Answer:
r = 3.8 × 10 ⁻¹⁴ m
Explanation:
given data
alpha nucleus = 5.1 MeV
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C
Kinetic energy of the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) = 9.564 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
solution
we know that when its kinetic energy is equal to the potential energy than alpha particle will deflect \
so
Kinetic energy = potential energy = k q₁q₂ ÷ r ..................1
here r is close distance the alpha particle
so r will be put here value
r = ( 9 × 10⁹ × 3.2 × 10⁻¹⁹ × 1.264 × 10⁻¹⁷ ) ÷ ( 9.564 × 10⁻¹³ )
r = 3.8 × 10 ⁻¹⁴ m
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. What is the direction of the electric field vector?
Answer:
the electric field is pointing horizontal direction and in south direction
Explanation:
In an electromagnetic wave, the magnetic field and electrical field are perpendicular to each other and these are perpendicular to the direction of the waves.
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?
Answer:
[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]
Explanation:
Hello,
In this case, since pressure is defined as the force applied over a surface:
[tex]P=\frac{F}{A}[/tex]
We can associate the force with the weight of the needle computed by using the acceleration of the gravity:
[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]
And the area of the the tip (circle) in meters:
[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]
Thus, the pressure exerted on the record turns out:
[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]
Which is truly a large value due to the tiny area on which the pressure is exerted.
Best regards.
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N . Rolling friction can be neglected. You may want to review (Pages 165 - 168) . Part A What is the magnitude of the force of the car on the truck
Answer:
a) 3344 N
b) 3344 N
Explanation:
This is the complete question
1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected. A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. B. What is the magnitude of the force of the truck on the car?
Mass of the car = 1100 kg
Mass of the truck = 2200 kg
Force exerted on the ground by the car = 5000 N
The total mass in the system = 1100 + 2200 = 3300 Kg
Total force in the system = 5000 N
Recall that the force in the system = mass x acceleration
therefore,
5000 = 3300 x a
Total acceleration in the system = 5000/3300 = 1.52 m/s^2
The force on the truck individually fro the car, will be the product of this acceleration and its mass
Force on the truck = 2200 x 1.52 = 3344 N
b) Force on the car From the truck will be equal to this force but will act in the opposite direction.
Force on the car from the truck is 3344 N
Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.2m behind the slits. Eleven bright fringes are seen, spanning a distance of 60mm .
What is the spacing (in mm) between the slits?
Answer:
0.3376 mm
Explanation:
The computation of the spacing in mm between the slits is shown below:
As we know that
[tex]d = \frac{m\lambda L}{\Delta y}[/tex]
where,
[tex]\lambda[/tex] = wavelength
L = distance from the scrren
[tex]\Delta y[/tex] = spanning distance
As there are 11 bright fingers seen so m would be
= 11 - 1
= 10
Now placing these values to the above formula
So, the spacing is
[tex]= \frac{(10)(633 \times 10^{-9})(3.2m)}{60 \times 10^{-3}}[/tex]
= 0.3376 mm
We simply applied the above formula.
Answer:
Explanation:
Maximum occurs when the path difference is an integral multiple of wavelength
Here [tex]\lambda[/tex] - Wavelength, [tex]d-[/tex] slit separation and [tex]m-[/tex] Order of pattern
Rearrange the equation for
[tex]\begin{aligned}d &=\frac{m \lambda}{\sin \theta} \\
\text { Here, } \sin \theta &=\frac{y}{L} \quad\left(\begin{array}{l}
\text { Here, } L-\text { separation between slit and screen } \\
y-\text { Distance between respective fringe from center on screen }\end{array}\right)[/tex]
[tex]d=\frac{m \lambda}{\left(\frac{y}{L}\right)} \\
&=\frac{m \lambda L}{y}[/tex]
Here, order
Due to the fact that there are 11 bright fringes seen, you take [tex]11-1=10[/tex]
since starts from 0,1,2,3
Substitute given values
[tex]\begin{aligned}d &=\frac{(10)\left(633 \times 10^{-9} \mathrm{m}\right)(3.2 \mathrm{m})}{60 \times 10^{-3} \mathrm{m}} \\&=\left(3.376 \times 10^{-4} \mathrm{m}\right)\left(\frac{1 \mathrm{mm}}{10^{-3} \mathrm{m}}\right) \\&=0.3376 \mathrm{mm}\end{aligned}[/tex]
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find
(a) the coefficient of static friction.
