Answers
Answer:
The starting velocity.
Explanation:
We must understand that this equation comes from the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 33 [m/s]
Vo = starting velocity [m/s]
a = acceleration = 3 [m/s²]
t = time = 30 [s]
So, these values can be assembly in the following way:
[tex]v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}[/tex]
Related Questions
What are 3 things you could you do this week to help you connect better with kids in
your classes?
Answers
Answer:
In my physics class, something that helps connect better with kids is keeping connected with them always make sure to ask if they understand what you're teaching if they are following because sometimes most kids are to afraid to admit that they are lost, another way to connect with kids is maybe posting surveys to be able to check in with each student especially during this hard times :)
Explanation:
1. start with fun activities.
2. Encourage single-tasking.
3. Designate a learning playing field.
Donuts are a simple or complex carbohydrate? *
Ok
Answers
Answer:
simple carbohydrate
Explanation:
Simple carbs are like cakes, donuts, and candy, etc.
Hope this helps :D
A north magnetic pole is facing another north magnetic pole with a distance xx. If the distance between the poles becomes 12x12x, what happens to the magnitude of the field energy between them
Answers
Answer:
Explanation:
When two north magnetic poles are placed close to each other , they will repel each other . If distance between them is increased due to mutual repulsion from x to 12 x , work is done by the magnetic field . This results in decrease of magnetic field energy .
Hence , when the distance between the poles becomes 12x from x , the magnitude of the field energy between them decreases .
A car travels 100 km due East in 2 hours. It then travels 50 km South in 1 hour. What is its average velocity?
Answers
The average velocity of the car is 37.27 km/h.
The given parameters;
Initial displacement of the car, x = 100 kmTime of motion, t = 2 hoursFinal displacement of the car, y = 50 kmtime of motion, t = 1 hourThe average velocity of the car is calculated as follows;
[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]
The total displacement of the car is calculated as follows;
[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]
The average velocity of the car is calculated as follows;
[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]
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Two trucks are traveling east on I-40 when the Walmart truck (61 mph, 57 ft long) decides to pass the FedEx truck (53 mph, 60 ft long). A 'pass' starts when the front of the Walmart truck is 45 ft behind the rear of the FedEx truck and ends when the rear of the Walmart truck is 45 ft in front of the front of the FedEx truck.
How long does it take for the Walmart truck to pass the FedEx truck?
How far does the Walmart truck travel while passing the FedEx truck?
Answers
Answer:
The answer is below
Explanation:
Let us assume that the slower truck which is the FedEx truck travels d mile. The pass started when the Walmart truck was 45 ft behind the FedEx truck and ended when the Walmart truck is 45 ft in front of the FedEx truck.
The additional distance covered by the truck Walmart = 45 ft + 60 ft (length of FedEx truck) + 45 ft = 150 ft
1 mile = 5280 ft
150 ft = 150 ft * 1 mile/5280 ft = 0.0284 mile
Therefore the Walmart truck covered an additional 0.0284 mile. Hence the distance covered by the walmart truck = d + 0.0284
Let us say it took t hours for the pass to be completed. Hence:
For FedEx; 53 mph = d / t. t =d / 53
For Walmart; 61 mph = (d + 0.0284) / t. t = (d + 0.0284) / 61
Since it took the trucks the same time. Hence:
d / 53 = (d + 0.0284) / 61
61d = 53d + 1.506
8d = 1.506
d = 1.506/8
d = 0.1882 mile = 993.75 ft
The FedEx truck traveled 993.75 mile while the Walmart truck traveled 1143.75 ft (993.75 + 150 ft)
The time (t) = d / 53 = 0.1882 / 53
t=0.00355 h = 12.8 seconds
Running along a rectangular track 50m × 40 m you complete one lap in 100 s. What is your average velocity for the lap ?
Answers
Answer:
1.8 m/s
Explanation:
Distance = 2*50 +2*40 [m] = 180 [m]
Time = 100 [s]
Velocity = Distance/Time = 1.8 m/s
Average velocity for the lap is 1.8 m/s
GIven:
Length of rectangular track = 50 m
Width of rectangular track = 40 m
Time taken to cover a lap = 100 seconds
Find:
Average velocity for the lap
Computation:
Perimeter of rectangle = Length of one lap
So,
Perimeter of rectangle = 2(l + b)
So,
Length of one lap = 2[50 + 40]
Length of one lap = 2[90]
Length of one lap = 180 meter
Average velocity = Distance / Time
Average velocity for the lap = 180 / 100
Average velocity for the lap = 1.8 m/s
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Choose the words that make each statement correct.
