The time that the ball took to take off at an angle of 29.0° is 1.13 s
When two lines meet at a point, they form an angle. The word "angle" refers to the length of the "opening" between these two rays. The symbol is used to denote it.
Radians, a unit of circularity or rotation, and degrees are the two most common units used to measure angles. In daily life, angles are present. For the design of highways, structures, and sports venues, engineers and architects use angles.
The motion along the x-direction is always uniform, so the position along the x-direction is given by
x(t) = (v₀cos∅)t
where,
v₀ is the initial speed,
cos∅ is the angle of launching,
t is the time.
We can re-write the formula as
v₀t = x(t)/cos∅ ...1
Along the y-direction the motion is accelerated (free-fall), so the equation for the vertical position is
y(t) = h = v₀sin∅t - 1/2gt² ...2
where,
h = 45.0 m is the height of the building
g = 9.8m/s² is the acceleration of gravity.
Substituting (1) into (2),
y(t) = h = x(t)tan∅ - 1/2gt²
And when this happens, the displacement along the x-axis is x = 47.8 m. Solving the equation for t and substituting the numbers, we find the time of flight:
[tex]t = \sqrt{\frac{2(h-xtan\theta)}{g} }[/tex]
[tex]t = \sqrt{\frac{2(45-47.8tan39^{\circ})}{9.8} }[/tex]
t = 1.13 s
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Full question
Suppose the ball is thrown from a 45m height building as in the PRACTICE IT problem at an angle of 29.0° below the horizontal. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
a) the time of flight s
an optical instrument consist of two lenses .if the object distance for lens 1 is 12cm then the image distance is 36cm.if the height of the object for lens 2 is 5cm then the image height is 2.5cm. evaluate the combined magnification of the two lenses
The magnification of first lens can be given as,
[tex]m_1=-\frac{v_1}{u_1}[/tex]Here, m1 is the magnification of first lens, v1 is the distance of image from first lens and u1 is the distance of object from first lens.
Substitute the known values in the equation,
[tex]\begin{gathered} m_1=-\frac{36\text{ cm}}{12\text{ cm}} \\ =-3 \end{gathered}[/tex]The magnification of second lens can be given as,
[tex]m_2=-\frac{v_2}{u_2}[/tex]Here, m2 is the magnification of second lens, v2 is the distance of image from second lens and u2 is the distance of object from second lens.
Substitute the known values in the equation,
[tex]\begin{gathered} m_2=-\frac{2.5\text{ cm}}{5\text{ cm}} \\ =-0.5 \end{gathered}[/tex]The combined magnification of two lenses can be given as,
[tex]m=m_1m_2[/tex]Substitute the known values,
[tex]\begin{gathered} m=(-3)(-0.5) \\ =1.5 \end{gathered}[/tex]Thus, the combined magnification of two lenses is 1.5.
1. A force of 20 N is applied to extend a horizontal spring by 5 cm. What is the spring constant?
2. A spring has a spring constant of 15 N/cm. Which of the following hanging mass would produce an extension of 3.0 cm?
The spring constant is 400 N/m and The mass which would produce an extension of 3.0 cm is 0.0459 kg.
Here we are dealing with a problem related to hook's law which states that the force required to expand or compress a spring by a few distances is corresponding to that distance. This can be expressed scientifically as F= -kx, where F is the force connected to the spring and x is the displacement of the spring, with negative esteem illustrating the displacement of the spring once it is extended; and here k is the spring constant.
For the first case, we are given the force applied which is 20 N, and the extended length which is 5cm =0.05 m,
So the spring constant will be, k= F/x= 20/0.05 = 400 N/m
For the second case we are given that the spring constant which is 15 N/m and the extension is 3.0 cm =0.03 m.
Since we know F= mg ( where g is the acceleration due to gravity )
So the hanging mass, will be
=>F= kx
=>mg=kx
=>m=kx/g
=15 x0.03/9.8
= 0.0459 kg
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The speed of sound traveling through a dense material is 600 m/s. If the wavelength is 15.0 m, what is the frequency of this sound?9,000 Hz80 Hz0.0025 Hz40 Hz
W0e are asked to determine the frequency of a sound given its velocity and wavelength. To do that we will use the following formula:
[tex]v=\lambda f[/tex]Where:
[tex]\begin{gathered} v=\text{ velocity} \\ \lambda=\text{ wavelength} \\ f=\text{ frequency} \end{gathered}[/tex]Now, we solve for the frequency by dividing both sides by the wavelength.
