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
what tools use cut wood
Answer:
hand saws
power saws
Circular Saw
Explanation:
that is all that i know
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
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
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.
When moving to a new apartment, you rent a truck and create a ramp with a 244 cm long piece of plywood. The top of the moving ramp lies on the edge of the truck bed at a height of 115 cm. You load your textbooks into a wooden box at the bottom of the ramp (the coefficient of kinetic friction between the box and ramp is = 0.2). Then you and a few friends give the box a quick push and it starts to slide up the ramp. A) What angle is made by the ramp and the ground?B) Unfortunately, after letting go, the box only tables 80cm up the ramp before it starts coming back down! What speed was the box initially traveling with just after you stopped pushing it?
Answer:
A) θ = 28.1º , B) v = 2.47 m / s
Explanation:
A) The angle of the ramp can be found using trigonometry
sin θ = y / L
Φ = sin⁻¹ y / L
θ = sin⁻¹ (115/244)
θ = 28.1º
B) For this pate we can use the relationship between work and kinetic energy
W =ΔK
where the work is
W = -fr x
the negative sign is due to the fact that the friction force closes against the movement
Lavariacion of energy cineta is
ΔEm = ½ m v² - mgh
-fr x = ½ m v² - m gh
the friction force has the equation
fr = very N
at the highest part there is no speed and we take the origin from the lowest part of the ramp
To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular
Axis y . perpendicular
N- Wy = 0
cos tea = Wy / W
Wy = W cos treaa
N = mg cos tea
we substitute
- (very mg cos tea) x = ½ m v²2 - mgh
v2 = m (gh- very g cos tea x)
let's calculate
v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)
v = RA (7.84 -1.729)
v = 2.47 m / s
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
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0215 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.15 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.
Answer:
h = 8.48*10^-3m
Explanation:
In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.
The total momentum of the system before the collision is equal to the total momentum after:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex] (1)
Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.
m1: mass of the marble = 0.0215kg
m2: mass of the pendulum concentrated at its end = 0.250kg
v1: horizontal speed of the arble before the collision = 5.15m/s
v2: horizontal speed of the pendulum before the collision = 0m/s
v: horizontal speed of both marble and pendulum after the collision = ?
You solve the equation (1) for v, and replace the values of the other parameters:
[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]
Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:
[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]
v = 0.40m/s
g: gravitational acceleration = 9,8m/s^2
[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]
Then, the height reached by marble and pendulum is 8.48*10^-3m
A student has made the statement that the electric flux through one half of a Gaussian surface is always equal and opposite to the flux through the other half of the Gaussian surface. This is:_______.
a. never true.
b. never false.
c. true whenever enclosed charge is symmetrically located at a center point, or on a center line or centrally placed plane
d. true whenever no charge is enclosed within the Gaussian surface.
e. true only when no charge is enclosed within the Gaussian surface.
Answer:
E.true only when no charge is enclosed within the Gaussian surface.
Explanation:
Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.
A d'Arsonal meter with an internal resistance of 1 kohm requires 10 mA to produce full-scale deflection. Calculate thew value of a series
A d’Arsonval meter with an internal resistance of 1 kΩ requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.
Answer:4kΩ
Explanation:Given;
internal resistance, r = 1kΩ
current, I = 10mA = 0.01A
Voltage of full scale, V = 50V
Since there is full scale voltage of 50V, then the combined or total resistance (R) of the circuit is given as follows;
From Ohm's law
V = IR
R = [tex]\frac{V}{I}[/tex] [substitute the values of V and I]
R = [tex]\frac{50}{0.01}[/tex]
R = 5000Ω = 5kΩ
The combined resistance (R) is actually the total resistance of the series arrangement of the series resistance([tex]R_{S}[/tex]) and the internal resistance (r) in the circuit. i.e
R = [tex]R_{S}[/tex] + r
[tex]R_{S}[/tex] = R - r [Substitute the values of R and r]
[tex]R_{S}[/tex] = 5kΩ - 1kΩ
[tex]R_{S}[/tex] = 4kΩ
Therefore the series resistance is 4kΩ
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
C.
