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
The most commonly used conductor in the laboratory is?
Answer:
A battery
Explanation:
A battery is the biggest conductor. I think.
Ann is driving down a street at 56 km/h. Suddenly a child runs into the street. If it takes Ann 0.749 s to react and apply the brakes, how far will she have moved before she begins to slow down? Answer in units of m.
The distance travelled by Ann before she begins to slow down is 11.65 m
How do I determine the distance travelled by Ann?First, we shall enlighten ourselves on what speed is. This is given below.
Speed is the distance an object travelled per unit time. It can be expressed as:
Speed = distance / time
Finally, we can obtain the distance Ann travelled as illustrated below.
From the question given above, the following data were obtained:
Speed = 56 Km/h = 56 / 3.6 = 15.56 m/sTime = 0.749 sDistance =?Speed = distance / time
15.56 = distance / 0.749
Cross multiply
Distance = 15.56 × 0.749
Distance = 11.65 m
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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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If an object travels at 10 m/s constantly for 1 minute, how far will it have travelled?
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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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The power in a lightbulb is given by the equation P=I^2 R, where I is the current flowing through the lightbulb and R is the resistance of the lightbulb. What is the current in a circuit that has a resistance of 25.0 Ω and a power of 30.0 W? A 0.830 A B 1,20 A C 0.910 A D 1.09 A
Answer:
Explanation:
The correct option will be the option (d) that is 1.09 A.
According to the equation given in the question:
That is; P=i^2R where I= current, R= resistance
P=30 W
R=25 ohm
30=i^2*25
I^2=30/25
I^2=1.2
I=1.09A
So, the correct answer of the given question will be 1.09A
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What are the scientific factors for time machine?
Explanation:
probably luck or the right material's
ANSWER NOW PLEASE
When you are standing still on the Earth, your speed relative to the sun is approximately how many km/s.
If you were standing still on the equator of Earth you would move at nearly 1,000 miles/hour which is approximately 1,600 km/hr.
What is speed?
Speed is measured as the ratio of distance to the time in which the distance was covered.
The Earth is round and spins once every 24 hours. This spinning of the earth causes the daily motion which we feel here. If you were standing still on the equator of Earth you would move at nearly 1,600 km/hr.
Speed is a scalar quantity as it has only direction and no magnitude.
Speed is defined mathematically as distance/ time.
speed = distance/time expressed in m/s.
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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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What is physical quantity?....
During camp, a simple way to estimate the height of a cliff is to drop a stone from the top and hear the splash when it hits the water below. The stone takes 3.6 seconds to drop. Assume sound speed is infinite. The height of the cliff is ___ meters.
The heigth of the cliff is 63.504 m.
What is height?
Height is the vertical distance between two points.
To calculate the height of the cliff, we use the formula below.
Formula:
S = ut+gt²/2........... Equation 1Where:
S = Height of the clifft = Timeu = Initial velocityg = Accceleration.From the question,
Given:
u = 0 m/st = 3.6 sg = 9.8 m/s²Substitute these values into equation 1
S = 0×3.6+9.8×3.6²/2S = 63.504 mHence, the heigth of the cliff is 63.504 m.
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Alexus was resting on the couch but then decided she needed some food. She ran at up to 15m/s, accelerating
at 3m/s² the whole time, until she got to the fridge. How far away is the fridge from the couch?
Alexus was resting on the couch but then decided she needed some food. She ran at up to 15m/s, accelerating at 3m/s² the whole time, until she got to the fridge . The fridge will be 37.5 m far away from the couch
The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Equation of kinematics are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.
given
final velocity = 15m/s
initial velocity = 0
a = 3m/s²
s = ?
a = acceleration = final - initial / time
3 = 15 - 0 / time
time = 15 /3 = 5 seconds
[tex]v^{2}[/tex] - [tex]u^{2}[/tex] = 2as
[tex]15^{2}[/tex] - 0 = 2 * 3 * s
s = 225 /6
= 37.5 m
The fridge will be 37.5 m away from the couch
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A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction of the initial acceleration vector and compare the initial acceleration vector's magnitude with respect to the crash acceleration magnitude.
Question 8 options:
The direction of the initial acceleration vector will point towards the wall, and its magnitude will be less than the acceleration vector of the crash.
The direction of the initial acceleration vector will point away from the wall, and its magnitude will be more than the acceleration vector of the crash.
The direction of the initial acceleration vector will point away from the wall, and its magnitude will be less than the vector of the crash.
The direction of the initial acceleration vector will point towards the wall, and its magnitude will be more than the acceleration vector of the crash.
The direction of the initial acceleration vector will point towards the wall, and its magnitude will be less than the acceleration vector of the crash, therefore the correct answer is option A.
