An engineer in a locomotive Sees a car stuck
on the track at a railroad crossing in front of
the train. When the engineer first sees the
car, the locomotive is 190 m from the crossing
and its spood is 18 m/s.
If the engineer's reaction time is 0.54 m.
what should be the magnitude of the mini-
mum deceleration to avoid an accident?
Answer in units of m/s
The magnitude of the minimum deceleration needed to avoid the accident is –0.89 m/s²
The solution of the question can be obtained by first approaching the relation between Speed, Distance and Time which is given by:
Speed = Distance/Time
Now according to question;
Speed = 18 m/s
Time = 0.54 s
Therefore, Distance = Speed×Time
Distance = 18 × 0.54
Distance = 9.72 m
Thus, the engineer travelled a distance of 9.72 m during the reaction time.
Now, there is a need to find distance between the engineer and the car. This can be obtained by:
Distance between the engineer and the car = Total distance - Distance during the reaction time
Distance between the engineer and the car = 190 – 9.72
Distance between the engineer and the car = 180.28 m
Finally, we need to determine the magnitude of the deceleration needed to avoid the accident. This can be obtained by using newton's third equation of motion, which is represented as:
v² = u² + 2as; where u is the initial velocity, v is the final velocity, a is the acceleration/deceleration and S is the distance.
According to the question;
Initial velocity (u) = 18 m/s
Final velocity (v) = 0 m/s
Distance (s) = 180.28 m
Deceleration (a) =?
Thus, using Newton's third equation of motion
0² = 18² + (2 × a × 180.28)
0 = 324 + 360.56a
=> –324 = 360.56a
=> a = –324 / 360.56
=> a = –0.89 m/s²
Therefore, the magnitude of the deceleration needed to avoid the accident is –0.89 m/s.
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WILL GIVE BRAINLIEST!!!!!!!
why would astronomers make the planets line up in a straight line on a model?
A horizontal force of 100 N is required to push a bookcase across a floor at a constant velocity.
The correct answer is :
Here 100 N force is applied to make the box move with constant velocity from rest. That means 100 N force is applied to overcome the limiting static friction and as soon as 100 N force is applied it starts moving.
Now,
Constant velocity means acceleration = 0
Net force acting on the box =mass × accelaration = mass × 0 = 0
Conceptually it is zero as it is balanced by kinetic friction which has equal value that of applied force. Because net force =Applied force - friction force and hence here friction force =applied force.
If there was any accelaration then there would exist a net force and then frictional force and applied force will be the same.
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If an object travels at 10 m/s constantly for 1 minute, how far will it have travelled?
PLEASE help me with this problem??!!
A kayaker needs to paddle north across a
100-m-wide harbor. The tide is going out, creating
a tidal current that flows to the east at 2.0 m/s
The kayaker can paddle with a speed of 3.0 m/s
Answer:
100
Explanation:
3-2=1
100/1=100
What is physical quantity?....
A bowling ball results in ______ friction compared to a sliding bowling ball.
A bowling ball results in kinetic friction compared to a sliding bowling ball.
What is the Conservation of momentum?According to Conservation of momentum, if a bowling ball hits some pins, the momentum that lost by the bowling ball is known to be equal to the momentum obtained by the pins.
The friction that is used in bowling is kinetic friction because the more oil that is placed down, the lower the friction that is found between the ball and that of the lane surface. The little friction, the stronger it is for the bowler to be able to send the ball in a curved path and thus the formula to find the kinetic friction is know to be : µk=F k/mg.
Therefore, A bowling ball results in kinetic friction compared to a sliding bowling ball.
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An object starts from rest and undergoes uniform acceleration. From 3.44s to 9.49s it travels 2.9m. What is the average velocity of the object during the time interval 18.6s to 22.92s (in m/s )?
First, see that in the time interval 3.44 to 9.49, the average velocity is [tex]2.9\text{m} / (9.49\text{s} - 3.44\text{s})=2.9\text{m}/6.05\text{s} \approx 0.479\text{m/s}[/tex]. So, as we have uniform acceleration, the velocity must be linearly increasing over this entire interval, so for the average to be 0.479 m/s over this interval, the velocity must be 0.479 m/s in the exact middle of this interval, or at 5.465s.
We now note that the object starts from rest, which means that at 0s, the velocity is 0 m/s. So, in 5.465 seconds, the velocity increases by 0.479 m/s. We again have that the object undergoes uniform acceleration, meaning that the acceleration over this interval is a constant [tex]\frac{0.479\text{m/s}}{5.465\text{s}} \approx 0.0876 \text{m/s}^2[/tex].
