sheet of paper on it. In the middle of paper draw a straight line and Mark a point B on it. Draw a perpendicular . Place a mirror on line such that the flat side of the mirror is along the line. Hold the mirror in the mirror holder.
Fix two steel pins P and Q on the straight line AB at least 10 cm apart. Look for the images of the pins P and Q and fix two pins P a such that P', Q', and images of P and Q are all in the same straight line. Remove the pins and draw small circles around the pinpricks.
Remove the mirror and Join P'Q' which produce the straight line to meet at B. Measure ∠ABN = i and ∠CBN = r. It is found that ∠i=∠r. This proves that the Angle of Incidence is equal to the Angle of Reflection.
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Suppose the carts collided on a surface that had a slight incline to it. Would you expect the momentum to be conserved
Momentum is conserved when carts are collided on a slanting plane.
To find the answer, we need to know about the conversation of momentum.
What's the conversation of momentum?Conservation of linear momentum says the total momentum before the collision and after the collision remains the same. Mathematically, m1u1+m2u2 = m1v1+m2v2How is the momentum conserved when collision occurs on a slanting plane?On a slanting plane, the velocity has two components, horizontal component horizontal component Vertical componentSo, its momentum has also similar two components. The momentum is conserved along horizontal direction and vertical direction separately.Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.
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what provides the centiperal force to a satellite revolving around the earth?
Answer:
the gravitational force
Question 17
The driver of a car traveling along a straight road with a
speed of 72KM ph observes a signboard which give the
speed limit to be 54KM ph. The signboard is 70m ahead
when the driver applies the brakes, calculate the
acceleration of the car which will cause the car to pass the
signboard at the stated speed limit?
So, the acceleration of the car is -1.25 m/s². In other word, the car is also decelerating by 1.25 m/s².
IntroductionHi ! I will help you to discuss about "deceleration in a straight line movement". Please note in advance that deceleration is acceleration which has a negative value. When an object decelerates, the object will continue to move until it reaches a certain speed (which is less than before) or until it stops. The higher the deceleration value, an object that is moving will stop faster and cover a shorter distance.
Formula UsedIn this opportunity, I will give you the following equation to express the relationship between final velocity and initial velocity, acceleration, and distance.
[tex] \boxed{\sf{\bold{(v_t)^2= (v_0)^2 + 2 \times a \times s}}}[/tex]
With the following condition:
[tex] \sf{v_t} [/tex] = final velocity of an object (m/s)[tex] \sf{v_0} [/tex] = initial velocity of an object (m/s)a = acceleration that happen (m/s²)s = the shift or distance of the object (m)Problem SolvingWe know that:
[tex] \sf{v_0} [/tex] = initial velocity of an object = 72 km/h = 20 m/s[tex] \sf{v_t} [/tex] = final velocity of an object = 54 km/h = 15 m/ss = the shift or distance of the object = 70 mNote :
1 m/s = 3.6 km/h. So 10 m/s = 36 km/hWhat was asked ?
a = acceleration that happen = ... m/s²Step by step :
[tex] \sf{(v_t)^2 = (v_0)^2 + 2 \times a \times s} [/tex]
[tex] \sf{15^2 = 20^2 + 2 \times a \times 70} [/tex]
[tex] \sf{225 = 400 + 140 \times a} [/tex]
[tex] \sf{140 a = -175} [/tex]
[tex] \sf{a = \frac{-175}{140}} [/tex]
[tex] \boxed{\sf{\bold{a = -1.25 \: m/s^2}}} [/tex]
ConclusionHere, we see that the acceleration is -1.25 m/s². In other words, the car is also decelerating by 1.25 m/s².
A dentist holds a small mirror 1.9
cm from a patient’s tooth. The image
formed is upright and 5
times as largeas the object. (a) Is
the image real or virtual? (b) What
is the focal length of the mirror? Is it
concave or convex? (c) If the mirror
is moved closer to the tooth, does the
image get larger or smaller? (d) For
what range of object distances does
the mirror produce an upright image?
