Answer:
Ludwig Boltzmann was an Austrian Physicist and he performed a simple but powerful experiment to gather evidence concerning the velocity distribution of a sample of gas particles.
His experiment revealed that the way Gases are distributed in its normal form across a range of temperature is dependent on the molar mass amen temperature of the gas. Gases with high temperature move faster due to the high number of colliding particles when compared to those with low temperature. Gases with lower molar mass move faster than those with higher molar mass.
g You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 78 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
F=49.48 N
Explanation:
Given that
Diameter , d= 30 mm
Holding pressure = 70 % P
P=Atmospherics pressure
We know that
P= 1 atm = 10⁵ N/m²
The force per unit area is known as pressure.
[tex]P=\dfrac{F}{A}[/tex]
[tex]F=P\times A[/tex]
[tex]F=0.7\times 10^5\times \dfrac{\pi}{4}\times 0.03^2\ N[/tex]
Therefore the force will be 49.48 N.
F=49.48 N
You then decide to throw another disc to your friend in the tree for fun. If you throw the disc at a speed of 10 m/s and your friend catches it 5.0 m above you, what was the speed of the disc just before your friend catches it
Answer:
Vf = 1.41 m/s
Explanation:
The velocity of the disc just before the friend catches it, can be found out by using 3rd equation of motion, as follows:
2gh = Vf² + Vi²
where,
g = - 9.8 m/s² (negative sign for upward motion)
h = height = 5 m
Vi = Initial Velocity = 10 m/s
Vf = Final Velocity = ?
Therefore,
2(-9.8 m/s²)(5 m) = Vf² - (10 m/s)²
Vf² = 100 m²/s² - 98 m²/s²
Vf = √(2 m²/s²)
Vf = 1.41 m/s
A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:_________
a) 10 m/s; 9.8 m/s2.
b) 9.0 m/s; 9.0 m/s2.
c) 9.0 m/s; 9.8 m/s2.
d) 1.0 m/s; 9.8 m/s2.
e) 1.0 m/s; 9.0 m/s2.
Answer:
a) 10 m/s; 9.8 m/s²
Explanation:
After leaving the bench the boy undergoes a semi-projectile motion from top to bottom. Neglecting the effects of air friction, we can safely assume that the horizontal velocity remains constant throughout the motion. Thus, the magnitude of velocity is 10 m/s. The only acceleration in the boy is the acceleration due to gravity, due to free fall motion. Therefore, the magnitude of acceleration is 9.8 m/s².
Therefore, One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:
a) 10 m/s; 9.8 m/s²
A spool is on a horizontal surface (with friction) and pulled to the right with a thread attached to the center of the spool, so that the spool rolls without slipping. The spool has a mass of M, moment of inertia I, and a radius of R.
Requird:
a. What is the relationship between the magnitude of the force of tension and the force of friction?
b. What expressions relates the magnitudes of the translational acceleration of the spool's center of mass a and the spool's rotational acceleration about the center of mass a (recall the spool rolls without slipping on the surface).
Answer:
The correct answer is -
A) [tex]F_{f} =(\frac{r}{R} ) F_{T}[/tex]
B) [tex]\alpha =\frac{a}{R}[/tex]
Explanation:
As it is mention that the spool has mass M, radius R and moment of inertia I. In the first part of the question as the spool is not moving and r < R which means there is net torque = 0
so, [tex]F_{f}{R} = F_{T}{r}[/tex]
and [tex]F_{f} =(\frac{r}{R} ) F_{T}[/tex]
In the second part of the question from the given information, we can express the angular acceleration
= [tex]\alpha =\frac{a}{R}[/tex] ( alpha = angular acceleration and a = translational acceleration)
A) The relationship between the magnitude of the force of tension and the force of friction Ff = (r/R)Ft
B) α = a/R
What is Rotational Acceleration?When As it is mentioned that the spool has mass M, Then radius R, and also a moment of inertia I. Then In the foremost part of the query as the spool is not moving and r < R which means there is net torque = 0
so, Ff R = Ftr
and then Ff = (r/R)Ft
In the second part of the query from the given notification, we can express the angular acceleration
Therefore, = α = a/R ( alpha = angular acceleration and a = translational acceleration)
Find more information about Rotational Acceleration here:
https://brainly.com/question/14001220
A 110.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 8900.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.