Finally, note again that as we are looking at uniform acceleration, by the same principle at the beginning, the average velocity of the object during the time interval from 18.6s to 22.92s is the same as the velocity at the exact middle of this interval, or at 20.76s. We have that acceleration is constant and 0.0876 m/s^2, and initial velocity is 0 at 0s. So, in 20.76 seconds, the object will have accelerated [tex]0.0876\text{m/s}^2 \cdot 20.76\text{s} \approx 1.82 \text{m/s}[/tex].
So, average velocity will be 1.82 m/s over the time interval 18.6s to 22.92s.
At the beginning of a 3.0-h plane trip, you are traveling due north at 192 km/h. At the end, you are traveling 250 km/h in the northwest direction (45° west of north).
Find the magnitude of the change in velocity.
Find the change in direction of your velocity. Enter the angle in degrees where negative indicates north of west and positive indicates south of west.
What is the magnitude of your average acceleration during the trip?
EDIT: ANSWERED
Magnitude of the change in velocity: 177.4 km/h
Change in direction of velocity: 4.9°
Magnitude of average acceleration during trip: 59.1 km/h2
The magnitude of the change in velocity is determined as 408.94 km/h.
The change in direction of the velocity is 64.4⁰ north of west.
The magnitude of the average acceleration during the trip is 0.0105 m/s².
Magnitude of change in velocityThe magnitude of change in velocity is the resultant velocity of the plane.
v² = a² + b² - 2ab cosθ
where;
θ is the angle between the two velocities = 45 + 90 = 135v² = (192²) + (250²) - 2(192 x 250) cos(135)
v² = 167,236
v = √167,236
v = 408.94 km/h
Vertical component of the velocityvyi = 192 km/h
vy2 = 250 x sin(45) = 176.77 km/h
vy(total) = 192 km/h + 176.77 km/h = 368.77 km/h
Horizontal component of the velocityvxi = 0
vx2 = - 250 km/h x cos(45) = -176.77 km/h
Change in direction of the velocityθ = arc tan (Vy/Vx)
θ = arc tan(368.77 / -176.77)
θ = -64.4 ⁰
θ = 64.4⁰ north of west.
Acceleration of the tripa = v/t
v = 408.94 km/h = 113.6 m/s
h = 3 h = 10,800 seconds
a = (113.6 m/s) / ( 10,800 s)
a = 0.0105 m/s²
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physics help needed
Answer: 1 > 3 > 2
Explanation: The range will increase with the velocity. If they are all launched at the same time the ones that are launched the hardest or with the most velocity will go the furthest horizontally. But because they are all launched from the same height and the force of gravity on all three projectiles is constant (the same for all three) they will all hit the ground at the same time but different distances from the starting point.
During weather that may produce freezing temperatures, Florida orange growers often spray water on their orchards to protect them. Which of the following best explains why this helps protect orange trees and fruit from freezing?
The statement that best explains why this helps protect orange trees and fruit from freezing is that the transfer of thermal energy in water as it freezes protects the orange trees by transferring heat energy from the orange trees and fruit. That is option A.
What is freezing?Freezing is the phase transition that involves the change of a liquid substance into a solid form as it's temperature becomes lower.
During winter seasons which is a weather that can freez the orange fruits, the orange grower sprays water on their orange trees to prevent it from freezing.
The mechanism that prevents the orange from freezing is the transfer of the thermal energy between the water and the orange fruits.
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2. Link walks 5 miles east, then 7 miles north, than 2 miles south.
a) What distance did he travel? What is his displacement?
b) Suppose Link made the whole journey in 4 hours. What was his average speed and
average velocity during his journey?
Link walks 5 miles east, then 7 miles north then 2 miles south
If you draw a simple diagram you’ll see that your displacement from the start is 5 miles East and 7 miles North and then 2 miles south
These distances are the sides of a right-angled triangle. The displacement you are looking for is the hypotenuse of the triangle. Use the Pythagorean theorem to calculate it.
