Answer:
Y = 40.94m
Explanation:
The initial speed of the sandbag is the same as the balloon and so is its position, so:
[tex]Y = Yo + Vo*t-\frac{g*t^2}{2}[/tex]
Replacing these values:
Yo = 40m Vo = 5m/s g = 9.81m/s^2 t = 0.25s
We get the position of the sandbag:
[tex]Y = 40+5*(0.25)-\frac{9.81*(0.25)^2}{2}[/tex]
Y = 40.94m
The weight W that a horizontal beam can support varies inversely as the length L of the beam. Suppose that aa 6 dash m6-m beam can support 14001400 kg. How many kilograms can aa 19 dash m19-m beam support?
Answer:
weight is 442.10 kg
Explanation:
given data
weight = 1400 kg
length = 6 m
to find out
how many kilogram 19 m beam support
solution
we know by inverse variation formula of weight that is
weight = [tex]\frac{k}{l}[/tex] .............1
here k is constant and l is length
so k will be
1400 = [tex]\frac{k}{6}[/tex]
k = 8400
so for 19 m weight will be by equation 1
weight = [tex]\frac{k}{l}[/tex]
weight = [tex]\frac{8400}{19}[/tex]
weight = 442.10 kg
The weight that a horizontal beam can support varies inversely as the length of the beam. Using this relationship, we can find the weight a 19-meter beam can support given that a 6-meter beam can support 1400 kg.
Explanation:The weight W that a horizontal beam can support varies inversely as the length L of the beam. This can be expressed mathematically as W = k/L, where k is a constant. In this case, we are given that a 6-meter beam can support 1400 kg. Let's use this information to find the value of k.
Using the given information, we can write the equation as 1400 = k/6. To solve for k, we can multiply both sides of the equation by 6, giving us k = 1400 x 6 = 8400.
Now that we have the value of k, we can use it to find how many kilograms a 19-meter beam can support. Using the equation W = k/L, we can plug in the values: W = 8400/19. Solving this equation gives us W = 442.105 kg (rounded to three decimal places). Therefore, a 19-meter beam can support approximately 442.105 kilograms.
Three displacements through a hedge maze. (b) The displacement vectors. (c) The first displacement vector and its components. (d) The net displacement vector and its components.d1=6.00 mθ1=40°d2=8.00 mθ2=30°d3=5.00 mθ3=0°,where the last segment is parallel to the superimposed x axis. When we reach point c, what are the magnitude and angle of our net displacement →dnet from point i?
Answer:
18.3 m , 25.4°
Explanation:
d1 = 6 m, θ1 = 40°
d2 = 8 m, θ2 = 30°
d3 = 5 m, θ3 = 0°
Write the displacements in the vector form
[tex]d_{1}=6\left ( Cos40\widehat{i}+Sin40\widehat{j} \right )=4.6\widehat{i}+3.86\widehat{j}[/tex]
[tex]d_{2}=8\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )=6.93\widehat{i}+4\widehat{j}[/tex]
[tex]d_{3}=5\widehat{i}[/tex]
The total displacement is given by
[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]
[tex]d=\left ( 4.6+6.93+5 \right )\widehat{i}+\left ( 3.86+4 \right )\widehat{j}[/tex]
[tex]d=16.53\widehat{i}+7.86\widehat{j}[/tex]
magnitude of resultant displacement is given by
[tex]d ={\sqrt{16.53^{2}+7.86^{2}}}=18.3 m[/tex]
d = 18.3 m
Let θ be the angle of resultant displacement with + x axis
[tex]tan\theta =\frac{7.86}{16.53}=0.4755[/tex]
θ = 25.4°
A ballistic pendulum can be used to measure the speed of a projectile, such as a bullet. The ballistic pendulum consists of a stationary 2.50- kg block of wood suspended by a wire of negligible mass. A 0.0100- kg bullet is fired into the block, and the block (with the bullet in it) swings to a maximum height of 0.650 m above the initial position (see class note for drawing). Find the speed with which the bullet is fired, assuming that air resistance is negligible.
Answer:
[tex]v_1 = 896.35 m/s[/tex]
Explanation:
As we know that bullet + pendulum system will move to the height of 0.650 m above the initial position
so here we can use energy conservation to find the speed just after the bullet hit the block
[tex]mgh = \frac{1}{2}mv^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.81)(0.650)}[/tex]
[tex]v = 3.57 m/s[/tex]
Now we can use momentum conservation to find the initial speed of the bullet
[tex]m_1v_1 = (m_1 + m_2)v[/tex]
[tex]0.0100 v_1 = (2.50 + 0.01)(3.57)[/tex]
[tex]v_1 = 896.35 m/s[/tex]
While John is traveling along an interstate highway, he notices a 194-mile marker as he passes through town. Later John passes another mile marker, 104. (a) What is the distance between town and John's current location? (b) What is John's current position?
