Answer:
735.75 N
Explanation:
We are given;
Area of small lift; A1 = 5 cm²
Area of large lift; A2 = 300 cm²
Mass required to lift large lift;m = 4500 kg
Since Force = mg, then, Force required to lift large lift;F2 = 4500 × 9.81 = 44145 N
By Pascal's law,
F1/A1 = F2/A2
We want to find minimum force that must be applied to the small piston;F1, so let's make F1 the subject.
Thus;
F1 = (F2 × A1)/A2
Plugging in the relevant values, we have;
F1 = (44145 × 5)/300
F1 = 735.75 N
v(t) = 12 sin(913t + 71°) volts. Find (a) angular frequency in radians per second, (b) frequency in Hz, (c) period, (d) maximum voltage, (e) minimum voltage, (f) Peak-to-Peak voltage , (g) rms voltage, (h) average voltage, (i) voltage expressed as a phasor, (j) the average power consumed by a 220 ohm resistor having this voltage , and (k) the voltage at t= 3ms
Answer:
Explanation:
v(t) = 12 sin(913t + 71°) volts
a ) 913° = (π / 180) x 913 radians
= 15.92 radians
a ) angular frequency ω = 15.92 radians / s
b ) ω = 2πn
n = ω / 2π
= 15.92 / 2 x 3.14
= 2.53 Hz
c ) Period = 1 / n
= 1 / 2.53 = .4 s .
d )
Maximum voltage = 12 volt
e) Minimum volts = - 12 volts
g ) rms volts = V / √2
= 12 / √2
= 8.48 V
h )
Average voltage = 0
j ) Average power
Vrms² / R
= 12 x 12 / 2 x 220
= .327 W.
k )
v(t) = 12 sin(913t + 71°)
v(t) = 12 sin(913x .003 + 71°)
= 12 sin(73.7°)
= 11.5 V .
A 0,2 kg ball is whirled at constant angular velocity in a vertical circle at the end of a 1.25m long string. If the maximum tension the string can stand is 6N, what is the maximum angular velocity a ball can have in radians per second?
Answer:
24 rad/s
Explanation:
.................
A 2020 kg car traveling at 14.2 m/s collides with a 2940 kg car that is initally at rest at a stoplight. The cars stick together and move 2.12 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Answer:
The answer is 3,064x[tex]e^{-4}[/tex]
Explanation:
When the collision happens, the momentum of the first car is applied to the both of them.
So we can calculate the force that acts on both cars as:
The momentum of the first car is P = 2020 kg x 14.2 m/s = 28,684 kg.m/sThe acceleration of both cars after the crash is going to be a = P / mtotal which will give us a = 28,684 / (2020+2940) = 5.78 m/s Since the second car was initially not moving, the final acceleration was calculated with the momentum of the first car.Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:
F = (2020+2940) x 5.78 = 28,684The friction force acts in the opposite direction and if they stop after moving 2.12 meters;
Friction force is Ff = μ x N where μ is the friction coefficient and the N is the normal force which is (2020+2940) x 10 if we take gravitational force as 10, equals to 49,600.F - Ffriction = m x V 28,684 - μ x 49,600 = 4960 x 5.78μ = 3,064x[tex]e^{-4}[/tex]Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 52.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.936 m/s2. Calculate her mass.
Answer:
55.56kg
Explanation:
Given:
F= 52N
a=0.936m/s²
Applyinc Newton's second law, that states: force is equal to mass times acceleration.
F = ma
m=F/a =>52 / 0.936
m=55.56kg
Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster
Explanation:
We have,
Speed of plane a is 900 km/h
Plane b is moving at a rate of [tex]\dfrac{250\ km}{5\ h}=50\ km/h[/tex]
It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.
Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.
A box is released from rest and allowed to slide down a ramp with friction. Which statement most accurately describes the energy transformations during this motion?
a) The box’s initial kinetic energy is transformed into potential energy.
b) The box’s initial potential energy and thermal energy is transformed into
kinetic energy.
c) The box’s initial kinetic energy is transformed into potential energy and
thermal energy.
d) The box’s initial potential energy is transformed into kinetic energy and
thermal energy.
Answer:
d) The box’s initial potential energy is transformed into kinetic energy and
thermal energy.