(11) in parallel
A potentiometer circuit consists of a
battery of e.m.f. 5 V and internal
resistance 1.0 12 connected in series with a
3.0 12 resistor and a potentiometer wire
AB of length 1.0 m and resistance 2.0 12.
Calculate:
(i) The total resistance of the circuit
The current flowing in the circuit
(iii) The lost volt from the internal
resistance of battery across the
battery terminals
(iv) The p.d. across the wire AB
(v) The e.m.f. of a dry cell which can be
balanced across 60 cm of the wire
AB.
Assume the wire has a uniform cross-
sectional area.
Answer:
fggdfddvdghyhhhhggghh
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
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]
Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals
Answer:
producers make its own energy frominorganic substances.
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]
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.
Based on what you know about electricity, hypothesize about how series resistors would affect current flow. What would you expect the effective resistance of two equal resistors in series to be, compared to the resistance of a single resistor?
Answer:
Effective resistance of two equal resistors in series is twice that of a single resistor and in essence will reduce the amount of current flowing in the circuit.
Explanation:
When two resistors are connected in series, their effective resistance is the sum of their individual resistances. For example, given two resistors of resistance values R₁ and R₂, their effective resistance, Rₓ is given by;
Rₓ = R₁ + R₂ --------------(1)
If these resistors have equal resistance values, say R, then equation 1 becomes;
Rₓ = R + R
Rₓ = 2R
This means that their effective resistance is twice of their individual resistances. In other words, when two equal resistors are in series, their effective resistance is twice the resistance of each single one of those resistors.
Now, according to Ohm's law, voltage(V) is the product of current (I) and resistance (R). i.e
V = IR
I = [tex]\frac{V}{R}[/tex]
We can deduce that current increases as resistance decreases and vice-versa.
So, if the two equal resistors described above are connected in series, the amount of current flowing will be reduced compared to having just a single resistor.
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that is 0.60 m from the center is closest to
Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is [tex]E = 7500 \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner circle is [tex]r_i = 0.80 \ m[/tex]
The radius of the outer circle is [tex]r_o = 1.20 \ m[/tex]
The charge on the spherical shell [tex]q_n = -500nC = -500*10^{-9} \ C[/tex]
The magnitude of the point charge at the center is [tex]q_c = + 300 nC = + 300 * 10^{-9} \ C[/tex]
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as
[tex]E = \frac{k * q_c }{x^2}[/tex]
substituting values
[tex]E = \frac{k * q_c }{x^2}[/tex]
where k is the coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
substituting values
[tex]E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}[/tex]
[tex]E = 7500 \ N/C[/tex]
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion, after which a horizontal force of 60 N is required to keep the block in moving with constant speed. Find the coefficient of static and kinetic friction between the block and the surface.
Answer:
μs = 0.30
μk = 0.24
Explanation:
In order to calculate the kinetic friction and static friction between the block and the surface, you take into account that the kinetic friction is important when the block is moving and the static friction when the block is at rest.
You use the following formula to find the coefficient of static friction:
[tex]F_1=\mu_s Mg[/tex] (1)
F1 = 75N
μs: coefficient of static friction = ?
M: mass of the block = 25.0kg
g: gravitational acceleration = 9.8m/s^2
You solve for μs in the equation (1):
[tex]\mu_s=\frac{F_1}{Mg}=\frac{75N}{(25.0kg)(9.8m/s^2)}=0.30[/tex]
For the coefficient of kinetic friction you have:
[tex]F_2=\mu_k Mg[/tex] (2)
F2 = 60N
μk: coefficient of kinetic friction = ?
You solve for μk in the equation (2):
[tex]\mu_k=\frac{F_2}{Mg}=\frac{60N}{(25.0kg)(9.8m/s^2)}=0.24[/tex]
Then, you have:
coefficient of static friction = 0.30
coefficient of kinetic friction = 0.24
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
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.
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
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
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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A passenger jet flies from one airport to another 1,233 miles away in 2.4 h. Find its average speed. = ____ m/s
Speed = (distance) / (time)
Speed = (1,233 mile) / (2.4 hour)
Speed = 513.75 mile/hour
Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)
Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)
Speed = 229.7 meter/second
Stress is a factor that contributes to heart disease risk.true or false
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 :)
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
A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
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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.