(i) After being released from rest in a uniform electric field, a pro- ton will move [(a) in the same direction as; (b) opposite the direction of] the electric field to regions of [(c) higher; (d) lower] electric potential.
(ii) After being released from rest in a uniform electric field, an electron will move [(e) in the same direction as; (f) opposite the direction of] the electric field to regions of [(g) higher; (h) lower] electric potential.
Answers
Answer:
i). (a) in the same direction as , (d) lower
ii). (f) opposite the direction of, (g) higher
Explanation:
An proton may be defined as a sub atomic particle and it has a positive electrical charge. Its mass is slightly less than that of a neutron. When a proton is placed in an electrical field that is uniformly charged, it is at rest. When the proton first moves out from rest from the uniform electric field, it will move in a direction which is same as that of the electric field and it will move to a region of higher potential.
An electron is defined as the subatomic particle having negative electric charge. When an electron is released form rest from an uniform electric field, it will move in the opposite direction of the uniform electric field and will move to the region of lower electric potential.
Using the graph, when (in seconds) would you expect the object to have the highest velocity? Explain your answer.
Answers
Answer:
blue
Explanation:
nswer the following about two objects, A and B, whose motion produced ihe following ... "A" starts with a greater (t) position. Since they ... since you are moving away from the origin
The discharge of a pump is 3 m above the inlet. Water enters at a pressure of 138 kPa and leaves at a pressure of 1380 kPa. The specific volume of the water is 0.001 m3/kg. If there is no heat transfer and no change in kinetic or internal energy, what is the work per unit mass
Answers
Answer:
The answer is "[tex]1.271 \ \frac{KJ}{kg}\\[/tex]"
Explanation:
[tex]\Delta e_{mech} =\frac{P_2-P_1}{P} + \frac{v_{2}^2-v_{1}^2}{2}+g(z_2-z_1)\\\\\Delta e_{mech} =\frac{ 1380 -138 \times 1000 }{1000} + 0+g(3-0)\\\\P = \frac{1}{v}= \frac{1}{0.001} = 1000 \frac{kg}{m} \\\\ \Delta e_{mech} =1242 +9.81(3)= 1271.43 \frac{J}{kg} \\\\\text{work per unit pass}= 1.271 \ \frac{KJ}{kg}\\[/tex]
A car is traveling 100 km/hr. How many hours will it take to cover a distance of 850 km?
Your answer:
.118 hours
8.5 hours
7.5 hours
23 hours
Answers
Answer:
8.5 hours
Explanation:
The Dark Matter Halo of our galaxy is The Dark Matter Halo of our galaxy is a halo component curiously absent in most others galaxies which astronomers have examined. the extension or the central bulge into space where dust makes the stellar material too dark to see. a disproven super-gravity concept once thought to be possible, but contradicted by evidence. a massive non-luminous cloud of material that surrounds the galaxy, providing the dominant source of gravity in our galaxy.
Answers
Answer:
you mean this halo right
Explanation:
Halo is an American military science fiction media franchise managed and developed by 343 Industries and published by Xbox Game Studios. The franchise and its early main installments were originally developed by Bungie. The central focus of the franchise builds off the experiences of Master Chief John-117, one of a group of supersoldiers codenamed Spartans, and his artificial intelligence (AI) companion, Cortana.
The original trilogy centers on an interstellar war between humanity and an alliance of aliens known as the Covenant. The Covenant, led by their religious leaders called the Prophets, worship an ancient civilization known as the Forerunners, who perished while defeating the parasitic Flood. The eponymous Halo Array are a group of immense, habitable, ring-shaped superweapons that were created by the Forerunners to destroy the Flood, but which the Covenant mistake for religious artifacts that, if activated, would transport them on a Great Journey to meet the Forerunners. They are similar to the Orbitals in Iain M. Banks' Culture novels, and to a lesser degree to author Larry Niven's Ringworld concept.[1][2][3][4]
The games in the series are critically acclaimed, with the original considered the Xbox's "killer app".[5] This led to the term "Halo killer" being used to describe console games that aspire, or are considered, to be better than Halo.[6] Fueled by the success of Halo: Combat Evolved, and by marketing campaigns from publisher Microsoft, its sequels went on to record-breaking sales.[7][8][9] The games have sold over 65 million copies worldwide, with the games alone grossing almost $3.4 billion.[10][11][12] Halo has since become one of the highest-grossing media franchises of all time. These strong sales led to the franchise's expansion to other media; the Halo Universe now spans multiple best-selling novels, graphic novels, comic books, short movies, animated movies and feature films, as well as other licensed products.