Displacement
= √ (5+2)^2 + 7^2)
= 9.899 miles
Distance
= 5+7+2
= 14 miles
Time to complete his whole journey = 4 hours
Average Speed= Distance travelled / Time
= 14÷4
= 3.5 miles/hour
Average velocity= Displacement / Time
= 9.899÷4
= 2.47475 miles/hour
The distance is 14 miles and the displacement is 9.899 miles, the Average Speed is 3.5 miles/hour and the Average Velocity is 2.47475 miles/hour
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An astronaut has left the space shuttle on a tether to test a new personal maneuvering device. She moves
along a straight line directly away from the shuttle. Her onboard partner measures her velocity before and
after certain maneuvers, and obtains the following results:
A. VL = 0.8 m/s,
B. U 1.6 m/s,
C. U = -0.4 m/s,
U2x = 1.2 m/s; (speeding up)
U2x = 1.2 m/s; (slowing down)
v₂x = -1.0 m/s; (speeding up)
D. U-1.6 m/s,. U2x = -0.8 m/s. (slowing down)
If t₁ = 2 s and t₂ = 4s in each case, find the average acceleration for each set of data.
(a) The acceleration is 0.2 m /s²
(b) The acceleration is - 0.2 m /s²
(c) The acceleration is - 0.3 m /s²
(d) The acceleration is 0.4 m /s²
Acceleration is outlined as The rate of amendment of rate with relevancy time.Acceleration could be a vector amount because it has each magnitude and direction. it's additionally the second by-product of position with relevancy time or it's the primary by-product of rate with relevancy time which is given by a = v₂ₓ - v₁ₓ / t₂ - t₁ .......(1) equation (1)(a) Putting v₁ₓ = 0.8 m/s , v₂ₓ = 1.2 m/s , t₁ = 2 s and t₂ = 4s in equation (1)
, we get a = 1.2 - 0.8 / 4 -2
a = 0.4 / 2
a = 0.2 m /s²
(b) Putting v₁ₓ = 1.6 m/s, v₂ₓ = 1.2 m/s , t₁ = 2 s and t₂ = 4s in equation (1) , we get a = 1.2 - 1.6 / 4 -2
a = - 0.4 / 2
a = - 0.2 m /s²
(c) Putting v₁ₓ = - 0.4 m/s, v₂ₓ = - 1.0 m/s, t₁ = 2 s and t₂ = 4s in equation (1) , we get a = -1.0 - (-0.4) / 4 -2
a = - 0.6 / 2
a = - 0.3 m /s²
(d) Putting v₁ₓ = -1.6 m/s, v₂ₓ = -0.8 m/s, t₁ = 2 s and t₂ = 4s in equation (1) , we get a = -0.8 - (-1.6) / 4 -2
a = 0.8 / 2
a = 0.4 m /s²
The traveller races in cases (a) and (c) and slows down in (b) and (d), however the common acceleration is positive in (a) and (d) and negative in (b) and (c). In alternative words, negative acceleration doesn't essentially indicate a deceleration down.
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The complete question is given below :
An astronaut has left the space shuttle on a tether to test a new personal maneuvering device. She moves along a straight line directly away from the shuttle. Her onboard partner measures her velocity before and after certain maneuvers, and obtains the following results:
(a) v₁ₓ = 0.8 m/s, v₂ₓ = 1.2 m/s; (speeding up)
(b)v₁ₓ = 1.6 m/s, v₂ₓ = 1.2 m/s; (slowing down)
(c) v₁ₓ = - 0.4 m/s, v₂ₓ = - 1.0 m/s; (speeding up)
(d) v₁ₓ = -1.6 m/s, v₂ₓ = -0.8 m/s; (slowing down)
If t₁ = 2 s and t₂ = 4s in each case, find the average acceleration for each set of data.
2. A radiographic technique calls for a 400 mA, 1/30s exposure. What is
the mAs?
Answer:
13.33 mA s
Explanation:
400 mA * 1/30 s = 13.33 mA s
please help look the pictures
The angular momentum can be indicated by formula mvr. The correct option is A.
What is angular momentum?The property of any rotating object given by moment of inertia times angular velocity is defined as angular momentum.
It is the property of a rotating body determined by the product of the rotating object's moment of inertia and angular velocity.
The rotational equivalent of linear momentum is the angular momentum formula. Both concepts are concerned with the rate at which anything moves.
It can be represented by symbol L.
L = mvr.
Thus, the correct option is A.
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A fugitive tries to hop on a freight train traveling at a constant speed of 5.2 m/s. Just as an empty box car passes him,the fugitive starts from rest and accelerates at a =1.2 m/s^2 to his maximum speed of 5.8 m/s, which he then maintains. How long does it take him to catch up to the empty box car?