[tex]\frac{v}{\lambda}=f[/tex]Now, we substitute the values:
[tex]\frac{600\frac{m}{s}}{15m}=f[/tex]Solving the operations:
[tex]40Hz=f[/tex]Therefore, the frequency is 40 Hz.
my opponent will say that recycling is in an issue here but that's just because they don't care about the sea turtles or other marine life affected by plastic in the ocean what is a logical fallacy
I’m not really sure if these answers are right so I’d really like some clarification! :)
Given a metal loop of wire with an area of 400 square centimeters and a magnetic field of 0.0875 Tesla through it. If the magnetic field is constant everywhere through the loop then which of these is the magnetic flux through the loop in SI units? [4. Define magnetic flux] 0.00350 0.0140 286
We will have that the value will be:
[tex]\Phi=(0.04m^2)(0.0875T)\Rightarrow\Phi=0.00350m^2T[/tex]A bullet is shot from a rifle with a speed of 600 m/s. If a target is 2000 meters away, how long will it take
to hit the target?
The time taken for the bullet moving with a speed of 600 m/s to hit a target 200 meters away is 3.33 s.
What is time?
Time is the measured or measurable period during which an action, process, or condition exists or continues.
To calculate the time it takes the bullet to hit the target, we use the formula below.
Formula:
v = d/t.......... Equation 1Where:
v = Velocityd = distancet = timeMake t the subject of the equation
t = d/v............ Equation 2From the question,
Given:
d = 2000 metersv = 600 m/sSubstitute these values into into equation 2
t = 2000/600t = 3.33 secondsHence, the time it will take to hut the target is 3.33 seconds.
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Please help!!
Question in the picture
Set of equation which is true for the situation is
c) 2SA = SB and 2BA=BB
What is pressure?
Force applied perpendicular to the surface of an object per unit area is called pressure
SI unit of pressure is one newton per square meter (N/m2).
Given, length = 3 foot, width= 1 foot and height = 1 foot
1foot = 0.305m
Volume of tank A = 3*0.305*0.305*0.305
= 0.085
density of water= 1000kg/m^3
Pressure= 1000kg/m^3 *gravity * height
=1000kg/m^3 * 9.8 m/s^2* 0.305
=2989 N
Force exerted by water at bottom= P * A
= 2989*0.085
=254.065 Nm^2
similarly, for tank B
Given ,length = 6 foot, width =1 foot and height= 1foot
Volume of tank B = 6*0.305*0.305*0.305
=0.17
pressure= 1000kg/m^3 * 9.8m/s^2 * 0.305
= 2989 N
Force exerted by water at bottom = P*A
= 2989 * 0.17
=508.13 Nm^2
Hence proved that pressure exerted by tank B is twice that of tank A.
And same goes for pressure exerted on the side of the tank.
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A vector is 148 m long and
points in a -96.3 degree
direction.
Find the x-component of the
vector.
The x-component of the vector is determined as -16.24 m.
What is the x-component of a vector?
The x-component of a vector is the vector component measured along x - direction. It is the vector acting in the x-coordinate.
The magnitude of x -component of a vector is given by the following formula as shown below;
Vₓ = V cosθ
where;
Vₓ is the x - component of the vectorV is the magnitude of the vector = 148 mθ is the angle of the vector = -96.3 degreesVₓ = 148 m x cos(-96.3)
Vₓ = -16.24 m
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Answer:
Y= -147
X= -16.24
Explanation:
Three point charges are located along the x-axis: q1 = +5.0 μC is at 20 cm, q2 = +4.0 μC is at 0 cm, and q3 = +10.0 μC is at –30 cm. What is the net electrostatic force on q2? 0.50 N to the left 500 N to the right 0.50 N to the right 8.5 N to the left
The correct answer is option C, 0.50 N to the left.