What is acceleration?The rate of change of the velocity with respect to time is known as the acceleration of the object.
As given in the problem, a test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction of the initial acceleration vector and compare the initial acceleration vector's magnitude with respect to the crash acceleration magnitude.
Thus, the initial acceleration vector will point in the direction of the wall and be smaller than the crash's acceleration vector, therefore the correct answer is option A.
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100 Points!
A projectile is fired in the earth's gravitational field with a horizontal velocity of v=9.00 m/s. How far does it go in the horizontal direction in 0.550s? Show your work.
B) How far does the projectile go in the vertical direction in 0.550s. Show your work
Answer:
A) 4.95 m
B) 1.48225 m
Explanation:
Constant Acceleration Equations (SUVAT)
[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]
When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.
Consider the horizontal and vertical motion of the projectile separately.
Part AThe horizontal component of velocity is constant, as there is no acceleration horizontally.
Resolving horizontally, taking → as positive:
[tex]u=9.00\quad v=9.00 \quad a=0\quad t=0.550[/tex]
[tex]\begin{aligned}\textsf{Using} \quad s & = \left(\dfrac{u+v}{2}\right)t:\\\\s&= \left(\dfrac{9+9}{2}\right)(0.550)\\s&= (9)(0.550)\\ \implies s&= 4.95\:\sf m\\\end{aligned}[/tex]
Part BAs the projectile is fired horizontally, the vertical component of its initial velocity is zero.
Acceleration due to gravity = 9.8 ms⁻²
Resolving vertically, taking ↓ as positive:
[tex]u=0\quad a=9.8\quad t=0.550[/tex]
[tex]\begin{aligned}\textsf{Using} \quad s & = ut+\dfrac{1}{2}at^2:\\\\s&= (0)(0.550)+\dfrac{1}{2}(9.8)(0.550)^2\\s&= 0+(4.9)(0.3025)\\\implies s&= 1.48225\:\sf m\\\end{aligned}[/tex]
What is the relationship between the distance traveled and time for an object that travels at a constant speed ?
HELP! Along the line through the centers of the Moon and the Earth, there is a point where the gravitational forces exerted by the Moon and the Earth, respectively, are equal. We consider an elementary mass located at this point. How can this state be rated? Stable equilibrium, unstable equilibrium, total mechanical energy minimum, total mechanical energy maximum or minimum
We consider an elementary mass located at this point to be in a stable equilibrium.
What is stable equilibrium?An equilibrium is said to be stable if small, externally induced displacements from that state produce forces that tend to oppose the displacement and return the body or particle to the equilibrium state.
Also a body is said to be in a stable equilibrium, if the opposing forces acting on a body are equal.
Examples of a stable equilibriuma weight suspended by a springa brick lying on a level surfacea cone resting on its baseThus, along the line through the centers of the Moon and the Earth, there is a point where the gravitational forces exerted by the Moon and the Earth, respectively, are equal. We consider an elementary mass located at this point to be in a stable equilibrium.
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I 'm driving down the street at 15 m/s, it takes me 20 minutes to get to my destination. What was the distance to my destination?
The distance to the destination is 18,000 meters or 18 kilometers. the formula for calculating the distance is speed × Time.
Using the formula:
Distance = Speed × Time
Speed = 15 m/s (meters per second)
Time = 20 minutes
First, we need to convert the time from minutes to seconds, as the speed is given in meters per second.
1 minute = 60 seconds
Time in seconds = 20 minutes × 60 seconds/minute = 1200 seconds
Now, we can calculate the distance:
Distance = 15 m/s × 1200 seconds
Distance = 18,000 meters
So, the distance to the destination is 18,000 meters, or 18 kilometers (since 1 kilometer = 1000 meters).
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pls pls help !!! <33
Answer: Vx is changing and Vx is greatest near the beginning of the launch.
Explanation: Think of Vx as a bullet fired from a gun. As soon as it is fired it starts to slow down so Vx is changing and at the beginning Vx has its greatest velocity because it can only go slower from there.
answers are B and C. horizontal velocity never changes so A cannot be the answer. that makes B one of the answers. I would also say that the horizontal velocity is greatest near the beginning of the launch as it just starts, since it stays at a constant pace after, so D is not an answer. hope this helps ^_^
Calculate the frequency of this radiation.
The frequency of a radiation that has a wavelength of 9.17 × 10-⁷m is 3.27 × 10¹⁴ Hz.
