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Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?
Answer:
The force on the second piston is 5246.5 N .
Explanation:
Area of first piston, a = 3.002 cm^2
Area of second piston, A = 315 cm^2
Force on first piston, f = 50 N
let the force of the second piston is F.
According to the Pascal's law
[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]
A long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point
The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
A study finds that the metabolic rate of mammals is proportional to m^3/4 , where m is the total body mass. By what factor does the metabolic rate of a 70.0-kg human exceed that of a 4.91-kg cat?
Answer:
The mass of human is 2898 times of the mass of cat.
Explanation:
A study finds that the metabolic rate of mammals is proportional to m^3/4 i.e.
[tex]M=\dfrac{km^3}{4}[/tex]
Where
k is constant
If m = 70 kg, the mass of human
[tex]M=\dfrac{70^3}{4}\\\\=85750[/tex]
If m = 4.91 kg, the mass of cat
[tex]M'=\dfrac{4.91^3}{4}\\\\=29.59[/tex]
So,
[tex]\dfrac{M}{M'}=\dfrac{85750}{29.59}\\\\=2897.93\approx 2898[/tex]
So, the mass of human is 2898 times of the mass of cat.
the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is
Answer:
The work done on the block by the spring as it accelerates the block is 4kx².
Explanation:
Let initial distance is x.
It was compressed three times farther and then the block is released, new distance is 3x.
The work done in compressing the spring is given by :
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]
So, the work done on the block by the spring as it accelerates the block is 4kx².
g A computer is reading data from a rotating CD-ROM. At a point that is 0.0189 m from the center of the disk, the centripetal acceleration is 241 m/s2. What is the centripetal acceleration at a point that is 0.0897 m from the center of the disc?
Answer:
the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²
Explanation:
Given the data in the question;
centripetal acceleration a[tex]_c[/tex]₁ = 241 m/s²
radius r₁ = 0.0189 m
radius r₂ = 0.0897 m
centripetal acceleration a[tex]_c[/tex]₂ = ? m/s²
since the rotational period will be the same for the two disk,
we use the centripetal acceleration formula a[tex]_c[/tex] = (4π²r/T²) to find the rotational period for the first disk.
a[tex]_c[/tex]₁ = (4π²r₁/T²)
make T² subject of formula
T² = 4π²r₁ / a[tex]_c[/tex]₁
we substitute
T² = ( 4 × π² × 0.0189 ) / 241
T² = 0.00309602528 s²
Now we use the same formula to find a[tex]_c[/tex]₂
a[tex]_c[/tex]₂ = ( 4π²r₂ / T² )
we substitute
a[tex]_c[/tex]₂ = ( 4 × π² × 0.0897 ) / 0.00309602528
a[tex]_c[/tex]₂ = 1143.8 m/s²
Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²
Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.
Answer:
Explanation:
First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.
Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.
For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.
The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.
First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.
A train moving with a uniform speed covers a distance of 120 m in 2 s. Calculate
(i) The speed of the train
(ii) The time it will taketo cover 240 m.
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 120 = (u \times 2) + \frac{1}{2} \times 0 \times {2}^{2} \\ 120 = 2u \\ { \tt{speed = 60 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 240 = (60t) \\ { \tt{time = 4 \: seconds}}[/tex]
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 26.5 m/s2 with a beam of length 5.89 m , what rotation frequency is required
Answer:
The angular acceleration is 4.5 rad/s^2.
Explanation:
Acceleration, a = 26.5 m/s2
length, L = 5.89 m
The angular acceleration is
[tex]\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2[/tex]
g four small masses 0.2 kg each are connected by light rods 0.4m long to form a square.what is the moment of interia axis
Complete Question
Four small masses of 0.2 kg each are connected by light rods 0.4m long to form a square. What is the moment of inertia of this object for an axis through the middle of the square and parallel to two sides.