Answer:
a) 90 miles is the distance between town and John's current location.
b) John's current position is the 104 mile marker.
Explanation:
See in the picture.
You are in an egg drop contest which you win by breaking your egg the hardest. You drop your egg off of a 3 story building (12m above the ground). You can throw it up with speed 11m/s, or straight down with speed 11m/s. Write a way to break it.
Answer: Ok, first start writing our movement equation.
First our acceleration will be the gravity; so a= g
integrating we can obtain the velocity v= g*t + v₀
where v₀ depends on if we trhow the egg upsides or downsides, being the positive direction downsides.
after, the position will be r = [tex]\frac{gt^{2} }{2}[/tex] + v₀*t - 12m
where the -12m is because you are upside a building.
first case: straight down with speed 11m/s
so the velocity is positive
then, if we do [tex]\frac{10*t^{2} }{2}[/tex] + 11*t - 12m = 0 and with bashkara obtain the positive root for the time t=1.6 seconds
putting this on our velocity equation you will get v = 10*1.6 + 11 = 27m/s is the velocity of the egg when it hits the ground.
for the other case, our equation will have the form of:
[tex]\frac{10*t^{2} }{2}[/tex] - 11*t - 12m = 0
and the positive root of the time is: t = 3 seconds.
putting it in the velocity equation gives you v= 10*3 - 11 = 29m/s
which is bigger than the first case, so the egg hits the ground with more velocity, ence more energy.
Final answer:
Throwing the egg straight down with an initial speed of 11 m/s from a 12 m height will result in a higher impact speed, calculated using the kinematic equation for final velocity, which factors in both the initial speed and acceleration due to gravity.
Explanation:
To maximize the impact of the egg and break it the hardest in the egg drop contest, we need to consider the effect of throwing it downward with an initial speed. If the egg is thrown straight down with an initial velocity of 11 m/s from a 12 m high building, it will gain speed due to gravity in addition to the initial speed given. The final velocity just before impact can be calculated using the kinematic equation:
v2 = u2 + 2gh
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s2), and h is the height.
Substituting the given values:
v2 = (11 m/s)2 + 2(9.8 m/s2)(12 m)
v2 = 121 m2/s2 + 235.2 m2/s2
v2 = 356.2 m2/s2
v = √356.2 m2/s2 ≈ 18.9 m/s
Therefore, throwing the egg straight down at 11 m/s will result in a higher impact speed upon hitting the ground compared to throwing it upwards or dropping it without initial velocity.
A charge of -6.50 nC is spread unifonnly over the surface of one face of a nonconducting disk of radius 1.25 cm. (a) Find the
magnitude and direction of the electric field this disk produces at a point P on the axis of the disk a distance of 2.00 cm from its center. (b) Suppose that the charge were all pushed away from the center and distributed unifonnly on the outer rim of the disk. Find the magnitude and direction of the electric field at point P. (c) If the charge is all brought tn the center of the disk, find the magnitude and direction of the electric field at point P. (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?
Answer:
(a) Magnitude = [tex]1.14\times 10^5\ N/C[/tex]
Direction = toward the center of the disc
(b) Magnitude = [tex]8.90\times 10^4\ N/C[/tex]
Direction = toward the center of the disc
(c) Magnitude = [tex]1.46\times 10^5\ N/C[/tex]
Direction = toward the center of the disc
(d) See explanation
(e) See explanation.
Explanation:
Given:
k = Boltzmann constant = [tex]9\times 10^9\ Nm^2/C^2[/tex]Q = Uniformly spread charge on the disc = [tex]-6.50\ \nC = -6.50\times 10^{-9}\ C[/tex]x = distance of the point P from the center of disc on its axis = 2 cm = 0.02 m r = radius of the disc = 1.25 cm = 0.0125 mAssume:
E = electric field at point P on the axisPart(a):
Using the formula for the electric field due to the uniform charged disc at a point P on its axis, we have
[tex]E = \dfrac{2 kQ}{r^2}\left ( 1-\dfrac{x}{\sqrt{x^2+r^2}} \right )\\\Rightarrow E = \dfrac{2 \times 9\times 10^9\times (-6.50\times 10^{-9})}{(0.0125)^2}\left ( 1-\dfrac{0.02}{\sqrt{(0.02)^2+(0.0125)^2}} \right )\\\Rightarrow E =-1.14\times 10^5\ N/C\\[/tex]
Hence, the magnitude of electric field at the given distance from the disc along its axis is [tex]1.14\times 10^5\ N/C[/tex] in the direction toward the center of the disc along the axis.