How to calculate frequency?The frequency of a radiation can be calculated by dividing the velocity of the radiation by the wavelength. That is;
f = v/(λ)
According to this question, some radiation detected by a thermal imaging camera during a training exercise has a wavelength of 9.17 × 10-⁷m. The frequency can be calculated as follows:
f = 3 × 10⁸m/s ÷ 9.17 × 10-⁷m
f = 3.27 × 10¹⁴ Hz
Therefore, the frequency of a radiation that has a wavelength of 9.17 × 10-⁷m is 3.27 × 10¹⁴ Hz.
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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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Someone pls tell me what the answer is
The distance of the ball that would land which was launched horizontally at a speed (s) of 3.4 m/s from a table that is 7.8 m tall (d) will land is 4.29m
Consider the ball moving horizontally off the table. When it moves, gravitational force acts on it pulling the ball downwards. In addition to that objects that are falling down would accelerate quickly i.e., let’s take an object falling at the speed of x, after time t it would travel at 2x, after 2t it would be 4x.
We need to find time to calculate distance by using speed equation.
For calculating time, let’s use the freefall formula,
d = 1/2 gt²
t = √(2d) / g
t = √2 (7.8)/ 9.8
t = 1.262 s
To calculate distance, we could use speed equation as we have both Time and Velocity
s = d / t
d = s t
d = 1.262 * 3.4
d = 4.29 m
Thus, the distance of the ball that would land is 4.29 m
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Select the correct answer.
Which physical property causes you to lean to one side when the bus you are traveling in takes a sharp turn?
A.
inertia
B.
mass
C.
speed
D.
velocity
Answer: I believe that the answer to your question is "Inertia", or A
Explanation:
Estimate the car's velocity at 4.0 s . Express your answer to two significant figures and include the appropriate units.
In camping, you drop a stone from the 1st spot to the lake. The splash is heard 4 seconds later. After walking, you drop a stone again at 2nd spot at the same lake. This time the splash is heard 12 seconds later. The height of the 2nd spot is ___ times the 1st spot.
A-3
B-12
C-6
D-9
The height of the 2nd spot is 3 times the height of the 1st spot (Option A)
How to determine the height of the first spotWe'll begin by obtaining the height of the first spot. This can be obtained as follow:
Time (t) = 4 seconds Velocity of sound (v) = 343 m/sHeight of first spot (x) =?v = 2x / t
343 = 2x / 4
Cross multiply
2x = 343 × 4
2x = 1372
Divide by 2
x = 1372 / 2
Height of first spot = 686 m
How to determine the height of the second spotTime (t) = 12 seconds Velocity of sound (v) = 343 m/sHeight of second spot (x) =?v = 2x / t
343 = 2x / 12
Cross multiply
2x = 343 × 12
2x = 4116
Divide by 2
x = 4116 / 2
Height of second spot = 2058 m
How to compare 2nd height with 1st height Height of 1st spot = 686 mHeight of 2nd spot = 2058 mComparison =?
Comparison = Height of 2nd spot / Height of 1st spot
Height of 2nd spot / Height of 1st spot = 2058 / 686
Height of 2nd spot / Height of 1st spot = 3
Cross multiply
Height of 2nd spot = 3 × Height of 1st spot
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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.
An object dropped from the window of a tall building hits the ground in 12.0 s. If its acceleration is 9.80 m/s2, the height of the window above the ground is
Answer:
58.8 metres
Explanation:
using the formula
h=ut+½gt
U=0
g=9.8m/s²
t=12.0a
Hence, h= 0.5 × 9.8 × 144
h = 58.8 metres
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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convert these into proper Vector notation Westward velocity of 42 km/h
Postion 6.5 measured in m that is North of the refrence point.
Downward acceleration measured in m/s2 that has a magnitude of 1.9.
Proper vector notations are V = 42i km/hr, r = 6.5j m, and a = -1.9j m/s².
In physics, a vector is a quantity with both magnitude and direction. It is typically represented by an arrow whose length is proportional to the magnitude of the quantity and whose direction is the same as that of the quantity.
A) Let us assume that the west direction is the positive X-axis.
In vector notation, the X-axis is denoted by "i". So, the westward velocity of 42 km/hr will be written as
V = 42i km/hr
B) Let us assume that the North direction is the positive Y- axis.
In vector notation, the Y-axis is denoted by "j". Also, the position vector is represented by the letter 'r'. So position 6.5 measured in m, that is North of the reference point will be written as
r = 6.5j m
C) Let us assmue thaat the downward direction is negative Y-axis.
In vector notation, the Y- axis is denoted by "j". so, the downward acceleration measured in m/s2 that has a magnitude of 1.9 will be written as
a = 1.9 × (-j) m/s²
a = -1.9j m/s²
So, proper vector notations are V = 42i km/hr, r = 6.5j m, and a = -1.9j m/s².
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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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