(b) the coefficient of kinetic friction between the block and the surface.
Answer:
(a) 0.31
(b) 0.245
Explanation:
(a)
F' = μ'mg.................... Equation 1
Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.
make μ' the subject of the equation above
μ' = F'/mg............. Equation 2
Given: F' = 75 N, m = 25 kg
constant: g = 9.8 m/s²
Substitute these values into equation 2
μ' = 75/(25×9.8)
μ' = 75/245
μ' = 0.31.
(b) Similarly,
F = μmg.................. Equation 3
Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.
make μ the subject of the equation
μ = F/mg.............. Equation 4
Given: F = 60 N, m = 25 kg, g = 9.8 m/s²
Substitute these values into equation 4
μ = 60/(25×9.8)
μ = 60/245
μ = 0.245
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
v = 1.7 m/s
A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva so you can better study its flow pattern in the laboratory. Your model must be able to move at 50 cm/s and you will place the model in honey instead of water. Honey has a density of 1400 kg/m3 and a viscosity of 600 Pa-s.
Required:
How long should your model be?
Answer:
Explanation:
For the problem, we should have same reynolds number
ρvd/mu = constant
1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600
d = 25.66 cm
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.
Complete Question
The diagram for this question is showed on the first uploaded image (reference homework solutions )
Answer:
The velocity at the bottom is [tex]v = 11.76 \ m/ s[/tex]
Explanation:
From the question we are told that
The total distance traveled is [tex]d = 1.2 \ m[/tex]
The mass of the block is [tex]m_b = 0.3 \ kg[/tex]
The height of the block from the ground is h = 0.60 m
According the law of energy
[tex]PE = KE[/tex]
Where PE is the potential energy which is mathematically represented as
[tex]PE = m * g * h[/tex]
substituting values
[tex]PE = 3 * 9.8 * 0.60[/tex]
[tex]PE = 17.64 \ J[/tex]
So
KE is the kinetic energy at the bottom which is mathematically represented as
[tex]KE = \frac{1}{2} * m v^2[/tex]
So
[tex]\frac{1}{2} * m* v ^2 = PE[/tex]
substituting values
=> [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]
=> [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]
=> [tex]v = 11.76 \ m/ s[/tex]
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V
A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?
C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.
Answer:
Explanation:
A )
electric field E = V / d where V is potential difference between plates separated by distance d .
putting the given values
E = 370 / .040 V / m
= 9250 V / m
B )
Force on charged particle of charge q in electric field E
F = q E
F = 2.4 x 10⁻⁹ x 9250
= 22200 x 10⁻⁹
= 222 x 10⁻⁷ N .
C ) since field is uniform , force will be constant
work done by electric field putting up this force
= force x displacement
= 222 x 10⁻⁷ x 40 x 10⁻³
= 888 x 10⁻⁹ J
D )
change in potential energy
= q ( V₁ - V₂ )
= 2.40 X 10⁻⁹ x 370
= 888 x 10⁻⁹ J .
(a) The magnitude of electric field in the region between the plates is 9,250 V/m.
(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is [tex]8.88 \times 10^{-7} \ J[/tex].
(d) the change of the potential energy is [tex]8.88 \times 10^{-7} \ J[/tex].
The given parameters;
distance between the two metal plates, d = 40 mmpotential difference between the plates, V = 370 V(a) The magnitude of electric field in the region between the plates is calculated as;
[tex]E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m[/tex]
(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;
F = Eq
F = 9,250 x 2.4 x 10⁻⁹
F = 2.22 x 10⁻⁵ N
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;
[tex]W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J[/tex]
(d) the change of the potential energy is calculated as;
[tex]\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J[/tex]
Learn more here:https://brainly.com/question/13014987
The electric field strength is 1.70 × 104 N/C inside a parallel-plate capacitor with a 0.800 m spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?