Part(b):
When all the charges on the disc are pushed away from the center of the disc to get uniformly distributed to its outer rim, the disc behaves like a uniformly charged ring.
Using the formula for the electric field due to the uniform charged ring at a point P on its axis, we have
[tex]E= \dfrac{kQx }{\left ( \sqrt{x^2+r^2} \right )^3}\\\Rightarrow E = \dfrac{ 9\times 10^9\times (-6.50\times 10^{-9})\times 0.02}{\left ( \sqrt{(0.02)^2+(0.0125)^2} \right )^3}\\&=-8.90\times 10^4\ N/C[/tex]
Hence, the magnitude of electric field at the given point from the rim along its axis is [tex]8.90\times 10^4\ N/C[/tex], and it is also toward the center of the along the axis.
Part (c):
When all the charge on the disc is pushed to the center of the disc, then the disc behaves as a point charge.
Using the formula of electric field due to a point charge at a fixed point, we have
[tex]E = \dfrac{kQ}{x^2}\\\Rightarrow E = \dfrac{9\times 10^9\times (-6.50\times 10^{-9})}{(0.02)^2}\\\Rightarrow E = -1.46\times 10^5[/tex]
Hence, the magnitude of electric field due to the point charge at the point P on the axis is [tex]1.46\times 10^5\ N/C[/tex]. which also points toward the center of the disc along the axis.
The direction of the electric field is found toward the negative x-axis because the electric field due to a point negative charge is radially inward.
Part (d):
The magnitude of electric field in part (a) is stronger than that in part (b) because the disc in part (a) is composed of infinite number of rings having radius varying from 0 cm to 2.5 cm whose components along the axis of electric field are added together which is greater than the electric field produced by a single ring in part (b).
Hence, the electric field in part (a) is greater than the electric field in part (b).
Part (e):
The electric field in part (c) is the strongest of the three fields. This is because the field due to a point charge on the axis has the only component along the axis. However, the field components in part (a) and part (b) are distributed along two axes whose axial component is considerable and the other component gets canceled out. This causes the field strength to be weaker in part (a) and part (c).
Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph . Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 73 mph .
Final answer:
Richard will save approximately 13.68 minutes by driving at 73 mph for 135 miles instead of at the speed limit of 65 mph. The mathematical concept used involves calculating travel time at different speeds and finding the difference between these times.
Explanation:
The student's question involves calculating how much time Richard will save by driving at 73 mph instead of the speed limit of 65 mph for a distance of 135 miles. This is a mathematics problem that falls under the category of rate, time, and distance, which is often covered in high school math classes.
To solve this, we need to calculate the time taken to travel 135 miles at both speeds and then find the difference in time. At 65 mph, the time taken (T1) is 135 miles / 65 mph. At 73 mph, the time taken (T2) is 135 miles / 73 mph. The time saved is T1 - T2.
Using the formula time = distance / speed, we get:
Time at 65 mph: T1 = 135 mi / 65 mph = 2.077 hours
Time at 73 mph: T2 = 135 mi / 73 mph = 1.849 hours
The time saved is T1 - T2 = 2.077 hours - 1.849 hours = 0.228 hours, which is about 13.68 minutes.
vector L is 303m long in a 205 direction. Vector M is 555m long in a 105 direction. Find the magnitude and direction of the vector.
Answer:
584.3 m at [tex]135.7^{\circ}[/tex] above the positive x-axis
Explanation:
We have to find the magnitude and direction of the resultant vector. In order to do that, we have to resolve each vector along the x- and y- direction first.
Resolving the vector L:
[tex]L_x = L cos \theta_L = 303 \cdot cos 205^{\circ} =-274.6m\\L_y = L sin \theta_L = 303 \cdot sin 205^{\circ} = -128.1 m[/tex]
Resolving the vector M:
[tex]M_x = M cos \theta_M = 555 \cdot cos 105^{\circ} =-143.6 m\\M_y = M sin \theta_M = 555 \cdot sin 105^{\circ} = 536.1 m[/tex]
Now we can find the components of the resultant vector by adding the components of each vector along each direction:
[tex]R_x = L_x + M_x = -274.6 +(-143.6)=-418.2 m\\R_y = L_y + M_y = -128.1 +536.1 = 408 m[/tex]
So the magnitude of the resultant vector is
[tex]R=\sqrt{R_x^2 + R_y^2}=\sqrt{(-418.2)^2+(408)^2}=584.3 m[/tex]
While the direction is given by:
[tex]\theta = tan^{-1} \frac{R_y}{R_x}=tan^{-1} \frac{408}{418.2}=44.3^{\circ}[/tex]
But since [tex]R_x[/tex] is negative and [tex]R_y[/tex] is positive, it means that this angle is measured as angle above the negative x-axis; so the direction of the vector is actually
[tex]\theta = 180^{\circ} - 44.3^{\circ} = 135.7^{\circ}[/tex]
The Stokes-Oseen formula for drag force, F, on a sphere of diameter D in a fluid stream of low velocity, v, density rho and viscosity µ is [tex]F = 3\pi \mu DV + \frac{9\pi}{16} \rho V^2 D^2[/tex]Is this formula dimensionally consistent?