Required:
a. What is the velocity of the truck right after the collision?
b. What is the change in mechanical energy of the car?
Answer:
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of the bodies before collision is equal to the sum of momentum of bodies after collision.
Momentum = Mass*velocity
BEFORE COLLISION
The momentum of a 110.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction = 110*25 = 2750kgm/s
The momentum of a 8900.0 kg truck with a speed of 20.000 m/s in an easterly direction = 8900*20 = 178000kgm/s
Sum of momentum before collision = 2750 + 178000 = 180,750 kgm/s
AFTER COLLISION
The momentum of the car will be 110*18 = 1980kgm/s
The momentum of the truck = 8900v where v is the velocity of the truck after collision.
Sum of momentum after collision = 1980 + 8900v
Applying the conservation law;
180750 = 1980 + 8900v
8900v = 180750-1980
8900v = 178770
v = 178770/8900
v = 20.09m/s
Velocity of the truck after collision is 20.09m/s
Note that the collision is inelastic i.e the body moves with different velocities after collision
b) The mechanical energy experienced by the bodies is kinetic energy.
Kinetic energy = 1/2mv²
Sum of the Kinetic energy before collision = 1/2(110)*25²+1/2(8900)*20²
= 34375 + 1780000
= 1,814,375Joules
Sum of kinetic energy after collision = 1/2*(110)*18²+1/2(8900)*20.09²
= 17820+1796056.045
= 1,813,876.045Joules
Change in mechanical energy = 1,813,876.045Joules - 1,814,375Joules
= -498.955Joules
Two identical metallic spheres, having unequal opposite charges are placed at a distance
of 5 metres apart in air. After bringing them in contact with each other, they are again
placed at the same distance apart. Now the force of repulsion between them is 20 N.
Calculate the final charge on each of them.
Answer:
Charge on the sphere is [tex]2.48*10^{-9} C[/tex]
Explanation:
distance apart r = 5 m
force of repulsion F = 20 N
The spheres had opposite unequal opposite charges, this means that on bringing them into contact, the sphere with the greater charge will have its charge cancel out the other charge on the other sphere. The resultant charge will then be evenly distributed between the two spheres. The result is that both spheres will now have like, equal amount of charge on them.
applying Coulumb's law,
[tex]F = \frac{kQ^{2} }{r^{2} }[/tex]
where
[tex]k = 9*10^{9} m^{3} s^{-4} A^{-2}[/tex]
substituting values into the equation, we have
[tex]20 = \frac{9*10^{9}*Q^{2} }{5^{2} }[/tex]
[tex]Q^{2} = \frac{20*25}{9*10^{9} }[/tex] = [tex]6.17*10^{-18}[/tex]
[tex]Q = \sqrt{6.17*10^{-18} }[/tex] = [tex]2.48*10^{-9} C[/tex]
A certain sea cow can paddle 2.0 m/s in still water. If she attempts to cross a river, from the south bank to the north with a current of 3.0 m/s flowing toward the east by paddling entirely perpendicularly to the flow of the river, in what direction will she be traveling relative to an observer on shore
Answer:
v = 3.6m / s , θ = 56º
Explanation:
This is a relative speed exercise, let's use the Pythagorean theorem
v = √ (v₁² + v₂²)
where v₁ is the speed of the sea still water and v₂ the speed of the current
let's calculate
v = √ (2² + 3²)
v = 3.6m / s
to find the direction we use trigonometry
tan θ = v₂ / v₁
θ = tan⁻¹ (v₂ / v₁)
let's calculate
θ = tan⁻¹ (3/2)
θ = 56º
what are the strengths and weaknesses of the four methods of waste management?
Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;
* It creates employment
* It keeps the environment clean
* The practice is highly lucrative
* It saves the earth and conserves energy
The weaknesses of the methods of waste management includes;
* The sites are often dangerous
* The process is mostly
* There is a need for global buy-in
* The resultant product had a short life
Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why?
Answer:
The ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil.
Explanation:
Oil is more less dense than water. Thus, the molecules that make up the oil are larger than those that that make up water, so they cannot pack as tightly together as the water molecules will do. Hence, they will take up more space per unit area and are we can say they are less dense.
So, we can conclude that the ball filled with water will reach the bottom of the container first this is because oil is less dense than water and so the glass ball filled with oil will be a lot less denser than the one which is filled with water.
The displacement (in meters) of a particle moving in a straight line is given by the equation of motion:
s = 4/t^2, where t is measured in seconds.
Required:
Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.
Answer:
At [tex]t = 1\; \rm s[/tex], the particle should have a velocity of [tex]-8\; \rm m \cdot s^{-1}[/tex].At [tex]t = 2\; \rm s[/tex], the particle should have a velocity of [tex]-1\; \rm m \cdot s^{-1}[/tex].At [tex]t = 3\; \rm s[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{27}\; \rm m \cdot s^{-1}[/tex].For [tex]a > 0[/tex], at [tex]t = a \; \text{second}[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{a^3}\; \rm m \cdot s^{-1}[/tex].
Explanation:
Differentiate the displacement of an object (with respect to time) to find the object's velocity.
Note that the in this question, the expression for displacement is undefined (and not differentiable) when [tex]t[/tex] is equal to zero. For [tex]t > 0[/tex]:
[tex]\begin{aligned}v &= \frac{\rm d}{{\rm d}t}\, [s] = \frac{\rm d}{{\rm d}t}\, \left[\frac{4}{t^2}\right] \\ &= \frac{\rm d}{{\rm d}t}\, \left[4\, t^{-2}\right] = 4\, \left((-2)\, t^{-3}\right) = -8\, t^{-3} =-\frac{8}{t^3}\end{aligned}[/tex].
This expression can then be evaluated at [tex]t = 1[/tex], [tex]t = 2[/tex], and [tex]t = 3[/tex] to obtain the required results.
If an individual is right-handed, under most circumstances it would be safe to assume that their right arm is stronger and more fatigue resistant, since it is probably used to carry out a majority of daily tasks.
a. True
b. False
Answer:
True
Explanation:
A right-handed individual prefers to perform his day to day task with his right hand. The coordination in the right hand is more developed than in the left hand of right-handed individuals, and allows a quicker and more accurate manipulation of objects. Muscles in the body are strengthened from constant use, and evolve to be more resistant to fatigue if they are used more frequently. This fact implies that right-handed individuals should have their right hand stronger, and more fatigue resistant when compared to their non-dominant other hand.
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107 N/m2). By what volume has 1.0 m3 of water from the surface of the lake been compressed if it is forced down to this depth? The bulk modulus of water is 2.3 × 109 Pa.
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
[tex]K = - V\cdot \frac{dP}{dV}[/tex]
Where:
[tex]K[/tex] - Bulk module, measured in pascals.
[tex]V[/tex] - Sample volume, measured in cubic meters.
[tex]P[/tex] - Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
[tex]-\frac{K \,dV}{V} = dP[/tex]
This resultant expression is solved by definite integration and algebraic handling:
[tex]-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP[/tex]
[tex]-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}[/tex]
[tex]\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}[/tex]
[tex]\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }[/tex]
The final volume is predicted by:
[tex]V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }[/tex]
If [tex]V_{o} = 1\,m^{3}[/tex], [tex]P_{o} - P_{f} = -10132500\,Pa[/tex] and [tex]K = 2.3\times 10^{9}\,Pa[/tex], then:
[tex]V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }[/tex]
[tex]V_{f} \approx 0.996\,m^{3}[/tex]
Change in volume due to increasure on pressure is:
[tex]\Delta V = V_{o} - V_{f}[/tex]
[tex]\Delta V = 1\,m^{3} - 0.996\,m^{3}[/tex]
[tex]\Delta V = 0.004\,m^{3}[/tex]
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Identify the following as combination, decomposition, replacement, or ion exchange reactions NaBr(aq) + Cl2(g) → 2 NaCl(aq) + Br2(g)
Answer:
Replacement
Explanation:
in replacements, like ions replace like. in this equation, we can see that Bromine replaced Chlorine. so, the answer is replacement.