It will take the fugitive 4.83 s to catch the empty box car
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration v is the final velocity u is the initial velocity t is the time How to determine the timeThe time taken for the fugitive to catch the car if he maintains his maximum speed can be obtained as follow:
Initial velocity (u) = 0 m/sAcceleration (a) = 1.2 m/s² Final velocity (v) = 5.8 m/sTime (t) =?a = (v – u) / t
Thus,
t = (v – u) / a
t = (5.8 – 0) / 1.2
t = 5.8 / 1.2
t = 4.83 s
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What are the scientific factors for time machine?
Explanation:
probably luck or the right material's
21
est
Show instructions
Question 26 of 27 | Page 26 of 27
Question 26 (1 point)
You are traveling down the road with a speed of 15 m/s when a deer runs out 20 m in front of your car. If at that instant you apply the brakes
and can decelerate your car at 4.5 m/s/s, will you hit the deer?
The car travels 25 meters before coming to rest and will hit the deer.
Given in the Question,
Initial speed = u = 15 m/s
Deceleration = a = 4.5 m/s²
According to the question, if the car stops before traveling 20m, it will avoid hitting the deer. So, we need to find the stopping distance for the car.
Deceleration is negative acceleration, so the sign of acceleration is will be -.
Also, the car comes to rest after applying the brakes. Therefore the final speed of the car will be zero.
Final speed = v = 0 m/s
Using the Third equation of motion,
v² - u² = 2as
Put in the values, we get
(0)² - (15)² = 2(-4.5)s
-2×4.5 × s = - 15× 15
s = 225/9
s = 25 m
So the car will come to rest after traveling 25 meters. But the Deer is present at 20 meters; therefore the car will hit the deer.
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ou throw a rock from the upper edge of a 87.0 m vertical dam with a speed of 21.0 m/s at 58.0∘ above the horizon. Neglect any effects due to air resistance. How much time 1 after throwing the rock will it return to its initial height?
You throw a rock from the upper edge of a 75.0 -m vertical dam with a speed of 25.0 m/s at 65.0∘ above the horizon.
What is horizon?The horizon is the line that, when seen from a position on or near the surface of a celestial body, appears to separate the surface from the sky of that body. All viewing directions are split according to whether it crosses the surface of the relevant body or not.
Since the true horizon is an imaginary line, it can only be seen with any degree of accuracy when it is situated along a generally flat surface, such as the oceans of the Earth. On Earth, the geography may also cause biological objects like trees and/or man-made objects like buildings to obstruct this line in some locations. The location where these obstructions overlap the sky is known as the visible horizon.
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In this circuit diagram, the resistance is 100 ohms, and the current is 0.1 amperes. The voltage is
The required voltage is 10 Volt for 100 ohms of resistance and 0.1 ampere of current.
electric current is defined as the rate at which charge passes a fixed unit cross sectional area.
voltage is defined as the physical quantity required to move charge from one end to other.
Resistance is defined as the the factor opposing electric current.
Ohms law states that voltage is directly proportional to electric current
V=IR
given:
current (I) = 0.1 A
resistance (R)=100 ohms
now from definition of voltage we have:
V=I.R
V=100×0.1
V=10volts
therefore the required voltage is 10 volts.
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What kinds of materials can be involved in static electric effects?
All types of materials, both metals and non metals
Only metals but not non-metals
Only non metals but not metals
Only certain metals but not all metals and not non-metals
Only certain metals and certain non-metals
The kinds of materials can be involved in static electric effects is All types of materials, both metals and non metals.
What is static electric effects?The static electric effects can be described as the effect that bring about the changes to the body as a result of the changes in the distribution that is been imposed on the electric charges that is found on the surface of the body.
It should be noted that sufficiently large surface charge density could be discovered as a result of the effect, however in the case above option A is correct.
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How large is a neutrino?
1 picometer
1 gigameter
1 yoctometer
1 nanometer
The Autobots are flying away from their home planet in a space cruiser that is accelerating constantly at 61.5 m/s2. If the ship started at rest, how far is it from the planet's surface after 10 min?
The space cruiser is 11070000 m far from the planet's surface
Data obatined from the questionThe following data were obtained from the question:
Initial velocity (u) = 0 mph = 0 m/sAcceleration (a) = 61.5 m/s² Time (t) = 10 minutes = 10 × 60 = 600 sDistance (s) =?How to determine the distanceWe can obtain the distance of the space cruiser from the planet's surface as follow:
s = ut + ½gt²
s = (0 × 600) + (½ × 61.5 × 600²)
s = 0 + (½ × 61.5 × 360000)
s = 0 + 11070000
s = 11070000 m
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What is the speed over the ground mosquito flying 2 m/s relative to the ar caught in a 2 m/s right angle crosswind
The speed over the ground is 0 m/s.