Answer:
the equation is dimensionally consistent
Explanation:
To verify that the formula is dimensionally consistent, we must verify the two terms of the sum and verify that they are units of force. We achieve this by putting the units of each dimensional term of the equation and verifying that the answer is in units of force
μ=viscosity=Ns/m^2
D=diameter
V=velocity
ρ=density=Kg/m^3π
First term
3πμDV=[tex]\frac{N.s.m.m}{m^{2}.s }[/tex]=N
the first term is dimensionally consistent
second term
(9π/16)ρV^2D^2=[tex]\frac{kg.m^2.m^2}{m^3.s^2} =\frac{kg.m}{s^2} =N[/tex]
.
as the two terms are in Newtons it means that the equation is dimensionally consistent
The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad A(t) = 415 - \sin\left(\dfrac{2\pi (t + 23.2)}{92.8}\right)A(t)=415−sin( 92.8 2π(t+23.2) )space, A, left parenthesis, t, right parenthesis, equals, 415, minus, sine, left parenthesis, start fraction, 2, pi, left parenthesis, t, plus, 23, point, 2, right parenthesis, divided by, 92, point, 8, end fraction, right parenthesis. The International Space Station reaches its perigee once in every orbit. How long does the International Space Station take to orbit the earth? Give an exact answer.
Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:
[tex]A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]
Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:
[tex]A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )[/tex]
- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min
The time of the International Space Station is mathematically given as
T = 5568sec
How long does the International Space Station take to orbit the earth?Generally, the equation for the altitude of the International Space Station is mathematically given as
[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]
Therefore
Using the equation
[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex] we decipher time to be
T = 2*p / w
In conclusion
T = 2*p / (2*p / 92.8)
T = 5568sec
Read more about Time
https://brainly.com/question/4931057
The height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula h = −16t2 + 136t, where h is the height in feet and t is the time in seconds. Find the time required for the object to return to its point of departure.
Explanation:
Given that, the height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula :
[tex]h=-16t^2+136t[/tex]
Where
h is the height in feet
t is the time in seconds
Let t is the time required for the object to return to its point of departure. At this point, h = 0
[tex]-16t^2+136t=0[/tex]
t = 8.5 seconds
So, the time required for the object to return to its point of departure is 8.5 seconds.
If you drop an object, it will accelerate downward at a rate of 9.8 meters per second per second. If you instead throw it downwards, its acceleration during the rest of the fall (after you release it and in the absence of air resistance) will be _____________ .
Still 9.8 m/s^2 downward.
Without air resistance, the acceleration doesn't depend on the initial velocity or the weight of the object. It only depends on what planet you're on.
A racehorse is running with a uniform speed of 69 km/hr along a straightaway. What is the time it takes for the horse to cover 400 meters?
21 seconds
2.1 minutes
0.35 hours
27.6 hours
Answer:
21 seconds
Explanation:
First, convert km/hr to m/s.
69 km/hr × (1000 m / km) × (1 hr / 3600 s) = 19.2 m/s
Distance = rate × time
400 m = 19.2 m/s × t
t = 20.9 s
Rounded to two significant figures, it takes the horse 21 seconds.
Answer:
The answer to your question is: 21 seconds
Explanation:
Data
speed = v = 69 km/h
distance = d = 400 m
time = t = ?
Formula
v = d / t
t = d / v
Process
Convert 400 m to km
1 km ----------------- 1000 m
x ------------------- 400 m
x = (400 x 1) / 1000
x = 0.4 km
Substitution
t = 0.4 / 69
t = 0.006 h
Covert time to minutes and seconds
60 min -------------------- 1 h
x ------------------- 0.006 h
x = (0.006 x 60) / 1
x = 0.36 min
Convert time to seconds
1 min ------------------ 60 s
0.36 min -------------- x
x = (0.36 x 60) / 1
x = 21.6 seconds ≈ 21 seconds
A tennis ball is dropped from 1.43 m above the
ground. It rebounds to a height of 0.945 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
2. (Let
down be negative.)
Answer in units of m/s
(This I already know -5.29 but I just don't know how to continue and do the rest of the parts)
Part 2
With what velocity does it leave the ground?
Answer in units of m/s.