Answer:
Single-replacement or replacement
Explanation:
The single-replacement reaction is a + bc -> ac + b, compare them.
NaBr + Cl2 -> 2 NACl + Br.
AB + C -> AC + B
As you can see they are the same ( even though the b is with the a and not with the c. The formula can be switched around a little with the order of b and c ) ((also like ions replace like ions in replacements, which they are in this))
A region of space contains a uniform electric field oriented along the y-axis. A frame of surface area A is placed perpendicular to the y-axis in the xz-plane. The magnitude of the electric flux through this frame is Φ0. A second frame is placed in the same electric field in such a way that the magnitude of the electric flux through it is Φ0/2. How is the plane of second frame oriented with respect to the plane of the first one?
Answer:
β = 30º
Explanation:
By definition, the vector flux across a surface, can be found integrating the dot product of the vector field (electric field E in this case) and the differential surface dA, across the entire surface.If the surface is placed perpendicular to the electric field, and this field is uniform, the total flux across the surface can be expressed as follows:Φ0 = E*A*cos 0º = E*A.
If the magnitude of the electric flux, is reduced to half of its original value, we can write the following equality:Φ0/2 = E*A*cos θ⇒ = Φ0 * cos θ (1)
where θ, is the angle between the electric field E and the vector
perpendicular to the plane traversed by E.
Rearranging terms in (1) we can solve for θ, as follows:
⇒ cos θ = 1/2 ⇒ θ = arc cos (1/2) = 60º
As this is the angle between the electric field, and the surface vector,
which is by definition, perpendicular to the plane, the angle between
the electric field and the plane can be found as follows:
β = 90º - θ = 90º - 60º = 30º
n a location where the train tracks run parallel to a road, a high-speed train traveling at 60 m/s passes a car traveling at 30 m/s in the opposite direction. How long does it take for the train to be 180 m away from the car? Group of answer choices
Answer:
Time taken for the train to travel 180 m away from the train is 2 sec.
Explanation:
velocity of the train = 60 m/s
velocity of the car = 30 m/s
Relative to the other, each vehicle passes the other with a velocity of
V = 60 m/s + 30 m/s = 90 m/s
If we take the car's speed as the reference speed, the speed of the train will be 90 m/s. Therefore, time taken for the train to travel 180 m away from the car is,
time = distance/speed = 180/90
time = 2 sec
Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.
Answer:
Q1_new = 515.68 µC
Q2_new = 246.82 µC
Explanation:
Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.
Then it is only necessary to calculate the charge on each capacitor:
Q1 = 5.85 µF * 250 V = 1462.5 µC
Q2 = 2.8 µF * 250 V = 700 µC
Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.
When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:
1462.5 µC - 700 µC = 762.5 µC
Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.
This 762.5 µC will be divided proportionately between the two capacitors.
Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC
Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC
A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s after this, the ball reaches the bottom of the incline. What was the magnitude of the ball's acceleration, assume it to be constant
Answer: The ball's acceleration is 2.35 m/s2
Explanation: Please see the attachment below
Answer:
The acceleration is [tex]a= 2.4 \ m/s^2[/tex]
Explanation:
From the question we are told that
The distance covered is [tex]d = 87 \ m[/tex]
The time taken is [tex]t = 8.6 \ s[/tex]
Time taken reach the bottom is [tex]t_b = 1 \ s[/tex]
According to the equation of motion
[tex]S = ut + \frac{1}{2} at^2[/tex]
since the ball started at rest u = 0 m/s
substituting values
[tex]87 = 0 + \frac{1}{2} * a * (8.6)^2[/tex]
=> [tex]a = \frac{2 * 87}{8.6^2}[/tex]
=> [tex]a= 2.4 \ m/s^2[/tex]
A body moving with uniform acceleration has two points (5,15) and (20,60) on the (v t) graphs of its motion. calculate the acceleration
Answer:
3m/s²
Explanation:
The slope on a velocity time graph represents the acceleration, so if you simply use the slope formula, you can find the acceleration between those two points.
m=rise/run
m=(60-15)/(20-5)
m=45/15
m=3 m/s ² squared (therefore this is your constant acceleration from those two points).
A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?
Answer:
(a) Fw = 101.01 N
(b) W = 282.82 J
(c) Fg = 382.2 N
(d) N = 368.61 N
(e) Net force = 0 N
Explanation:
(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:
[tex]F_g-F_w=0[/tex] (1)
Fg: gravitational force over the object
Fw: worker's force
However, in an incline you have that the gravitational force on the object, due to its weight, is given by:
[tex]F_g=Wsin\theta=Mg sin\theta[/tex] (2)
M: mass of the ice block = 39 kg
g: gravitational constant = 9.8m/s^2
θ: angle of the incline
You calculate the angle by using the information about the distance of the incline and its height, as follow:
[tex]sin\theta=\frac{0.74m}{2.8m}=0.264\\\\\theta=sin^{-1}(0.264)=15.32\°[/tex]
Finally, you solve the equation (1) for Fw and replace the values of all parameters:
[tex]F_w=F_g=Mgsin\theta\\\\F_w=(39kg)(9.8m/s^2)sin(15.32\°)=101.01N[/tex]
The worker's force is 101.01N
(b) The work done by the worker is given by:
[tex]W=F_wd=(101.01N)(2.8m)=282.82J[/tex]
(c) The gravitational force on the block is, without taking into account the rotated system for the incline, only the weight of the ice block:
[tex]F_g=Mg=(39kg)(9.8m/s^2)=382.2N[/tex]
The gravitational force is 382.2N
(d) The normal force is:
[tex]N=Mgcos\theta=(39kg)(9.8m/s^2)cos(15.32\°)=368.61N[/tex]
(e) The speed of the block when it slides down the incle is constant, then, by the Newton second law you can conclude that the net force is zero.
A block of mass M is suspended from a ceiling by a spring with spring stiffness constant k. A penny of mass m is placed on top of the block. What is the maximum amplitude of oscillations that will allow the penny to just stay on top of the block
Answer:
Explanation:
Given that:
Mass of block M
Mass of penny m
spring stiffness constant k
The frequency of oscillation of the block [tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m} }[/tex]
The angular velocity is [tex]\omega =2\pi f[/tex]
[tex]=\sqrt{k/m}[/tex]
when the penny is resting on the block
The acceleration of the penny = acceleration of the block
If R is the reaction of the block on the penny
[tex]R-mg=a_{max}m\\\\=-\omega^2A_{max}m\\\\R=mg-\omega^2A_{max}m[/tex]
The penny will leave the block if R = 0
[tex]mg=\omega^2A_{max}m\\\\g=\omega^2A_{max}\\\\A_{max}=\frac{g}{\omega^2} \\\\=\frac{g}{(k/M)} \\\\A_{max}=gM/k[/tex]
Therefore the amplitude [tex]A_{max}<gM/k[/tex] for the penny to remain on the block
You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to be 2.40 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. What frequency is required?