Answer:
[tex]I=0.032kgm^2[/tex]
Explanation:
From the question we are told that:
Mass[tex]m=0.2kg[/tex]
Length [tex]l=0.4m[/tex]
Generally the equation for Inertia is mathematically given by
[tex]I=md^2[/tex]
[tex]I=0.8*0.20(\frac{0.40}{2})^2[/tex]
[tex]I=0.032kgm^2[/tex]
A charged particle is injected into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field lines. Ignoring the particle's weight, the particle will
Answer:
The charged particle will follow a circular path.
Explanation:
Formula for the magnetic force is;
F = qvb sin θ
Where;
where;
q = the charge
v = the velocity
B = the magnetic field
θ = the angle between the velocity and magnetic field
We are told that velocity vector is perpendicular to the magnetic field lines. Thus, angle is 90.
So sin θ = sin 90 = 1
Thus,
F = qvB
Now, since the velocity vector is perpendicular to the magnetic field line,it also means from flemmings right hand rule, that the magnetic force is as well perpendicular to both of them.
Therefore, we have:
- a force that is always perpendicular to the velocity and as well constant in magnitude since magnitude of velocity or magnetic field does not change.
What this statement implies is that the force is acting as a centripetal force, and therefore, the charged particle will be kept in a uniform circular motion.
A ball has a mass of 4.65 kg and approximates a ping pong ball of mass 0.060 kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00 m / s, determine the final velocities of both masses after the collision. Use equations 9.21 and 9.22 from the textbook. The book is on WebAssign.
Answer:
the final velocity of the ball is 4.87 m/s
the final velocity of the ping ball is 9.87 m/s
Explanation:
Given;
mass of the ball, m₁ = 4.65 kg
mass of the ping ball, m₂ = 0.06 kg
initial velocity of the ping ball, u₂ = 0
initial velocity of the ball, u₁ = 5 m/s
let the final velocity of the ball = v₁
let the final velocity of the ping ball, = v₂
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
4.65(5) + 0.06(0) = 4.65v₁ + 0.06v₂
23.25 + 0 = 4.65v₁ + 0.06v₂
23.25 = 4.65v₁ + 0.06v₂ ------ (1)
Apply one-directional velocity equation;
u₁ + v₁ = u₂ + v₂
5 + v₁ = 0 + v₂
5 + v₁ = v₂
v₁ = v₂ - 5 -------- (2)
substitute equation (2) into (1)
23.25 = 4.65(v₂ - 5) + 0.06v₂
23.25 = 4.65v₂ - 23.25 + 0.06v₂
46.5 = 4.71 v₂
v₂ = 46.5/4.71
v₂ = 9.87 m/s
v₁ = v₂ - 5
v₁ = 9.87 - 5
v₁ = 4.87 m/s
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of
ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.
magnitude m/s
direction ° counterclockwise from the +x-axis
Answer:
a) v = 517.99 m / s, b) θ = 296.3º
Explanation:
This is an exercise in kinematics, we are going to solve each axis independently
X axis
the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ= 3670 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
v₀ₓ = 3670 - 5.10 670
v₀ₓ = 253 m / s
Y axis
the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after
t = 670 s
v_y = v_{oy} + a_y t
v_{oy} = v_y - a_y t
v_oy} = 4378 - 7.30 670
v_{oy} = -513 m / s
to find the velocity modulus we use the Pythagorean theorem
v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]
v = [tex]\sqrt{253^2 +513^2}[/tex]
v = 517.99 m / s
to find the direction we use trigonometry
tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]
θ'= tan⁻¹ [tex]\frac{voy}{voy}[/tex]
θ'= tan⁻¹ (-513/253)
tea '= -63.7
the negative sign indicates that it is below the ax axis, in the fourth quadrant
to give this angle from the positive side of the axis ax
θ = 360 - θ
θ = 360 - 63.7
θ = 296.3º
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car B is 100 from car A, car A begins to accelerate toward car B with a constant acceleration of 5 m/s/s. Let right be positive.