Part 3
If the tennis ball were in contact with the
ground for 0.00638 s, find the acceleration
given to the tennis ball by the ground.
Answer in units of m/s
2.
Answer:
Part 2: 4.30 m/s
Part 3: 1503 m/s^2
Explanation:
First you already had -5.29 m/s
Part 2
With what velocity does it leave the ground ?
it is the same process as question 1, but in this case the height is 0.945 m
[tex]v =\sqrt{2gh}[/tex]
[tex]v =\sqrt{2*9.8*0.945} \\v =4.30\ m/s[/tex]
Part 3
This is a simple formula of :
[tex]v_{f}-v_{i}=at \\where \\v_{f}: final\ speed \\v_{i}: initial\ speed \\a: acceleration\\t: time[/tex]
Final speed: is the speed the ball leaves the ground = 4.30 m/s
Initial speed: is the speed the ball hits the ground = -5.29 m/s
4.30 - (-5.29) = 0.00638a
a = 1503 m/s^2
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.80 ✕ 105 m/s2 for 9.60 ✕ 10−4 s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)
Answer:
Final velocity will be 556.8 m/sec
Explanation:
We have given the average rate of change of velocity that is [tex]a=5.8\times 10^5m/sec[/tex]
Time [tex]t=9.6\times 10^{-4}sec[/tex]
Initial velocity of bullet will be zero that is u=0
Now according to first law of motion
v=u+at, here v is final velocity, u is initial velocity and t is time
So [tex]v=0+5.8\times 10^5\times 9.6\times 10^{-4}=556.8m/sec[/tex]
Rohit is trying to tune an instrument. He strikes a tuning fork and then plays a note that almost matches it. Rohit hears the volume of the two waves beating, which sounds like "wah-wah-wah-wah."
The number of "wah"s every second is the difference between the frequencies of the tuning fork and the note he is playing.
Answer:
The two waves produced have different frequencies.
Explanation:
Rohit is trying to tune an instrument. He strikes a tuning fork and then plays a note that almost matches it. Rohit hears the volume of the two waves beating, which sounds like "wah-wah-wah-wah."
The options can be the following
The two waves produced have different speeds.
The two waves produced have different amplitudes.
The two waves produced have different wavelengths.
The two waves produced have different frequencies.
A wave is a disturbance which transfers energy through a medium without displacing the medium itself.
The two waves have different frequencies to be able to produce such a sound. That's called a beat frequency. the waves interfere constructively.
The ________ proposes that the bodies of our solar system formed at essentially the same time from a rotating cloud of gases and dust. a. Plate Tectonics theory b. Nebular hypothesis c. Big Band theory d. Heliocentric theory
Answer:
B. Nebular Hypothesis
Explanation:
It is important to read the statement carefully so we can rule out options. The text is referring to solar system formation, it means we should discard all options which are not related to that. Big bang theory is a concept to explain the origin of the whole universe, not the solar system only, so it is discarded. Plate tectonics describes inner earth layers that are essential parts of the planet movemements like earthquakes and they are not related to solar system either. Discarded. Helioentric theory is the idea that defines the sun as the center of the solar system, but again it does not explain anything about its origin, so let me explain a litte in detail about the right choice: Nebular Hypothesis.
Nebular Hypothesis
This is the way astronomers accept the solar system was created as well as other other plantetary systems. Million of years after Big Bang, there were gases and small dust particles all over the universe and they started to join and concentrate into bigger and bigger amount of gases turning into enourmous clouds made of all the necessary elements to begin the formation of stars, planets and eventually the solar system.
You started to run at 10 km/h when you left your house and you arrived at school 30 minutes later. Assuming that your average acceleration was 30 km/h/h, how fast were you running when you arrived?a) 15 km/hb) 20 km/hc) 25 km/hd) 30 km/h
I was running at 25 km/h speed when i arrived the school. Thus, option c) 25 km/h is correct.
To determine how fast you were running when you arrived at school, we need to use the formula that relates initial velocity, acceleration, and time:
v = u + at
Where:
v is the final velocityu is the initial velocitya is the accelerationt is the timeGiven that:
Initial velocity, u = 10 km/hAcceleration, a = 30 km/h²Time, t = 0.5 hours (since 30 minutes is 0.5 hours)We can substitute these values into the formula:
v = 10 km/h + (30 km/h² * 0.5 h)
This simplifies to:
v = 10 km/h + 15 km/h
Thus, v = 25 km/h.
Therefore, you were running at 25 km/h when you arrived at school. The correct answer is c) 25 km/h.
In a crash test, a 2,500 kg car hits a concrete barrier at 1.3 m/s2. If the car strikes the barrier with a force of 3,250 N, how much resistance force does the barrier provide? N This pair of forces is an example of Newton's law of motion.