Answer:
f = 1.69*10^5 Hz
Explanation:
In order to calculate the frequency of the sinusoidal voltage, you use the following formula:
[tex]V_L=\omega iL=2\pi f i L[/tex] (1)
V_L: voltage = 12.0V
i: current = 2.40mA = 2.40*10^-3 A
L: inductance = 4.70mH = 4.70*10^-3 H
f: frequency = ?
you solve the equation (1) for f and replace the values of the other parameters:
[tex]f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz[/tex]
The frequency of the sinusoidal voltage is f
A 4.00 kg ball is moving at 4.00 m/s to the EAST and a 6.00 kg ball is moving at 3.00 m/s to the NORTH. The total momentum of the system is:___________.A. 14.2 kg m/s at an angle of 48.4 degrees SOUTH of EAST.B. 48.2 kg m/s at an angle of 24.2 degrees SOUTH of EAST.C. 48.2 kg m/s at an angle of 48.4 degrees NORTH of EAST.D. 24.1 kg m/s at an angle of 24.2 degrees SOUTH of EAST.
E. 24.1 kg m/s at an angles of 48.4 degrees NORTH of EAST.
Answer:
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EASTExplanation:
Momentum = mass*velocity of a body
For a 4.00 kg ball is moving at 4.00 m/s to the EAST, its momentum = 4*4 = 16kgm/s
For a 6.00 kg ball is moving at 3.00 m/s to the NORTH;
its momentum = 6*3 = 18kgm/s
Total momentum = The resultant of both momentum
Total momentum = √16²+18²
Total momentum = √580
total momentum = 24.1kgm/s
For the direction:
[tex]\theta = tan^{-1} \frac{y}{x}\\\theta = tan^{-1} \frac{18}{16}\\ \theta = tan^{-1} 1.125\\\theta = 48.4^{0}[/tex]
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST
A 50-cm-long spring is suspended from the ceiling. A 410 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 16 cm before coming to rest at its lowest point. It then continues to oscillate vertically. Part A What is the spring constant
Answer:
25.125 N/m
Explanation:
extension on the spring e = 16 cm 0.16 m
mass of hung mass m = 410 g = 0.41 kg
equation for the relationship between force and extension is given by
F = ke
where k is the spring constant
F = force = mg
where m is the hung mass,
and g is acceleration due to gravity = 9.81 m/s^2
imputing value, we have
0.41 x 9.81 = k x 0.16 = 0.16k
4.02 = 0.16k
spring constant k = 4.02/0.16 = 25.125 N/m
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and position. A motorcyclist travels around a curved path that has a radius of 250 ft . While traveling around the curved path, the motorcyclist increases speed by 1.05 ft/s2 . Part A - Finding the time interval for the motorcyclist to reach a given acceleration If the motorcyclist starts from rest, determine the time needed to reach an acceleration of 4.25ft/s2. Express your answer to three significant figures and include the appropriate units.
Answer:
a^2 = ar^2 + al^2 where ar is the radial acceleration and al is the
linear acceleration - since vectors ar and al are at right angles
ar^2 = a^2 - al^2 = 4.25^2 - 1.05^2
ar = 4.12 ft/s^2
ar = V^2 / R where ar is the radial acceleration
So V^2 = ar * R = 4.12 * 250 = 1030 ft^2/s^2
V = 32.1 m/s the linear speed of the cycle
Also, V = al t or t = V / al = 32.1 / 1.05 = 30.6 sec
An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron's probability of tunneling through the barrier increase
Answer:
factor that the electron's probability of tunneling through the barrier increase 2.02029
Explanation:
given data
kinetic energy = 10.1 eV
height = 18.2 eV
width = 1.00 nm
wavelength = 546 nm
solution
we know that probability of tunneling is express as
probability of tunneling = [tex]e^{-2CL}[/tex] .................1
here C is = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
here h is Planck's constant
c = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-10.1) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c = 2319130863.06
and proton have hf = [tex]\frac{hc}{\lambda } = {1240}{546}[/tex] = 2.27 ev
so electron K.E = 10.1 + 2.27
KE = 12.37 eV
so decay coefficient inside barrier is
c' = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
c' = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-12.37) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c' = 1967510340
so
the factor of incerease in transmisson probability is
probability = [tex]e^{2L(c-c')}[/tex]
probability = [tex]e^{2\times 1\times 10^{-9} \times (351620523.06)}[/tex]
factor probability = 2.02029
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be 70.0 km directly north of his starting point.
a. What was the length of the third leg of his trip?b. What was the direction of the third leg of his trip?