Given,
The charges,
q₁=5×10⁻⁶ C
q₂=4.0×10⁻⁶ C
q₃=10.0×10⁻⁶ C
The distance between the 1st charge and the second charge, r₁=20 cm=0.20 m
The distance between 2nd charge and the 3 rd charge, r₂=30 cm=0.30 m
As all the charges have the same sign, the nature of the force is repulsive. Thus the force applied by the q₂ is directed to the left. And the force applied by the charge q₃ is pointed to the right.
From Coulomb's law, the net electrostatic force on q₂ is given by,
[tex]\begin{gathered} F=\frac{kq_1q_2}{r^2_1}-\frac{kq_2q_3}{r^2_2} \\ =kq_2(\frac{q_1}{r^2_1}-\frac{q_3}{r^2_2}) \end{gathered}[/tex]On substituting the knwon values,
[tex]\begin{gathered} F=9\times10^9\times4\times10^{-6}(\frac{5\times10^{-6}}{0.20^2}-\frac{10\times10^{-6}}{0.30^2}) \\ =0.5\text{ N} \end{gathered}[/tex]Thus the magnitude of the net electrostatic force on the charge q₂ is 0.5 N
As the force to the left is greater than the force to the right, the net force will be directed to the left.
Therefore the correct answer is option A, 0.50 N to the left.
can you help me with this ap physics class from my school
We know that the components of any vector along the x and y axes are given by:
[tex]\begin{gathered} v_x=v\cos\theta \\ v_y=v\sin\theta \\ \text{ where:} \\ v\text{ is the magnitude of the vector.} \\ \theta\text{ is the angle from the }x\text{ axis to the vector. } \end{gathered}[/tex]In this case we know the tha magnitude of the vector is 6.96 m/s; we also know that the angle from the x-axis to the vector is 51.5° (the sum of angles θ and the 26.5° angle); then we have:
[tex]\begin{gathered} v_x=6.96\cos51.5=4.33 \\ v_y=6.96\sin51.5=5.45 \end{gathered}[/tex]Therefore, the components of the vector are:
[tex]\begin{gathered} v_x=4.33 \\ v_y=5.45 \end{gathered}[/tex]A satellite orbiting Earth at an orbital radius r has a velocity v. Which represents the velocity if the satellite is moved to an orbital radius of 5r?(1 point)
The equation which represent the velocity, if the satellite is moved to an orbital radius of 5r is V = V/[tex]\sqrt\\5[/tex].
What is Orbital velocity?Orbital velocity or orbital speed is the minimum velocity a body must maintain to stay in an orbit. Due to the inertia of a moving body, the body has the tendency to move on in a straight path.
It is given that the radius is 5r
To find out the equation which represent the velocity if the satellite is moved to an orbital radius of 5r.
Mathematically, the orbital velocity is calculated by the formula:
V = [tex]\sqrt{GM}[/tex]/r
where, V = orbital velocity,
G = gravitational constant
M = mass of a satellite
r = radius
Substituting the value of r in this equation, we have:
V = [tex]\sqrt{GM}[/tex]/r
V = 1/ [tex]\sqrt{5}[/tex] × [tex]\sqrt{GM}[/tex]/r
V = [tex]\sqrt{GM}[/tex]/r
Therefore,
V = 1/ [tex]\sqrt{5}[/tex] × V
V = V/[tex]\sqrt{5}[/tex]
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A shopper in a supermarket pushes a loaded cart with a horizontal force of 14 N. The cart
has a mass of 32 kg.
The acceleration of gravity is 9.8 m/s
2
.
Disregarding friction, how far will the cart
move in 4 s, starting from rest?
Answer in units of m.
Answer:
ΔS = 3.5 m
Explanation:
First, apply the 2º Newton's Law to find the resulting acceleration.