Answer: 3250 N in the opposite direction
Explanation:
According to Newton's third law of motion:
"Each action has an equal and opposite reaction"
To understand it better:
When two bodies interact between them, appear equal forces and opposite senses in each of them. For example, if a body A exerts a force on another body B, it performs on A another force with the same magnitude and in the opposite direction.
Hence, this law states that for each force acting on a body, ther is a force of equal intensity but in the opposite direction on the body that produced it.
In this sense if the car hit the concrete barrier with a force of 3250 N, the concrete barrier will exert a force of -3250 N on the car (where the negative sign indicates the force is acting in the opposite direction).
Answer:3250 , third law
Explanation:
A ball is thrown against a wall and bounces back toward the thrower with the same speed as it had before hitting the wall. Does the velocity of the ball change in this process?A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.B) The velocity is the same because the speed is the same.C) Without knowing the mass of the ball, the velocity is not determined.D) The velocity is now twice as large because the ball is going into the opposite direction.
Answer:
A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.
Explanation:
Final answer:
The velocity of a ball changes after bouncing off a wall because while its speed remains the same, its direction is reversed. Therefore, the ball's velocity and momentum experience a change, but the overall system momentum is conserved. So the correct option is A.
Explanation:
When considering whether the velocity of a ball changes after it bounces off a wall, it is important to understand that velocity is a vector quantity, meaning it has both magnitude (speed) and direction. If a ball is thrown against a wall and bounces back with the same speed but in the opposite direction, its velocity has indeed changed due to a change in direction. The correct answer here is A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed. The mass of the ball is irrelevant to the consideration of velocity in this context, as velocity does not depend on mass.
Additionally, during this process, the ball experiences a change in momentum. Momentum is also a vector, and the direction of the ball's momentum is reversed upon bouncing off the wall while maintaining the same magnitude, assuming an elastic collision with no energy loss. The change in the direction of the ball's velocity also indicates a reversal in momentum. However, the total system momentum, including the wall and the ball, remains conserved.
Flexible buildings (made of wood or steel) have a ________ resonant period than a stiffer building (one of brick or concrete).
Answer:longer
Explanation:
Answer:
Longer
Explanation:
All structures have a characteristic period, or reverberation, which is the quantity of seconds it takes for the structure to normally vibrate to and fro. The ground likewise has a particular full recurrence. Hard bedrock has higher frequencies milder silt. In the event that the time of ground movement coordinates the normal reverberation of a structure, it will experience the biggest motions conceivable and endure the best harm. Wood and steel are more favored in risky areas.High rises wherever should be strengthened to withstand solid powers from high breezes, however in tremor zones, there are extra contemplation.A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
1- Find an expression for the maximum tension T max for which the small box rides on top of the large box without slipping.Express your answer in terms of the variablesM ,m , ms, and appropriate constants.2-A horizontal rope pulls a 10 kg wood sled across frictionless snow. A 6.0 kg wood box rides on the sled. What is the largest tension force for which the box doesn't slip? Assume that Mk= 0.50.
Answer:
T = g μ_s ( M+m )
78.4 N
Explanation:
When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .
Net force on the lower box
= T - mg μ_s = M a ( a is the acceleration created by net force in M )
Considering force on the upper box
mg μ_s = ma
a = g μ_s
Put this value of a in the equation above
T - m gμ_s = M g μ_s
T = mg μ_s + M g μ_s
= g μ_s ( M+m )
2 )
Largest tension required
T = 9.8 x .50 x ( 10+6 )
= 78.4 N
The maximum tension at which the small box doesn't slip off the large box can be calculated using the formula T_max = μs * mg. Substituting the given values into the formula, the largest tension is found to be 29.4 N for the small box not to slip off a 10 kg wood sled on frictionless snow.
Explanation:The maximum tension, T_max, for which the small box remains stationary with respect to the larger box can be found using the concept of static friction. This static friction, f_s, can be expressed as the product of the static friction coefficient, μs, and the normal force on the small box, which is its weight, mg. So, f_s = μs * mg.
Given that the maximum tension, T_max, equals f_s, we can derive the formula as T_max = μs * mg. This is an expression for the maximum tension in terms of the given variables.
For the numerical part of the question, given that the mass of the box m = 6 kg, and the coefficient of kinetic friction μk = 0.50, we can substitute these values into the derived formula: T_max = μk * m * g = 0.50 * 6.0 kg * 9.8 m/s^2 = 29.4 N. This is the largest tension force for which the box doesn't slip.
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While following the directions on a treasure map a pirate walks 37.0 m north and then turns and walks 8.5 m east what is the magnitude of his resultant displacement?