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
A roller coaster car with a mass 700 Kg starts from rest at h1 above the ground and slides along a track. The car encounters a loop of radius 12 m. The bottom of the loop is a height h2 = 5 m from the ground. What would be the max height of release h1 for the roller coaster car if the amount of thermal energy produced between the point of release and the top of the loop should not exceed 15% of the initial mechanical energy and the normal force at the top should be no more than 580 N? answer in m
Given that,
Mass of roller coaster car = 700 kg
Radius = 12 m
Height from the ground h₂= 5 m
Normal force = 580 N
We need to calculate the speed of roller coaster at top
Using balance equation
[tex]N+mg=\dfrac{mv^2}{r}[/tex]
Put the value into the formula
[tex]580+700\times9.8=\dfrac{700v^2}{12}[/tex]
[tex]v^2=\dfrac{12(580+700\times9.8)}{700}[/tex]
[tex]v=\sqrt{\dfrac{12(580+700\times9.8)}{700}}[/tex]
[tex]v=11.29\ m/s[/tex]
We need to calculate the value of maximum height
Using conservation of energy
[tex]mgh_{1}=\dfrac{1}{2}mv^2+0.15(mgh_{1})+mgh_{2}[/tex]
[tex]mgh_{1}-0.15(mgh_{1})-mgh_{2}=\dfrac{1}{2}mv^2[/tex]
Put the value into the formula
[tex]9.8(h_{1}-0.15h_{1}-5)=\dfrac{1}{2}\times(11.29)^2[/tex]
[tex]0.85h_{1}-5=\dfrac{(11.29)^2}{9.8}[/tex]
[tex]h_{1}=\dfrac{13.00+5}{0.85}[/tex]
[tex]h_{1}=21.18\ m[/tex]
Hence, The max height of release h₁ for the roller coaster car is 21.18 m
If you slide down a rope, it's possible to create enough thermal energy to burn your hands or your legs where they grip the rope. Suppose a 30 kg child slides down a rope at a playground, descending 2.5 m at a constant speed.
How much thermal energy is created as she slides down the rope?
Answer:
Q = 735 J
Explanation:
In this exercise we must assume that all the mechanical energy of the system transforms into cemite energy.
Initial energy
Em₀ = U = m g h
final energy
[tex]Em_{f}[/tex] = Q
Em₀ = Em_{f}
m g h = Q
let's calculate
Q = 30 9.8 2.5
Q = 735 J
Which statement describes an essential characteristic of data in an experiment?
Answer:
the data must be reliable
Explanation:
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 8.1 rad/s. The drawing shows the position of the block when the spring is unstrained. This position is labeled x= 0 m. The drawing also shows a small bottle located 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 m, and is then thrown to the left. In order for the block to knock over the bottle,it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.
Answer:
v₀ = 0.5058 m/s
Explanation:
From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m
Now, the potential energy of the block at x = 0.08 m is ½kx²
where;
k is the spring constant given by; k = ω²m
ω is the angular velocity of the oscillation
m is the mass of the block.
Thus, potential energy of the spring at the bottle(x = 0.08 m) is;
U = ½ω²m(0.08m)²
Also, potential energy of the spring at the bottle(x = 0.05 m) is;
U = ½ω²m(0.05m)²
and the kinetic energy of the block at x = 0.05 m is;
K = ½mv₀²
Thus;
½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²
Inspecting this, ½m will cancel out to give;
ω²(0.08)² = ω²(0.05)² + v₀²
Making v₀ the subject, we have;
v₀ = ω√((0.08)² - (0.05)²)
So,
v₀ = 8.1√((0.08)² - (0.05)²)
v₀ = 0.5058 m/s