∑F = m × a
F = m × a
14 = 32 × a
a = 14 /32 = 7/16
a = 0.4375 m/s²
Then, we must apply the equation of accelerated movement:
ΔS = V₀*t + (1/2)*a*t²
Starting from the rest and using the value of acceleration:
ΔS = 0*4 + (1/2)*(0.4375)*(4)²
ΔS = (1/2)*(0.4375)*(16)
ΔS = 3.5 m
An electron moves between the ends of a 9.00 volt battery. How much work is done on the electron? A)0.78 x10^-15 J B)9.00 x10^-17 J C)1.44 x10^-18 J D)25.0 x 10^-20 J E)15.3 x 10^-20 J
When an electron moves between the ends of a 9.00-volt battery, the work done on the electron is 1.44 * 10⁻¹⁸ Jouoles; option C
What is the work done on an electron?The work done in moving an electron between two points is defined as the product of the potential difference between the points and the charge on an electron.
Mathematically;
Work done = ΔV * qwhere;
ΔV is the potential difference between the points
q is the charge on the electron.
The potential difference between the two points is the work done in moving a unit positive charge from one point to the other in the circuit.
The charge, q on an electron is 1.609 * 10⁻¹⁹ Coulombs.
For the given electron that moves between the ends of a 9.00-volt battery;
ΔV = 9.000 Volts
q = 1.609 * 10⁻¹⁹ Coulombs
Work done = 9.000 * 1.609 * 10⁻¹⁹
Work done = 1.44 * 10⁻¹⁸ Jouoles
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A 75 kg student drives a 650 kg car. What is the applied force on the car if * 5 points the car starts from rest and accelerates to 60 m/s in 8 seconds?
Answer:
4500 N
Explanation:
Bc f is equal to mass times acceleration and the acceleration is 60
If a 75 kg student drives a 650 kg car, then the car starts from rest and accelerates to 60 m/s in 8 seconds, then the applied force on the car would be 5437.5 Newtons.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
The mathematical expression for Newton's second law is as follows
F = m × a
As given in problem an If a 75 kg student drives a 650 kg car, the car starts from rest and accelerates to 60 m/s in 8 seconds.
The acceleration of the car = 60 / 8
= 7.5 m / s²
The force applied on the car = 725 × 7.5
= 5437.5 Newtons
Thus, the applied force on the car would be 5437.5 Newtons.
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The highest freefall jump on record is from a height of almost 38,000 m. At this height, the acceleration of gravity is slightly weaker than the accelera
of gravity on the surface of Earth. Although the skydiver used a parachute to slow himself down below a certain altitude, how would the increasing
acceleration impact his velocity and rate of displacement covered?
BIU
S2
X
Increasing his acceleration will impact his velocity and rate of displacement covered in that as the speed increases due to the increased rate of acceleration, the rate of air resistance also increases.
What is air resistance?Air resistance is a force created by air. When an item moves through the air, the force operates in the opposite direction.
When a diver descends, the force of air resistance acts to counteract the force of gravity. As the skydiver falls faster and faster, the quantity of air resistance grows until it equals the magnitude of gravity's force.
A balance of forces is achieved when the force of gravity equals the force of air resistance, and the skydiver no longer accelerates. The skydiver reaches what is known as terminal velocity.
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A 10 kg rubber block sliding on a concrete floor (µ = 0.65) Calculate the frictional force.
Answer:
Therefore the friction Force is 65 N.
Explanation:
Look at the image above.
The frictional force of the rubber block is 63.7N.
What is Frictional force?Friction is defined as a force that opposes the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are many types of friction such as dry friction which is defined as the force that opposes the relative lateral motion of two solid surfaces in contact.
This friction force is directed in the opposite direction to the motion of the object itself because the friction described so far between surfaces in relative motion is called kinetic friction.
For above given information,
mass of the block, m =10kg
coefficient of friction, µ=0.65
f=coefficient of the friction ×normal= µ*R
R= normal =mg = 9.8×10=98N
So, f =0.65×98=63.70N
≈65N(when g is taken as 10m/s²)
Thus, the frictional force of the rubber block is 63.7N.
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Jerry is riding his motorcycle at a speed of 45 m/s. He accelerates at a speed of -8 m/s² for 4seconds. How fast will he be going
Given
Initial speed, u=45 m/s
Acceleration
[tex]a=-\frac{8m}{s^2}[/tex]Time taken, t=4 sec
To find
The final speed
Explanation
Let the speed he be going in be v
Thus,
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \Rightarrow-8=\frac{v-45}{4} \\ \Rightarrow-32+45=v \\ \Rightarrow v=13\text{ m/s} \end{gathered}[/tex]Conclusion
The velocity he is going in is 13 m/s.