Answer:
Here, "v" is the velocity of electron and "V" is the potential.
An accelerating voltage of 2.25 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 36.4 cm away. (a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field
Answer:
s= 8.28×10⁻¹⁶m
Explanation:
given
V= 2.25×10³V
from conservation of energy
mv²/2=qΔV
v=√(2qΔV/m)
v= √(2×1.6×10⁻¹⁹×2.25×10³/9.1×10⁻³¹)
=√7.9×10¹⁴m/s
=2.8×10⁷m/s
the deflection of electron beam is
S= gt²/2
recall t= d/v
s=g([tex]\frac{d}{v}[/tex])²/2
s= [tex]\frac{1}{2}[/tex]×9.8×(0.364/2.8×10⁷)²
s= 8.28×10⁻¹⁶m
in a _system supply and demand forces affect the production and consumption decisions. There is little to no _control in such a system
Answer:
in a free market system supply and demand forces affect the production and consumption decisions. There is little to no government control in such a system .
Explanation:
A free market is an economic system in which prices are based on competition between private actors and are not affected by other factors besides supply and demand, that is, where there are no external variables that condition the market.
Free market economy systems are characterized by limited government intervention, which characterizes democratic, liberal states and the modern global economy, in which the market in its private face makes most of the economic decisions, leaving the government a minimum amount of necessary regulations.
Three flat layers of transparent material are stacked upon one another. The top layer has index of refraction n1, the middle has n2 and the bottom one has n3. If n1 > n2 > n3, and if a ray of light strikes the top layer at an angle of incidence, in which layer is the angle of refraction the greatest? Why?
a. the bottom layer
b. the top layer
c. Once the ray enters the touching layers, the angle of refraction remains constant.
d. the middle layer
Answer:
a. the bottom medium
Explanation:
it has the least index of refraction and hence most rarer.
A vertical spring-mass system undergoes damped oscillations due to air resistance. The spring constant is 2.15 ✕ 104 N/mand the mass at the end of the spring is 10.7 kg.
(a) If the damping coefficient is b = 4.50 N · s/m, what is the frequency of the oscillator?Hz
(b) Determine the fractional decrease in the amplitude of the oscillation after 7 cycles.
Answer:
7.13Hz
Explanation:pls see attached file
How do you use these muscles in your everyday life? What daily activities do you complete that mimic the movements of these exercises
Answer:
If ur talking abkut hamstrings then it would be running that mimics them xplanation:
This was on a gym class quizz and I got it wrong but turned out this was the right answer
Answer:
In this activity, I exercised my hips, thighs, knees, calves, ankles, and legs. In some exercises, I specifically worked on only one type of muscle or on a combination of muscles. For example, the lunges mainly exercised the muscles of the inner thighs while the dead lift worked the muscles of the leg as well as the back and shoulders. I haven't consciously exercised my leg muscles before, but I have often noticed their tightening during my daily body movements, like when I climb the stairs or run to catch the school bus.
Explanation:
Hope this helped:)
What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)
Answer:
1.99*10-4sec
Explanation:
Signal propagation speed=0.82∗2.46∗108m/s
d=2000 m
Tp=20000/0.82∗2.46∗108 sec
ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8
= 1.99* 10^-4seconds
Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is
N
100
m
What is the elastic potential energy stored from the spring's compression?
Choose 1 answer:
-3.0J
-0.045 J
0.090 J
0.045 J
Answer:
0.045 J
Explanation:
From the question,
The elastic potential energy stored in a spring is given as,
E = 1/2ke²...................... Equation 1
Where E = elastic potential energy, k = spring constant, e = compression.
Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m
Substitute these values into equation 1
E = 1/2(100)(0.03²)
E = 50(9×10⁻⁴)
E = 0.045 J
Hence the right option is 0.045 J
Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m.The spring constant k is 100 N/m.
What is the elastic potential energy stored from the spring’s compression?
Answer: 0.045 J
A piece of tape is pulled from a spool and lowered toward a 100-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.
Requried:
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance.