T
a ball is thrown upward with a beginning speed of 40m/s. The graph below shows how the speed of the ball changes until it reaches its maximum height.
use the graph to find
a) the time when the ball reaches its maximum height
b) the acceleration of the ball
c) the maximum height the ball went
Answers
Answer:
a) 4.0816s
b) -9.8 ms^-1
c) 81.63265m
A train 350 m long is moving on a straight track with a speed of 84.1 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 15.8 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
Answers
Answer:
t = 25.0 s
Explanation:
Assuming that the engineer applies the brakes just over the crossing, the train moves exactly 350 m at a constant acceleration, with a final speed (when the last car of the train leaves the crossing) of 15.8km/h.Since we know the initial and final speeds, and the horizontal distance traveled (the length of the train) we can use the following kinematic equation to get the acceleration:[tex]v_{f}^{2} - v_{o}^{2} = 2*a* \Delta x (1)[/tex]
Since we need to find the time in seconds, it is advisable to convert vf and vo to m/s first, as follows:[tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]
[tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]
Replacing (2) and (3) in (1), since Δx =350m, we can solving for a:[tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]
In order to get the time, we can simply use the definition of acceleration, and rearrange terms:[tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Answers
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
What genetic test would you get if there was a specific genetic disease in your family
Answers
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the proton at the end of the motion
Answers
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, [tex]v_p_i[/tex] = 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
[tex]W =K.E_f - K.E_i[/tex]
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
[tex]K.E_f =EQd + \frac{1}{2}m_pv_p_i^2[/tex]
[tex]m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C[/tex]
[tex]K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\[/tex]
[tex]K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J[/tex]
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
What is the maximum height achieved if a 0.400 kg mass is thrown straight upward with an initial speed of 40.0 m⋅s−1? Ignore the effect of air resistance
Answers
The maximum height : 81.63 m
Further explanationGiven
0.4 kg mass
vo = initial speed = 40 m/s
Required
the maximum height
Solution
We can use the law of conservation energy(ME=PE+KE) or use parabolic motion
For parabolic motion :
h max = (vo²sin²θ)/2g
θ = 90°(straight upward)
Input the value :
h max = (40²sin²90°)/2 x 9.8
h max = 81.63 m
What fitness component is plank exercises?
Answers
Answer:
The plank (also called a front hold, hover, or abdominal bridge) is an isometric core strength exercise that involves maintaining a position similar to a push-up for the maximum possible time.Explanation:
please mark me has brainllestAt what speed, in m/s, would a moving clock lose 1.3ns in 1.0 day according to experimenters on the ground?
Answers
Answer:
v=0.14c
Explanation:
Emma is working in a shoe test lab measuring the coefficient of friction for tennis shoes on a variety of surfaces. The shoes are pushed against the surface with a force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine has to pull with a force of only 200 N to keep the material moving.
a. What is the coefficient of static friction between the shoe and the material?
b. What is the coefficient of dynamic friction between the shoe and the material?
c. Draw a Free Body Diagram for the above.
Answers
Answer:
Explanation:
Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .
a )
Given,
Normal force N = 400 N
Force of friction = 300 N
μ = coefficient of static friction = ?
Putting the values ,
300 = 400 μ
μ = .75
b )
Normal force N = 400 N
Force of friction = 200 N
μ = coefficient of kinetic friction = ?
Putting the values ,
200 = 400 μ
μ = .50
c ) see attached file .
A sports car of mass m has the same kinetic energy as an SUV with a mass 3m as each is driven along the same road. Which vehicle, if either, has the larger momentum and what is the difference in their momenta, if any
Answers
Answer:
Explanation:
Kinetic energy ( KE ) = 1/2 m v²
= m²v² / 2 m = p² / 2m where p is momentum
KE = p² / 2m
p² = 2m KE
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
p₁ = √ ( 2 m x KE )
p₂ = √ ( 6 m x KE )
p₂ - p₁ = √ ( 6 m x KE ) - √ ( 2 m x KE )
= √KE m ( √6 - √2 )
Kinetic energy ( K.E )
[tex]= \frac{1}{2} m v^2\\\\= \frac{m^2 v^2}{2 m} \\\\= \frac{p^2}{2m}[/tex]
where p is momentum
[tex]K.E =\frac{p^2}{2m}\\\\p^2 = 2m. KE[/tex]
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
[tex]p_1 =\sqrt{(2m*K.E)}\\\\p_2 = \sqrt{(6m*K.E)} \\\\p_2 - p_1 = \sqrt{(6m*K.E)} -\sqrt{(2M*K.E} \\\\p_2 - p_1 = \sqrt{K.E m(\sqrt{6}-\sqrt{2}) }[/tex]
The difference is shown above.