If Hubble's constant had a value of 95 km/s/Mpc, what would be the age of the Universe?
The age of the Universe is 13.8 ± 0.1 billion years.
What is Hubble's constant?The ‘hubble constant, h0’ is used to determine the size of the universe. It is a measurement that is used in describing the universe's expansion. It was a constant introduced by Edwin Hubble. He found out that the farther the distance of the galaxies from the Earth, the faster they would appear to be moving.
One of the most important numbers in cosmology Hubble's constant because it is needed to estimate the size and age of the universe which is 13.8 ± 0.1 billion years.
The age of the universe using Hubble's constant which is 13.8 ± 0.1 billion years.
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A grinding wheel 0.35 m in diameter rotates at 2500 rpm. What is the linear speed of apoint on the edge of the grinding wheel?
Linear velocity (V) = r* w
where:
r = radius = diameter/2= 0.35 /2 = 0.175 m
w= angular velocity = rpm = 2500 = 2500/60 = 125/3 rps
1 revolution = 360° = 2pi rad/s = 125/3 2pi rad/s
= 261.8 rad/s
v = 0.175 m * 261.8 m/s = 45.8 m/s
A lamp draws 0.50 A from a 120 V generator. (a) How much power does the generator deliver to the lamp? (b)How much electric energy does the lamp convert to light and heat in a period of 5.0 minutes?
Given:
The lamp draws the current is
[tex]I=0.50\text{ A}[/tex]The voltage supplied by the generator is
[tex]V=120\text{ V}[/tex]The time for which energy converted is
[tex]\begin{gathered} t=5\text{ min} \\ t=5\times60\text{ s} \\ t=300\text{ s} \end{gathered}[/tex]Explanation:
(a)
The power delivered by the generator is given by
[tex]P=VI[/tex]Plugging all the values in the above equation, we get
[tex]\begin{gathered} P=120\text{ V}\times0.50\text{ A} \\ P=60\text{ watt} \end{gathered}[/tex]The power delivered by the generator is
[tex]60\text{ watt}[/tex](b)
Energy is converted by the electric lamp is given by
[tex]E=Pt[/tex]Plugging all the values in the above relation, we get
[tex]\begin{gathered} E=60\text{ watt}\times300\text{ s} \\ E=18000\text{ J} \end{gathered}[/tex]Thus, the electric energy convert by the lamp is
[tex]18000\text{ J}[/tex]Divya and Melissa go to a restaurant. They decide to try something they have not eaten before and order chicken sandwiches with sriracha sauce. Both Divya and Melissa find it difficult to eat the sandwiches as they feel the sandwiches are too spicy. Which of the following processes help them identify the spicy taste?
The processes that help Divya and Melissa identify the spicy taste of the chicken sandwiches with sriracha sauce is sensation.
Sensation is the process in which one feels something after being in physical contact with something. As Divya and Melissa eats the chicken sandwiches with sriracha sauce, their senses tells that the food is spicy.
Acculturation is the process in which one adopts and adjusts to a new cultural environment. Homogenization is the process in which two non-soluble liquids are mixed uniformly. Adaptation is the process of changing one's trait to a new environment.
Therefore, the processes that helps them identify the spicy taste is sensation.
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A block is pulled to the right by a rope as shown above. Which of the following statements are true about the block?I. The normal force is equal to the weight.II. The normal force is mgcosθ.III. The horizontal component of F is Fcosθ.answer choice-I only.II and III onlyIII only.II only.I and III only.
Answer:
III only
Explanation:
The net vertical force is equal to 0 because the block is not moving up. Then, we can write the following equation
Fnet = N + Fsinθ - mg = 0
Where Fsinθ is the vertical component of the force F. Solving for the normal force N, we get:
N = mg - Fsinθ
Therefore, the normal force is not equal to the weight and the normal force is not mgcos(θ)
However, the horizontal component of F is Fcos(θ). So, the answer is
III only.
At the beginning of a new school term, a
student moves a box of books by attaching a
rope to the box and pulling with a force of
F = 84.4 N at an angle of 64°, as shown in the
figure.