Answer:
The magnitude of the electric force is [tex]F_e = 0.00098 \ N[/tex]
Explanation:
From the question we are told that
The mass of the paper is [tex]m= 100 mg = 100 *10^{-6} \ kg[/tex]
The position is [tex]d = 8.0\ mm = 0.008 \ m[/tex]
Generally the magnitude of the electric force at the point of equilibrium between the electric force and the gravitational force is mathematically represented as
[tex]F_e = F_g = mg[/tex]
Where [tex]F_g[/tex] is gravitational force
substituting values
[tex]F_e = 100 *10^{-6} * 9.8[/tex]
[tex]F_e = 0.00098 \ N[/tex]
Now generally the gravitational force acts downward (negative y axis ) hence the reason the electric force is same magnitude but opposite in direction (upward + y - axis )
On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?
Answer:
T = 2.06h
Explanation:
In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex] (1)
T: time for a complete orbit = ?
r: radius of the orbit
G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2
Mm: mass of the moon = 7.34*10^22 kg
The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:
[tex]r=R_m+160km\\\\[/tex]
Rm: radius of the moon = 1737.1 km
[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]
Then, you replace all values of the parameters in the equation (1):
[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]
In hours you obtain:
[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]
The time that the Apollo takes to complete an orbit around the moon is 2.06h
A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 N. 825 N. frictional force acting on the box as it moves to the right what is the net force in the Y direction
Answer:A
Explanation:
Explanation:
Given that,
Gravitational force = 600 N
Frictional force = 25 N
Pulled by the Force = 250 N
We know that,
The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.
The balance equation is along y-axis
The box will not move in y-axis therefore, the net force in the y-axis will be zero.
Hence, The net force in the y-direction will be zero.
Passengers in a carnival ride move at constant speed in a circle of radius 5.0 m, making a complete revolution in 4.0 s. As they spin, they feel their backs pressing against the wall holding them in the ride. A. What is the direction of the passengers' acceleration? a. No direction (zero acceleration) b. Directed towards center c. Directed away from center d. Directed tangentially B. What is the passengers' linear speed in m/s? C. What is the magnitude of their acceleration in m/s^2? D. What is their angular speed in rad/s?
Answer:
A. b) Directed towards center
B. [tex]v = 7.854\ m/s[/tex]
C. [tex]a_c = 12.337\ m/s^2[/tex]
D. [tex]w = 1.57\ rad/s[/tex]
Explanation:
The "force" that they feel pressing their backs against the wall is because the reaction to the centripetal acceleration .
A.
This acceleration has its direction towards the center of the circle. (option b)
B.
Their linear speed can be calculated with the equation:
[tex]v = (\theta/t)*r[/tex]
Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:
[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]
C.
The centripetal acceleration is given by the equation:
[tex]a_c = v^2/r[/tex]
[tex]a_c = 7.854^2/5[/tex]
[tex]a_c = 12.337\ m/s^2[/tex]
D.
Their angular speed is given by the equation:
[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]
a. Using the information below, calculate the cyclotron period of an electron that is launched into a magnetic field of strength 1 Gauss with a speed 200 m/s.
Electron Mass = 9.11 x 10^-31 kg
Proton Mass = 1.67 x 10^-27 kg
Elementary Charge = 1.602 x10^9 Nm/C
b. Using the same information from above, calculate the net work done on the charged particle by the magnetic field as it makes one full rotation.
Answer:
Explanation:
In cyclotron charged particle moves in a circular path in a magnetic field .
for rotation
mv² / R = Bqv where m is mass and q be charge of the particle which moves on circular path of radius R with velocity v .
v = BqR / m
Time period of rotation
T = 2πR / v
= 2πR m / BqR
= 2π m / Bq
For electron
T = 2π x 9.1 x 10⁻³¹ / (1 x 10⁻⁴ x 1.602 x 10⁻¹⁹)
= 35.67 x 10⁻⁸ s
b )
work done on the charged particle will be zero because force on charged particle is perpendicular to its movement so work done will be zero
Determining the Mass of a Mystery Mystery object Object distance (m) Brick mass (kg) Brick distance (m) Brick torque (Nm) Object mass (kg) Fobject (N) Fbricks (N) Fpivot point (N) A. 1.00 20 B. 1.00 20 C. 1.00 20 D. 1.00 20
Answer:
Explanation:
according to resultant of two parallel forces,
Fpivot = Fobject + Fbricks
so that, the net force is zero
A particle accelerator fires a proton into a region with a magnetic field that points in the x-direction. (a) If the proton is moving in the y-direction, what is the direction of the magnetic force on the proton
Answer:
The magnitude of the magnetic field will act in a direction towards me.