1) How much time elapses before the two cars meet? 2) How far does car A travel before the two cars meet? 3) What is the velocity of car B when the two cars meet?
4) What is the velocity of car A when the two cars meet?
Answer:
Let's define t = 0s (the initial time) as the moment when Car A starts moving.
Let's find the movement equations of each car.
A:
We know that Car A accelerations with a constant acceleration of 5m/s^2
Then the acceleration equation is:
[tex]A_a(t) = 5m/s^2[/tex]
To get the velocity, we integrate over time:
[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]
Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:
[tex]V_a(t) = (5m/s^2)*t[/tex]
To get the position equation we integrate again over time:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
[tex]V_b(t) =20m/s[/tex]
To get the position equation, we can integrate:
[tex]P_b(t) = (20m/s)*t + P_0[/tex]
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
[tex]P_b(t) = (20m/s)*t + 100m[/tex]
Now we can answer this:
1) The two cars will meet when their position equations are equal, so we must have:
[tex]P_a(t) = P_b(t)[/tex]
We can solve this for t.
[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]
We only care for the positive solution, which is:
[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]
Car A reaches Car B after 11.48 seconds.
2) How far does car A travel before the two cars meet?
Here we only need to evaluate the position equation for Car A in t = 11.48s:
[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s
[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]
A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the instant you make measurements on the glider, it is moving at 0.835 m/sm/s and is 4.00 cmcm from its equilibrium point.
Required:
a. Use energy conservation to find the amplitude of the motion.
b. Use energy conservation to find the maximum speed of the glider.
c. What is the angular frequency of the oscillations?
(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is
W = - (1/2 kx ² - 1/2 k (0.0400 m)²)
(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)
By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so
W = ∆K = 0 - 1/2 m (0.835 m/s)²
Solve for x :
- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²
==> x ≈ 0.0493 m ≈ 4.93 cm
(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is
W = - 1/2 k (0.0400 m)²
If v is the glider's maximum speed, then by the work-energy theorem,
W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²
Solve for v :
- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²
==> v ≈ 1.43 m/s
(c) The angular frequency of the glider's oscillation is
√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz
The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s
From the given information;
the mass of the glider = 190 gForce constant k = 160 N/mthe horizontal speed of the glider [tex]v_x[/tex] = 0.835 m/sthe distance away from the equilibrium = 4.0 cm = 0.04 mUsing energy conservation E, the amplitude of the motion can be calculated by using the formula:
[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]
[tex]\mathbf{E =0.194 \ J}[/tex]
Similarly, we know that:
[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]
Making amplitude A the subject, we have:
[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]
[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]
[tex]\mathbf{A =0.049 \ cm}[/tex]
Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:
[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]
[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]
The angular frequency of the oscillation can be computed by using the expression:
[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]
[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]
ω = 29.02 rad/s
Learn more about energy conservation here:
https://brainly.com/question/13010190?referrer=searchResults
what does it mean to do science
Answer:
Doing science could be defined as carrying out scientific processes, like the scientific method, to add to science's body of knowledge.
7. The gravitational potential energy of a body depends on its A speed and position B. mass and volume. C. weight and position D.speed and mass
Answer:
Option "D" is the correct answer to the following question.
Explanation:
The gravitational potential energy of an item is determined by its mass, elevation, and gravitational acceleration. As a result, angular momentum and energy are preserved. The gravitational potential energy, on the other hand, varies with distance. When a consequence, kinetic energy varies during each orbit, resulting in a faster speed as a planet approaches the Sun.
Answer:
SPEED AND MASS
Explanation:
TOOK THE TEST
A car of mass 500 kg increases its velocity from 40 metre per second to 60 metre per second in 10 second find the distance travelled and amount of force applied
Answer:
it is answer of u are question
(a) If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve on wet concrete where the coefficient of kinetic friction is 0.5 and the coefficient of static friction is 0.7.