The acceleration of gravity is 9.8 m/s².
The box of books has a mass of 14 kg and
the coefficient of kinetic friction between the
bottom of the box and the floor is 0.3.
64°
14 kg
p=0.3
What is the acceleration of the box?
Answer in units of m/s².
Answer:
1.22 m/s² (3 s.f.)
Explanation:
Draw a diagram modelling the given situation (see attachment).
F = Friction.Given values:
Pulling force = 84.4 NMass (m) = 14 kgAcceleration due to gravity (g) = 9.8 m/s²Coefficient of friction (μ) = 0.3As the pulling force is at an angle to the plane (ground), resolve the force into components parallel and perpendicular to the plane.
Resolving vertically (↑) to find the Normal Reaction, R:
[tex]\begin{aligned}\implies R+82.2 \sin 64^{\circ}&=14g\\R&=14g-82.2 \sin 64^{\circ}\end{aligned}[/tex]
The frictional force takes its maximum value when an object starts to move (or is on the point of moving):
[tex]\boxed{F_{\text{max}}= \mu R}[/tex]
where R is the Normal Reaction and μ is the coefficient of friction.
Using F = μR to find the Frictional Force, F:
[tex]\begin{aligned}\implies F &= 0.3\left(14g-82.2 \sin 64^{\circ}\right)\\ & =0.3\left(14(9.8)-82.2 \sin 64^{\circ}\right)\\& =0.3\left(137.2-82.2 \sin 64^{\circ}\right)\\ & = 41.16-24.66\sin 64^{\circ}\end{aligned}[/tex]
Newton's second law states that the overall resultant force acting on a body is equal to the mass of the body multiplied by the body’s acceleration:
[tex]\boxed{F_{\text{net}}=ma}[/tex]
Resolving horizontally (→) using Newton's second law of motion to find acceleration:
[tex]\begin{aligned}\textsf{Using} \quad F_{\text{net}}&=ma\\\\\implies 82.2 \cos64^{\circ}-(41.16-24.66\sin 64^{\circ})&=14a\\17.0383694...&=14a\\a&=\dfrac{17.0383694...}{14}\\a&=1.21702638...\\a&=1.22\;\sf m/s^2\;(3\;s.f.)\end{aligned}[/tex]
Therefore, the acceleration of the box is 1.22 m/s² (3 s.f.).
If a car accelerated from 55 mi/hr to 16 mi/hr. in 3.2s how far did the car go while it was accelerating?
The cargo while it was accelerating at a distance of 0.602 meters.
Acceleration is the name we give to any system wherein the speed changes. on the grounds that speed is a pace and a direction, there are only two approaches so that you can boost up.
Velocity is the rate of change of displacement. Acceleration is the price of alternate velocity. velocity is a vector quantity because it consists of both significance and direction. Acceleration is likewise a vector quantity as it's far simply the charge of alternate velocity.
initial velocity = 55 m/hr
= 55/60× 60 m/sec
= 0.0152 m/sec
final velocity = 16 mi/hr
= 16/3600
= 0.0044 m/sec
time = 3.2 sec
V = u - at
a = ( U - V )/t
= (0.0152 - 0.0044)/t
= 0.0108 m/sec²
S = ut + 1/2at²
= 0.0152 × 3.2 sec + 1/2 × 0.0108 m/sec² × (3.2)²
= 0.0486 + 0.553
= 0.602 meter.
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In your readings you learned that “correlation does not imply causality.” Which of the following is the best interpretation of this statement?
A. Just because there is a relationship between two behaviors does not mean that one causes the other
B. Many factors influence the relationship between two behaviors.
C. A relationship between two behaviors does not mean that one or both behaviors are testable.
D. Correlated behaviors are the result of operational conditioning.
SOMEONE PLEASE HELP!
Just because there is a relationship between two behaviors does not mean that one causes the other. Option A.
Natural observation is a research method that observes objects in their natural environment. This approach is commonly used by psychologists and other social scientists. It is a process of deep reflection, questioning, and evidence evaluation. A study in which researchers look for relationships between people's demographic neighborhood and their level of prejudice.