Explanation:
When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field. In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.
You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your normal weight is w= 635 N. If the elevator has an initial acceleration of magnitude 2.45 m/s2, what does the scale read? Express your answer in newtons.
Answer: 479. 425 N
Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.
It is given by N= m(g+a)
When it is accelerating downward, the scale reading is less than the true weight.
It so given by N = m(g-a)
The answer to the above questions is in the attached photo
Answer:
the scale will read 476.414 N
Explanation:
Weight = 635 N
mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)
mass m = 635 ÷ 9.81 = 64.729 kg
initial acceleration of the elevator a = 2.45 m/s^2
the force produced by the acceleration of the elevator downwards = ma
your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss
apparent weight = weight - ma
apparent weight = 635 - (64.729 x 2.45)
apparent weight = 635 - 158.586 = 476.414 N
To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch
Answer:
By a factor of about 0.23
Explanation:
Pressure is force over an area: P=F/A
Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.
By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.
PressureFriction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.
Pressure exists as force over an area: P=F/A
Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
let's divide P₁ by P₂
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that, we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,
To learn more about Pressure refer to:
https://brainly.com/question/912155
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What is meant civilized?
Answer:
at an advanced stage of social and cultural development. "a civilized society"
Explanation:
polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)
A 30 W engine generates 3600 J of energy. How long did it run for?
Answer:
so the time taken will be 120 seconds
Explanation:
power=30W
work done=3600J
time=?
as we know that
[tex]power=\frac{work done}{time taken}[/tex]
evaluating the formula
power×time taken=work done
[tex]time taken=\frac{work done}{power}[/tex]
[tex]time taken=\frac{3600J}{30W}[/tex]
[tex]Time taken=120seconds[/tex]
i hope this will help you :)
A block is attached to a horizontal spring and it slides back and forth in simple harmonic motion on a frictionless horizontal surface. At one extreme end of the oscillation cycle, where the block comes to a momentary halt before reversing the direction of its motion, another block is placed on top of the first block without changing its zero velocity. The simple harmonic motion then continues. What happens to the amplitude and the angular frequency of the ensuing motion of the two-block system
Answer:
A = A₀ , w = w₀/√2
Explanation:
This is a problem that we must solve with Newton's second law. They indicate that at the end of the initial movement where the speed is zero, add a mass to the block, we assume that it has the same mass, therefore the total mass is m_total = 2 m. Let's write Newton's second law at this point
[tex]F_{e}[/tex] = m_total a
the elastic force is
F_{e} = - k x
acceleration is
a = d²x / dt²
we substitute
- k x = m_total d²x / dt²
d²x / dt² + (k / m_total) x = 0
we substitute
d²x / dt² + (k /2m) x = 0
the solution to this differential equation is
x = A cos (wt + Ф)
where
w = √ (k / 2m)
to find the constant Ф we use the velocity
v = dx / dt = - Aw sin (wt + Ф)
At the most extreme point and when the new movement begins (t = 0) they indicate that v = 0
0 = - A w sin Ф
for this expression to be zero the sine must be zero therefore Ф = 0
when replacing
x = A cos (wt)
w = 1 /√2 √ (k / m)
if we want to relate to the initial movement (before placing the block)
w₀ = √ (k / m)
w = w₀ /√ 2
The amplitude of the movement is the distance from the equilibrium point to where the movement begins, in this case it is the same as in the initial movement
A = A₀
the subscript is used to refer to the oscillations before placing the second block
we substitute to have the final equation
x = A₀ cos (w₀ t /√2)
A = A₀
w = w₀/√2