(b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate where the coefficient of kinetic friction is 0.3 and the coefficient of static friction is 0.55?
(c) If the truck has four-wheel drive, and the cabinet is wooden, what is it's maximum acceleration (in m/s2)?
Answer:
a) a = 27.44 m / s², b) a = 5.39 m / s², c) a = 156.8 m / s², cabinet maximum acceleration does not change
Explanation:
a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.
Let's apply Newton's second law
we set a regency hiss where the x axis is in the direction of movement of the truck
Y axis y
N- W = 0
N = W = m g
X axis
2fr = m a
the expression for the friction force is
fr = μ N
fr = μ m g
we substitute
2 μ m g = m /2 a
a = 4 μ g
a = 4 0.7 9.8
a = 27.44 m / s²
b) let's look for the maximum acceleration that can be applied to the cabinet
fr = m a
μ N = ma
μ m g = m a
a = μ g
a = 0.55 9.8
a = 5.39 m / s²
as the acceleration of the platform is greater than this acceleration the cabinet must slip
c) the friction force is in the four wheels as well
With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all
applying Newton's second law
4 fr = (m/4) a
16 mg = (m) a
a = 16 g
a = 16 9.8
a = 156.8 m / s²
cabinet maximum acceleration does not change
Suppose that a ball decelerates from 8.0 m/s to a stop as it rolls up a hill, losing 10% of its kinetic energy to friction. Determine how far vertically up the hill the ball reaches when it stops. Show your work.(2 points)
Answer:
The maximum height is 0.33 m.
Explanation:
initial velocity, u = 8 m/s
final velocity, v = 0 m/s
10% of kinetic energy is lost in friction.
The kinetic energy used to move up the top,
KE = 10 % of 0.5 mv^2
KE = 0.1 x 0.5 x m x 8 x 8 = 3.2 m
Let the maximum height is h.
Use conservation of energy
KE at the bottom = PE at the top
3.2 m = m x 9.8 x h
h = 0.33 m
The height traveled vertically up the hill by the ball when it stops is 0.327 meter.
Given the following data:
Velocity = 8.0 m/sKinetic energy = 10% lost to friction.Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To determine how far (height) vertically up the hill the ball reaches when it stops:
By applying the law of conservation of energy, we have:
Kinetic energy lost at the bottom = Potential energy gained at the top.
Mathematically, the above expression is given by the formula:
[tex]0.1 \times \frac{1}{2} mv^2 = mgh\\\\0.1 \times \frac{1}{2} v^2 = gh\\\\h=\frac{0.1v^2}{2g}[/tex]
Substituting the given parameters into the formula, we have;
[tex]h=\frac{0.1 \times 8^2}{2\times 9.8} \\\\h=\frac{0.1 \times 64}{19.6} \\\\h=\frac{6.4}{19.6}[/tex]
Height, h = 0.327 meter.
Read more on kinetic energy here: https://brainly.com/question/17081653
When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)
P = F v
where P is power, F is the magnitude of force, and v is speed. So
50.0 hp = 37,280 W = F (24.6 m/s)
==> F = (37,280 W) / (24.6 m/s) ≈ 1520 N
explain why sound wave travel faster in liquid than gas
Answer:
Because gas contains free molecules but not liquid.
Please mark as brainliast
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.
Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?
Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the spring to a distance of 0.200 mm , how far up the slope will an identical ice cube travel before reversing directions
Answer:
The correct answer will bs "2.41 m".
Explanation:
According to the question,
M = 50 g
or,
= 0.050 kg
[tex]\Theta = 25^{\circ}[/tex]
k = 25.9 N/m
Δx = 0.200 m
Let the traveled distance be "x".
By using trigonometry, the height will be:
⇒ [tex]h = l Sin \Theta[/tex]
hence,
⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]
[tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]
By putting the values, we get
[tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]
[tex]l=2.41 \ m[/tex]
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
You are on an airplane that is landing. The plane in front of your plane blows a tire. The pilot of your plane is advised to abort the landing, so he pulls up, moving in a semicircular upward-bending path. The path has a radius of 450 m with a radial acceleration of 17 m/s^2.