Psychologists use this because it is the most objective method known to produce reliable insights. The science of behavior and mental processes. It attempts to describe and explain aspects of human thinking feeling perception and behavior. Naturalistic observation is a method of observation that involves observing people's behavior in environments in which they normally occur. Laboratory Observation observing behavior in a more artificial environment focusing on a small number of carefully defined behaviors.
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I need helppp
What is the force applied to the small piston,f,in Lb that is required to deliver a working force,F, of 1000 lb? The area of the small piston,a, is 0.1963 sq in. and the area of the large piston,A, is 6.28 sq in.
The area of the small piston, a, is 0.1963 sq in. and the area of the large piston, A, is 6.28 sq in. The force applied to the small piston (f)= 31.21 lb that is required to deliver a working force (F)= 1000 lb
What is force?Force is a physical quantity that means if some kind of external or internal affair is happen on a object and the object is move a bit that means a force is act on it. Its a vector quantity.
How can we calculate the force?To calculate the force of the small piston f we are using the formula,
p=P
Or, f/a=F/A
Here we are given,
a= The area of the small piston = 0.1963 sq in
A= The area of the big piston = 6.28 sq in.
F= The force of the big piston = 1000 lb.
We have to calculate the force of the small piston = f
Now we put the values in above equation we get,
f / a = F / A
Or, f = a F/A
Or, f = [tex]\frac{0.1963 \times 1000}{6.28}[/tex]
Or, f = 31.21 lb
Now from the calculation we can say that, The force applied to the small piston (f)= 31.21 lb
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a javelin is thrown at 28.5m/s from flat ground at a 43.2 degree angle. How long does it take to reach its maximum height? IN SECONDS
The time taken for the javelin thrown at 28.5 m/s from flat ground at a 43.2 degree angle to reach the maximum height is 2.0 s
How do I determine the time taken to reach the maximum height?From projectile motion, we understand that the time taken to reach maximum height is given as:
t = uSineθ / g
Where
t is the time to reach maximum heightu is the initial velocity θ is the angle of projectiong is acceleration due to gravityNow, using the above formua, we can obtain the time taken for the javelin to reach the maximum height as follow:
Initial velocity (u) = 28.5 m/s
Angle of projection (θ) = 43.2 degrees
Acceleration due to gravity (g) = 9.8 m/s²
Time taken to reach the peak (t) = ?
t = uSineθ / g
t = (28.5 × Sine 43.2) / 9.8
t = 2.0 s
Thus, the time taken to reach maximum height is 2.0 s
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Please help me understand this I go to get it
Answer:
Vx = 150 * cos 53 = 90.3 m/s horizontal speed
Vy = 150 * sin 53 = 120 m/s vertical speed
t1 = 120 / 9.8 = 12.2 sec time to reach vertical height (vy = 0 at apex)
H1 = 1/2 g t^2 = 9.8 / 2 * 12.2^2 = 729 m height risen above cliff
729 + 405 = 1134 m total height above striking point
H = 1/2 g T^2 where T is time to fall 1134 m
T = (2 * 1134 / 9.8)^1/2 = 15.2 sec
T + t1 = 15.2 + 12.2 = 27.4 sec total time in air
Sx = Vx * 27.4 = 90.3 * 27.4 = 2470 m total horizontal distance
Help! I've tried solving it a few times!
An electron travels 2.68 m in 6.11 × 10^-8 s. How fast does it travel in cm/s? Answer in units of cm/s.
If an electron travels 2.68 m in 6.11 × 10^-8 s, it will travel at a speed of 43.86 * [tex]10^{8}[/tex] cm / s
v = d / t
v = Velocity
d = Distance
t = Time
d = 2.68 m
t = 6.11 * [tex]10^{-8}[/tex]
v = 2.68 / 6.11 * [tex]10^{-8}[/tex]
v = 0.4386 * [tex]10^{8}[/tex] m / s
v = 0.4386 * [tex]10^{8}[/tex] * 10² cm / s
v = 43.86 * [tex]10^{8}[/tex] cm / s
One meter is equal to 100 centimeters. So in order to convert a meter value into centimeter, multiply the meter value by 100 or 10².
Therefore, the electron travels at 43.86 * [tex]10^{8}[/tex] cm / s
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