Answer:
follow me for legitimate answer… oh? U don’t like the feeling of people stealing ur points huh?! Well theirs a taste of your own medicine
Explanation:
A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acceleration of the stone and it's centripetal force?
force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N
Answer:
Acceleration: [tex]100\; {\rm m\cdot s^{-2}}[/tex] assuming that the radius of the rotation is [tex]1\; {\rm m}[/tex].
Centripetal force: [tex]15\; {\rm N}[/tex].
Explanation:
In a circular motion, if the tangential velocity is [tex]v[/tex] and the radius of the motion is [tex]r[/tex], the centripetal acceleration of the motion would be [tex]a = (v^{2} / r)[/tex].
In this question, it is implied that for this circular motion, [tex]v = 10\; {\rm m\cdot s^{-1}}[/tex] while [tex]r = 1\; {\rm m}[/tex]. Thus, the (centripetal) acceleration would be:
[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(10\; {\rm m\cdot s^{-2}})^{2}}{1\; {\rm m}} \\ &= 100\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].
Note that the unit of mass in this question is gram, whereas the standard unit for mass should be [tex]{\rm kg}[/tex] (so as to leverage the fact that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].) Apply unit conversion: [tex]m = 150\; {\rm g} = 0.150\; {\rm kg}[/tex].
Using that fact that [tex](\text{net force}) = (\text{mass}) \, (\text{acceleration})[/tex]:
[tex]\begin{aligned} (\text{net force}) &= (\text{mass}) \, (\text{acceleration}) \\ &= 0.150\; {\rm kg} \times 100\; {\rm m\cdot s^{-2}} \\ &= 15\; {\rm kg \cdot m \cdot s^{-2}} \\ &= 15\; {\rm N}\end{aligned}[/tex].
A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take a pulse to travel from one support to the other
The time taken by the pulse to travel from one support to the other is 0.208 s.
Given:The mass of the cord is m = 0.65 kg.
The distance between the supports is, d = 8.0 m.
The tension in the cord is T = 120 N.
The time taken by the pulse to travel from one support to the other is given as,
[tex]v=\frac{d}{t}[/tex]
[tex]t=\frac{d}{v}[/tex]
Here, v is the linear velocity of a pulse. Its value is,
[tex]v=\sqrt{\frac{T d }{m} }[/tex]
[tex]v=\sqrt{\frac{120 * 8}{0.65} }[/tex]
[tex]v= 38.43 m/s[/tex]
Then,
[tex]t=\frac{8}{38.43}[/tex]
[tex]t=0.208 s[/tex]
Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.
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The engine of a 1560-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to a speed of 100 km/h at full power on a level road. Is your answer realistic
The time required to accelerate this car from rest to a speed of 100 km/h at full power on a level road is 7.73 seconds.
Mass of the engine = 1500 kg
Power rating = 75 kW = 75,000 W
Final speed = 100 km/hr = = 27.78 m/s
v₁ = 0
Power = Work done ÷ Time
Work done = Final energy - Initial energy
=1/2 x 1560 x 27.78^2 = 1/2 x 1560 x 0^2\
= 578703.72 J
Thus,
75,000 = 578703.70 ÷ time=7.72 seconds.
Work is done every time a force moves something over a distance. By multiplying the force by the distance traveled in the direction of the force, you can calculate the energy transferred or the work done. Energy transmitted = work completed = force x distance traveled in the direction of the force.
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What is the net force (magnitude and direction) of the system below?