Answer: [tex]38\ m[/tex]
Explanation:
For this exercise you can use the Pythagorean theorem to find the magnitude of the resultant displacement.
Then:
[tex]d^2=\triangle x^2+\triangle y^2[/tex]
You can observe that the square of the displacement is equal to the sum of the square of the horizontal displacement and the square of the vertical displacement.
Since the pirate walks 37.0 meters north and then turns and walks 8.5 meters east:
[tex]\triangle x=37.0\ m\\\triangle y=8.5\ m[/tex]
Substituting values and solving for "d", you get:
[tex]d=\sqrt{(37.0\ m)^2+(8.5\ m)^2}\\\\d=38\ m[/tex]
After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground. It is moving with a speed of 9 m/s when it reaches a maximum height of 7 m above the ground. What is the speed of the ball when it leaves Sarah's hand?
Answer:
13.82 m/s
Explanation:
You can see it in the pic.
Answer:
vo = 13.74 m/s
Explanation:
The goal is to find the initial velocity vo. According to the exercise, the maximum height that the ball will reach and the initial height will be respectively, yo = 1.5 m and ymax = 7 m, tmax will be equal to the time it takes for the ball to reach the maximum height. Having the equation:
vx = vo * cos∝o
clearing vo:
vo = vx/cos∝o
vy = vo * sin∝o + gtmax
0 = sin∝o * (vx/cos∝o) + gtmax
0 = vx * tan∝o + gtmax
gtmax = -vx * tan∝o
clearing tmax:
tmax = - (vx * tanao/g)
ymax = i + (vo * sin∝o) * tmax + (gtmax^2/2)
Replacing:
ymax = i - (vx^2 * tan^2∝o/2 * g)
clearing the angle a or:
∝o = arctan (((2 * g * (i - ymax)) / vx^2)^1/2)
substituting values:
∝o = arctan (((2 * (- 9.8) * (1.5 - 7)) / (9^2))) = 49.08 °
We find the initial speed with the following formula:
vo = vx/cos∝o = 9/cos49.08 ° = 13.74 m/s
A charge of −3.77 µC is located at the origin, and a charge of −4.92 µC is located along the y axis at 2.86176 m. At what point along the y-axis is the electric field zero? The value of the Coulomb constant is 8.99 × 109 N · m2 /C 2 .
Answer:
1.336 m
Explanation:
q = - 3.77 micro C at y = 0 m
Q = - 4.92 micro C at y = 2.86176 m
Let the electric field is zero at P which is at a distance y from origin.
Electric field at P due to q is equal to the electric field at P due to Q.
[tex]E_{q}=E_{Q}[/tex]
[tex]\frac{kq}{OP^{2}}=\frac{kQ}{AP^{2}}[/tex]
[tex]\frac{3.77}{y^{2}}=\frac{4.92}{\left ( 2.86176-y \right )^{2}}[/tex]
[tex]\frac{\left ( 2.86176-y \right )}{y}}=1.142[/tex]
2.86176 - y = 1.142 y
2.142 y = 2.86176
y = 1.336 m
Thus, the electric field is zero at y = 1.336 m.
A pellet gun is fired straight downward from the edge of a cliff that is 12.7 m above the ground. The pellet strikes the ground with a speed of 27.9 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?
Answer:
h₂ = 27.22m : height the pellet reaches
Explanation:
Pellet kinetics
Pellet moves with uniformly accelerated movement in
freefall
v f₁²=v₀²+2g*h₁ (formula 1) : ball movement down
v f₂²=v₀²-2g*h₂ (formula (formula 2) : ball movement up
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
g: acceleration due to gravity in m/s²
h :height in m
Known data
g=9.8 m/s²
h₁=12.7m
v f₁= 27.9 m/s
Initial velocity calculation (v₀)
We replace the data in formula 1:
27.9²=v₀²+2*9.8* 12.7
27.9²-2*9.8* 12.7=v₀²
529.49=v₀²
[tex]v_{o} =\sqrt{529.49}[/tex]
[tex]v_{o} =23.1\frac{m}{s}[/tex]
Calculation of the height that the pellet reaches when it goes up
Data:
[tex]v_{o} =23.1\frac{m}{s}[/tex]
v f₂=0 :When the pellet reaches the maximum height the final speed is zero
We replace the data in formula (2)
0²=23.1²-2*9.8*h₂
19.6h₂ =23.1²
h₂ =23.1²÷19.6
h₂ =27.22m
A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. What happens to the acceleration of the truck as its trailer leaks sand at a constant rate through a hole in its bottom?
Answer:
Acceleration will increase.
Explanation:
The relation between force, mass and acceleration according to the Newton's second law of motion is given as:
F = ma
We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:
[tex]a=\frac{F}{m}[/tex]
Since, F is constant, the acceleration of the truck is inversely proportional to the mass.