The speed of an object or body is the magnitude of the change of its position over time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.
As it is given that the mosquito is flying at 2 m/s relatives to the air caught in a 2 m/s right angle crosswind.
As we know that speed over the ground is the difference between the flight speed and resistance speed.
So, the general equation for the speed over the ground is :
v = Flight Speed of the mosquito - Resistance Speed of the crosswind
This implies, [tex]v= 2 {~}m / s - 2{~} m / s[/tex]
[tex]v = 0 {~}m / s[/tex]
Hence, the speed over the ground is 0 m/s.
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a half meter ruler is pivoted at its midpoint and balances whaen a weight of 20N is placed at the 10 cm mark and a weight W is placed at the 45 cm mark on the ruleer. Calculate the weight W
Answer:
35
Explanation:
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-fall acceleration on the moon is 1/6 of its value on earth. Suppose he hit the ball with a speed of 25 m/s at an angle 34° above the horizontal.
How long was the ball in flight?
How far did it travel?
Ignoring air resistance, how much farther would it travel on the moon than on earth?
The time of flight of the ball is 17.12 seconds.
The horizontal distance or range of the ball is 354.8 m.
Time of flight of the ball
The time of flight of the ball is calculated as follows;
T = (2u sinθ)/g
where;
u is the initial velocityg is acceleration due to gravity on moonT = (2 x 25 sin34) / (¹/₆ x 9.8)
T = 17.12 s
Horizontal displacement of the golf ballThe range of the golf ball is calculated as follows;
R = Uxt
R = (U cosθ)t
R = (25 cos34) x 17.12
R = 354.8 m
Thus, the time of flight of the ball is 17.12 seconds.
The horizontal distance or range of the ball is 354.8 m.
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A model shows that the moon has grown to twice its size yet has remained in the same place and one or two sentences explain how this would impact the gravity between earth and the moon?
The attractive force of Earth on the moon will be said to be double when there is a doubling of the mass of the moon.
Since the moon has an attracting force, it is one that will remain the same on Earth. In gravitational forces, the two concerned objects always feel the same force.
What would occur to the gravitational pull between Earth and moon if the moon were twice as large?The explanation of how this would affect the gravity between earth and the moon is that this can result to the earth tilt a little bit harder to change, which could imply that there will be a more stable climate and ice ages may not occur as often.
Therefore, The attractive force of Earth on the moon will be said to be double when there is a doubling of the mass of the moon. Since the moon has an attracting force, it is one that will remain the same on Earth. In gravitational forces, the two concerned objects always feel the same force.
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A stone of mass 6.74kg is dropped from rest at a height of 6.02m. What impulse does gravity impart to this stone from the instant is dropped until it hits the ground, assuming negligible air resistance?
The impulse the gravity imparts on the stone from the instant it was dropped until it hits the ground is 72.66 Ns
What is impulse?
This is defined as the change in momentum of an object.
Impulse = change in momentum
Impulse = final moment – Initial momentum
Impulse = force × time
How to determine the timeInitial velocity (u) = 0 m/sAcceleration due to gravity (g) = 9.8 m/s²Height (h) = 6.02 mTime (t) =?h = ½gt²
6.02 = ½ × 9.8 × t²
6.02 = 4.9 × t²
Divide both side by 4.9
t² = 6.02 / 4.9
Take the square root of both side
t = √(6.02 / 4.9)
t = 1.1 s
How to determine the impulseTime (t) = 1.1 sMass (m) = 6.74 kgAcceleration due to gravity (g) = 9.8 m/s²Force (F) = mg = 6.74 × 9.8 = 66.052 NImpulse =?Impulse = Force × time
Impulse = 66.052 × 1.1
Impulse = 72.66 Ns
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Image has question details
The electric force on one of the masses is 0.6 N.
The acceleration of the mass is 0.35 m/s².
Electric force between the masses
The electric force between the masses is calculated as follows;
F = kq²/r²
where;
K is Coulomb's constantr is the distance between the chargesq is the chargeF = (9 x 10⁹ x (9.8 x 10⁻⁶)²)/(1.2²)
F = 0.6 N
Acceleration of the massThe acceleration of the mass is calculated as follows;
F = ma
a = F/m
a = (0.6 N) / (1.7 kg)
a = 0.35 m/s²
Thus, the electric force on one of the masses is 0.6 N.
The acceleration of the mass is 0.35 m/s².
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