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A 45.0-kg girl stands on a 13.0-kg wagon holding two 18.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 6.5 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time
Answers
Answer:
v = 4.0 m/s
Explanation:
Assuming no external forces acting during the instant that the girl throws the weights, total momentum must be conserved.Since all the masses at rest initially, the initial momentum must be zero.So, due to momentum must keep constant, final momentum must be zero too, as follows:[tex]p_{f} = m_{w} * v_{w} + m_{g+w} *v_{g+w} = 0 (1)[/tex]
Assuming the direction towards the back of the wagon as positive, and replacing the masses in (1), we can solve for vg, as follows:[tex]v_{g+w} =- \frac{m_{w} *v_{w}}{m_{g+w} } = - \frac{36.0kg *6.5m/s}{58.0kg } = -4.0m/s (2)[/tex]
This means that the girl (along with the wagon on she is standing) will move at a speed of 4.0 m/s in an opposite direction to the one she threw the weights.It takes 500 W of power to move an object 96 m in 12 s. What force is being applied to the object?
Answers
Answer:
Explanation:
Power, by definition, is the amount of work per unit of time. We arent given work in this question, but we can find it because work is how much force per unit of distance.
[tex]P=\frac{W}{t_f-t_i} =\frac{F*d}{t_f-t_i}[/tex]
Plug in all the values, and its algebra at this point
[tex]500=\frac{F*96}{12}[/tex]
6000 = 96F
F = 62.5 Newtons
The force which is being applied to the object with the power of 500 Watt is 62.5 N.
What is Power?
Power can be defined as the rate of doing work. It is the work done in unit time. The SI unit of power is Watt (W) which is equal to joules per second (J/s). Sometimes, the power of motor vehicles and other machines is given in terms of Horsepower (hp), This unit is approximately equal to 745.7 watts of power.
The power of a system can be calculated as the product of force applied and distance travelled by the system per unit time taken.
Therefore, the force of the object with power 500W can be calculated as:
P = f × d/ t
where, P = Power of the object,
f = Force applied,
d = distance travelled,
t = time taken to cover the distance
f = (P × t)/ d
f = (500 × 12) / 96
f = 6000/ 96
f = 62.5 N
Therefore, the force which is being applied to the object is 62.5N (Newton).
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Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a current of 7.68 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 65.0o relative to a magnetic field whose magnitude is 0.59 T. The magnitude of the net magnetic force experienced by the two-wire unit is 4.11 N. What is the current I
Answers
Answer:
[tex]4.77\ \text{A}[/tex]
Explanation:
F = Magnetic force = 4.11 N
[tex]I_n[/tex] = Net current
[tex]I_2[/tex] = Current in one of the wires = 7.68 A
B = Magnetic field = 0.59 T
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]65^{\circ}[/tex]
[tex]l[/tex] = Length of wires = 2.64 m
[tex]I[/tex] = Current in the other wire
Magnetic force is given by
[tex]F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}[/tex]
Net current is given by
[tex]I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}[/tex]
The current I is [tex]4.77\ \text{A}[/tex].
A truck travels on a straight road at a velocity of 17 meters per second. Over 20
seconds, it accelerates uniformly to 27 meters per second. What distance did the truck
travel during this acceleration?
Answers
Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m
The equations of motion can be used to obtain the distance covered as 440 m.
We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;
v^2 = u^2 + 2as and v = u + at
v = final velocity
u = initial velocity
a = acceleration
s = distance
To obtain the acceleration;
27 = 17 + 20(a)
27 - 17 = 20a
a = 0.5 ms-2
Now, to obtain the distance;
v^2 = u^2 + 2as
v^2 - u^2/as = s
s = (27)^2 - (17)^2/2(0.5)
s = 440 m
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Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accelerates at 2 m/s2. Instantly the speeder becomes aware that he is being chased and starts to accelerate at 1 m/s2. How much time (in s) passes until the police catch the speeder
Answers
Answer:
The time that passes until the police catch the speeder is 82.6204 seconds.