The net force (magnitude and direction) of the system given in the question is 40 N horizontal to the right
How to determine the net forceCase 1 (Net force between up and downward force)
Force up (Fu) = 50 NForce down (Fd) = 30 NNet force 1 (F1) = ?F1 = Fu - Fd
F1 = 50 - 30
F1 = 20 N up
Case 2 (Net force between right and left)
Force right (Fr) = 60 NForce left (Fl) = 20 NNet force 2 (F2) = ?F2 = Fr - Fl
F2 = 60 - 20
F2 = 40 N right
SUMMARY
Net force between up and down = 20 N upNet force between right and left = 40 N rightFrom the above, the net force between right and left (i.e 40 N) is greater than the net force between up and down (i.e 20 N)
Thus, the net force of the system will be 40 N horizontal to the right
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A force of 15 n is applied to a spring, causing it to stretch 0.3 m. what is the spring constant for this particular spring? n/m
Answer:
-50N/m
Explanation:
Force , F = 15N
Displacement , x = 0.3m
Spring constant , K = ?
K = -F/x
K = -15N/0.3m
K = -50N/m
Answer: 50 N/m
Explanation:
Edge 2022
What was the initial speed of a car if its speed is 40 m/s after 5 seconds of accelerating at -4 m/s²?
A. 50 m/s
B. 60 m/s
C. 25 m/s
D. 20 m/s
Answer:
[tex]\huge\boxed{\sf V_i=60 \ m/s}[/tex]
Explanation:
Given Data:Final speed = [tex]V_f[/tex] = 40 m/s
Time = t = 5 s
Acceleration = a = -4 m/s²
Required:Initial velocity = [tex]V_i[/tex] = ?
Formula:[tex]\displaystyle a = \frac{V_f-V_i}{t}[/tex]
Solution:Put the givens in the above formula
[tex]\displaystyle -4=\frac{40 - V_i}{5} \\\\Multiply \ -5 \ to \ both \ sides\\\\-4 \times 5 = 40 - V_i\\\\-20 =40-V_i\\\\Subtract \ 40 \ to \ both \ sides\\\\-20-40=-V_i\\\\-60\ m/s=-V_i\\\\60 \ m/s = V_i\\\\V_i=60 \ m/s\\\\\rule[225]{225}{2}[/tex]
Write an argument of any one
Do you think parallel universes exist?
or
Do you think portals exist?
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
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A toy car starts from the rest and accelerates at 1.50m/s2 [E] for 2.25s. What is the final velocity, of the car
3.375m/s is the final velocity of the car.
How do you find final velocity?The final velocity depends on how large the acceleration is and the distance over which it acts.
Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.
According to the question,
A toy car starts from the rest and accelerates
So the acceleration = 1.50m/s²
Time = 2.25s
[tex]x=x_{0} + vt[/tex]
[tex]x = 0 + ( 1.50m/s^2*2.25s)[/tex]
[tex]x = 3.375m/s[/tex]
The final velocity, of the car is 3.375 m/s.
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A uniform stick has a mass of 4.42 kg and a length of 1.23 m. It is initially lying flat at rest on a frictionless horizontal surface and is struck perpendicularly by a puck imparting a horizontal impulsive force of impulse 12.8 N s at a distance of 46.4 cm from the center. Determine the subsequent motion of the stick.
10.7 rad/s is the final angular velocity of the stick.
Given:
Mass of the stick = 4.42 kg
Length of the stick = 1.23m
Force of impulse (I) = 12.8 N s
The linear velocity of the stick, [tex]v=\frac{I}{m}[/tex]
[tex]v=\frac{12.8 N.s (\frac{1 kg m/s^2}{1 N}) }{4.42 kg}[/tex]
[tex]v[/tex] [tex]= 2.89 m/s[/tex]
Therefore, the final linear velocity of the stick is 2.89 m/s
∴[tex]w=\frac{12 Ir}{ml^{2} }[/tex]
[tex]w=\frac{12 ( 12.8 N.s ) ( 46.4 cm)}{(4.42 kg) (1.23 m)^2}[/tex]
[tex]w= \frac{12 (12.8 N.s) (46.4 cm) (\frac{10^-^2 m}{1 cm} )}{(4.42 kg) (1.23m)2}[/tex]
[tex]w=10.7 rad/s[/tex]
Therefore, 10.7 rad/s is the final angular velocity of the stick.
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