There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.
Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.
So, the acceleration of the truck will increase.
As the truck loses mass due to leaking sand, while a constant driving force is maintained, the acceleration of the truck increases according to Newton's second law of motion.
Explanation:The question is about the changes in the acceleration of a truck as it loses mass due to leaking sand. From a physics perspective, the truck's acceleration will actually increase as it loses mass. This is because the force propelling the vehicle remains constant while its mass decreases. According to Newton's second law of motion, which states that force equals mass times acceleration (F=ma), if the force remains constant and mass decreases, the acceleration must increase to keep the equation balanced.
For example, with initial values, if the force propelling the truck is 30 newtons and the truck (with its sand) has a mass of 15 kilograms, the acceleration would be 2 m/s² (from F=ma, or 30N=15kg*a). Now suppose the sand leaks out and the truck’s mass drops to 10kg. If the driving force stays the same at 30N, when plugging these values back into F=ma, the new acceleration would be 3 m/s². Thus, the acceleration of the truck has increased as it has lost mass.
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A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. A vector from the station to the point where the boat was later found is B = 30.0 km, 15.0° north of east.
How far did the boat travel from the point where the distress call was made to the point where the boat was found? In other words, what is the magnitude of vector C?
A)65.3 km
B)39.7 km
C)26.5 km
D)54.0 km
E)42.5 km
Answer:
d = 39.7 km
Explanation:
initial position of the boat is 45 km away at an angle of 15 degree East of North
so we will have
[tex]r_1 = 45 sin15 \hat i + 45 cos15 \hat j[/tex]
[tex]r_1 = 11.64 \hat i + 43.46\hat j[/tex]
after some time the final position of the boat is found at 30 km at 15 Degree North of East
so we have
[tex]r_2 = 30 cos15\hat i + 30 sin15 \hat j[/tex]
[tex]r_2 = 28.98\hat i + 7.76 \hat j[/tex]
now the displacement of the boat is given as
[tex]d = r_2 - r_1[/tex]
[tex]d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)[/tex]
[tex]d = 17.34 \hat i - 35.7 \hat j[/tex]
so the magnitude is given as
[tex]d = \sqrt{17.34^2 + 35.7^2}[/tex]
[tex]d = 39.7 km[/tex]
Two boxes (24 kg and 62 kg) are being pushed across a horizontal frictionless surface, as the drawing shows. The 42-N pushing force is horizontal and is applied to the 24-kg box, which in turn pushes against the 62-kg box. Find the magnitude of the force that the 24-kg box applies to the 62-kg box.
Final answer:
The magnitude of the force the 24-kg box applies to the 62-kg box on a frictionless surface can be calculated using Newton's second law (F=ma), considering both boxes as a single system to find the combined acceleration, then applying that acceleration to determine the force on the 62-kg box.
Explanation:
To find the magnitude of the force that the 24-kg box applies to the 62-kg box, we need to apply Newton's second law of motion which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F=ma). Since the 42-N pushing force is applied to the 24-kg box, this force is responsible for accelerating both the 24-kg and 62-kg boxes, which are considered to be one system in this context. Therefore, the acceleration of the system can be calculated using a = F_total/m_total, where F_total is the total force (42 N) and m_total is the combined mass of both boxes (24 kg + 62 kg).
Once the acceleration is known, the force exerted by the 24-kg box on the 62-kg box can be found by calculating the force needed to accelerate the 62-kg box at the previously determined acceleration. This is found using the same equation F=ma, where m is the mass of the 62-kg box and a is the acceleration of the system.
Question 4 (4 points)
Velocity is change in displacement divided by time. Acceleration is ___ divided by
time.
O 1) change in speed
2) change in velocity
3) change in distance
4) change in position
Onection 5 11 nainte!
Answer:
2. Change in velocity
Explanation:
The units that acceleration use are distance/time^2 [tex](m/s^2)[/tex] (in metric units).
The acceleration is the change of velocity over time, so
[tex]\frac{dv}{dt} =\frac{dx}{dt}*\frac{1}{dt}[/tex]
In another way, if the velocity is constant, the acceleration is zero, but if the velocity change over the time (as a car do when the red light change), the acceleration takes value!.
Velocity changes in displacement divided by time. Acceleration is the change in velocity divided by time.
Answer: Option 2
Explanation:
Acceleration is defined as the rate of velocity changes of an object in a given time period. This can express in the equation as follows,
[tex]\bold{a = \frac{v}{t} m / s^{2}}[/tex]
There two types in it as linear and radial acceleration. When velocity changes in the opposite direction to an object then it is called deceleration. But these are the same as a change in velocity.