Required:
What is the plane's speed?
Answer:
v = 87.46 m/s
Explanation:
The radial acceleration is the centripetal acceleration, whose formula is given as:
[tex]a_c = \frac{v^2}{r}[/tex]
where,
[tex]a_c[/tex] = centripetal acceleration = 17 m/s²
v = planes's speed = ?
r = radius of path = 450 m
Therefore,
[tex]17\ m/s^2 = \frac{v^2}{450\ m}\\\\v^2 = (17\ m/s^2)(450\ m)\\\\v = \sqrt{7650\ m^2/s^2}[/tex]
v = 87.46 m/s
A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before coming to rest.
Explanation:
v = 45m/s
t = 5s
Therefore,
d = v*t
= 45*5
= 225m
Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.
Answer:
Atmospheric pressure at Badwater is 1.01022 atm
Explanation:
Data given:
1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa
Elevation (h) = 86m
gravity (g) = 9.8 m/s2
Density of air P = 1.225 kg/m3
Therefore pressure at bad water Pb = Pi + Pgh
Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)
Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa
hence:
Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm
A 1,200kg roller coaster car starts rolling up a slope at a speed of 15m/s. What is the highest point it could reach
Answer: 11.36 m
Explanation:
Given
Mass of roller coaster is m=1200 kg
Initial speed of roller coaster is v=15 m/s
Energy at bottom and at the top is same i.e.
[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]
Thus, the highest point reach by the roller coaster is 11.36 m
Answer:
11.36m
Explanation:
An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke's Law) located at the end of the track. The spring constant is 52.1 N/m, the mass of the object 0.250 kg and the speed of the object is 1.70 m/s immediately before the collision.
(a) Determine the spring's maximum compression if the track is frictionless.
?? m
(b) If the track is not frictionless and has a coefficient of kinetic friction of 0.120, determine the spring's maximum compression.
??m
(a) As it gets compressed by a distance x, the spring does
W = - 1/2 (52.1 N/m) x ²
of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so
W = ∆K
- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.118 m
(b) Taking friction into account, the only difference is that more work is done on the object.
By Newton's second law, the net vertical force on the object is
∑ F = n - mg = 0
where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives
n = mg = 2.45 N
and from this we get the magnitude of kinetic friction,
f = µn = 0.120 (2.45 N) = 0.294 N
Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:
W (friction) = - (0.294 N) x
W (spring) = - 1/2 (52.1 N/m) x ²
==> W (total) = W (friction) + W (spring)
Solve for x :
- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.112 m
For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:
a) The spring's maximum compression when the track is frictionless is 0.118 m.
b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.
a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex] (1)
Where:
[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg
[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s
k: is the spring constant = 52.1 N/m
x: is the distance of compression
After solving equation (1) for x, we have:
[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]
Hence, the spring's maximum compression is 0.118 m.
b) When the track is not frictionless, we can calculate the spring's compression by work definition:
[tex] W = \Delta E = E_{f} - E_{i} [/tex]
[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex] (2)
Work is also equal to:
[tex] W = F*d = F*x [/tex] (3)
Where:
F: is the force
d: is the displacement = x (distance of spring's compression)
The force acting on the object is given by the friction force:
[tex] F = -\mu N = -\mu m_{o}g [/tex] (4)
Where:
N: is the normal force = m₀g
μ: is the coefficient of kinetic friction = 0.120
g: is the acceleration due to gravity = 9.81 m/s²
The minus sign is because the friction force is in the opposite direction of motion.
After entering equations (3) and (4) into (2), we have:
[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]
[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]
[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]
Solving the above quadratic equation for x
[tex] x = 0.112 m [/tex]
Therefore, the spring's compression is 0.112 m when the track is not frictionless.
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