Explanation:
A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.
The position is calculated by the expression:
x = x0 + v0*t + 1/2*a*t²
where:
x0 is the initial position. v0 is the initial velocity. a is the acceleration. t is the time interval in which the motion is studied.First, let’s look at the police car’s equations of motion. In this case:
x0= 0 v0= 50 m/s a= 2 m/s²So: x = 50 m/s*t + 1/2*2 m/s²*t²
Now for the speeder’s car’s equations of motion you know:
x0= 3 km= 3,000 m v0= 55 m/s a= 1 m/s²So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
When the police catch the speeder they are both in the same position. So:
50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
Solving:
0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²
0= 3,000 + 55 *t + 1/2*t² - 50*t - 1*t²
0= 3,000 + 55 *t - 50*t - 1*t² + 1/2*t²
0= 3,000 + 5*t - 1/2*t²
Applying the quadratic formula:
[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]
x1= -72.6209
and x2= 82.6209
Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.
A 20- kg and a 50-kg ball rest at the top of a ramp. Which statement best describes the potential energy of the balls?
Answers
Answer:The 50-kg has the greater potential energy because it has greater mass
Explanation:
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
Answers
Answer:
The added mass will mean a longer period of oscillation.
Explanation:
The period of oscillation here is given by the formula;
T = 2π√(m/k)
Where m is mass and k is spring constant
From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.
Thus, the added mass will mean a longer period of oscillation.
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answers
Answer:
distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
Explanation:
Given that;
mass of block m = 0.200 kg
distance travelled d = 0.796 m
time t = 2.00 s
m₂ = 0.400 kg
If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?
Now, using the second equation of motion;
d = ut + ([tex]\frac{1}{2}[/tex] × at²)
as the object started from rest, u=0
so, we substitute
0.796 = 0×2 + ([tex]\frac{1}{2}[/tex] × a(2)²)
0.796 = 0 + ([tex]\frac{1}{2}[/tex] × 4a)
0.796 = 2a
a = 0.796 / 2
a = 0.398 m/s²
using first equation of motion
[tex]V_{f}[/tex] = u + at
we substitute
[tex]V_{f}[/tex] = 0 + 0.398 × 2
[tex]V_{f}[/tex] = 0.796 m/s
now, average velocity is given as;
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
now, distance as the block moves in 2s will be;
D = [( 0.796 m/s + 0 ) / 2 ] × 2
D = 0.796 m
Therefore, distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
The distance traveled by the object with the uniform motion can be given by the second equation of the motion.
The distance traveled by the big block with weight in 2 seconds is.
What is second equation of motion?
The distance traveled by the object with the uniform motion can be given by the second equation of the motion. It can be given as,
[tex]s=ut+\dfrac{1}{2} at^2[/tex]
Given information-
The mass of the small block is 0.200 kg.
The total distance traveled by the block is 0.796.
Initial velocity of small block is zero.
Total time taken by the block to travel this distance is 2 seconds.
Put the values in the above equation as,
[tex]0.796=0\times 2+\dfrac{1}{2} a\times 2^2\\0.796=2a\\a=0.398[/tex]
Thus the acceleration of the small block is 0.398 meter per second.
Now the mass is doubled which is, 0.400 kg. As the acceleration does not depends on the mass, thus the acceleration for both cases is 0.398.
The velocity of the big block can be given as,
[tex]V=u+0.398\times2\\V=0+0.796\\V=0.796[/tex]
The velocity of the big block is 0.796.
The average velocity of the big block is given by,
[tex]V_{avg}=(\dfrac{0.796+0}{2} })\\V_{avg}=0.398[/tex]
The distance traveled by the object is the ratio of the velocity of the body to the time taken by it. Thus the distance traveled by the big block in 2 seconds is,
[tex]d={{0.398} \times2}\\d=0.796[/tex]
Thus the distance traveled by the big block with weight in 2 seconds is.
Learn more about the second equation of motion here;
https://brainly.com/question/8898885
A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path
Answers
Answer:
4units
Explanation:
To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;
D = √(x2-x1)²+(y2-y1)²
Given the coordinate points
(3,5) and (7,5)
Substitute
D = √(7-3)²+(5-5)²
D = √(7-3)²+0²
D = √4²
D = √16
D = 4
Hence the total distance the beam will travel along this path is 4units