Explanation:
Let us analyze the situation first. Initially we have a box at rest placed at some height on a ramp. So, the three energies associated with the box will be:
Kinetic Energy = 0 (due to zero velocity)
Thermal Energy = 0 (due to not temperature change)
Potential Energy = Greater than zero value (due to height)
Now, when the box is released to slide down the ramp with friction, the energies become:
Kinetic Energy = Increasing (due to increase in velocity)
Thermal Energy = Increasing (due to increase in temperature, because of friction)
Potential Energy = Decreasing (due to decrease in height)
So, from Law of Conservation of Energy, we can write:
Loss of Potential Energy = Gain in Kinetic Energy + Gain in Thermal Energy
So, the correct option is:
d) The box’s initial potential energy is transformed into kinetic energy and thermal energy.
The label on a battery-powered radio recommends the use of a rechargeable nickel-cadmium cell (nicads), although it has a 1.25-V emf, whereas an alkaline cell has a 1.58-V emf. The radio has a 3.65 Ω resistance. How much more power is delivered to the radio by alkaline cell, which has an internal resistance of 0.200Ω than by an nicad cell, having an internal resistance of 0.0.040Ω?
Answer:
0.2 W more power than nicad cell is delivered by alkaline cell
Explanation:
FOR NICKEL-CADMIUM CELL (nicads):
First we find the current supplied to radio by the cell. For this purpose, we use the formula:
I = E/(R+r)
where,
I = current supplied
E = emf of cell = 1.25 V
R = resistance of radio = 3.65 Ω
r = internal resistance of cell = 0.04 Ω
Therefore,
I = (1.25 V)/(3.65 Ω + 0.04 Ω)
I = 0.34 A
Now, we calculate the power delivered to radio by following formula:
P = VI
but, from Ohm's Law: V = IR
Therefore,
P = I²R
where,
P = Power delivered = ?
I = current = 0.34 A
R = Resistance of radio = 3.65 Ω
Therefore,
P = (0.34 A)²(3.65 Ω)
P = 0.41 W
FOR ALKALINE CELL:
First we find the current supplied to radio by the cell. For this purpose, we use the formula:
I = E/(R+r)
where,
I = current supplied
E = emf of cell = 1.58 V
R = resistance of radio = 3.65 Ω
r = internal resistance of cell = 0.2 Ω
Therefore,
I = (1.58 V)/(3.65 Ω + 0.2 Ω)
I = 0.41 A
Now, we calculate the power delivered to radio by following formula:
P = VI
but, from Ohm's Law: V = IR
Therefore,
P = I²R
where,
P = Power delivered = ?
I = current = 0.41 A
R = Resistance of radio = 3.65 Ω
Therefore,
P = (0.41 A)²(3.65 Ω)
P = 0.61 W
Now, fo the difference between delivered powers by both cells:
ΔP = (P)alkaline - (P)nicad
ΔP = 0.61 W - 0.41 W
ΔP = 0.2 W
Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 5.00 m above the ground and measure that it hits the ground 0.804 s later. Part APart complete What is the acceleration of gravity near the surface of this planet
Answer:
g = 15.5 m/s²
Explanation:
In order to find the acceleration due to gravity near the surface of this planet can be calculated by using 2nd equation of motion. The 2nd equation of motion is given as:
h = Vi t + (0.5)gt²
where,
h = height covered by the wrench = 5 m
Vi = Initial Velocity = 0 m/s
t = Time Taken to hit the ground = 0.804 s
g = acceleration due to gravity near the surface of the planet = ?
Therefore,
5 m = (0 m/s)(0.804 s) + (0.5)(g)(0.804 s)²
g = (5 m)/(0.3232 s²)
g = 15.5 m/s²
Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared.
Answer:
[tex]10.07m/s^2[/tex]
Explanation:
Information we have:
velocities:
initial velocity: [tex]v_{i}=0mph[/tex] (starts from rest)
final velocity: [tex]v_{f}=50mph[/tex]
time: [tex]t=2.22s[/tex]
Since we need the answer in [tex]m/s^2[/tex], we nees to convert the speed to meters per second:
[tex]v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s[/tex]
We find the acceleration with the following formula:
[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]
substituting the known values:
[tex]a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2[/tex]
the acceleration is 10.07[tex]m/s^2[/tex]
Answer:
The acceleration of the Cheetah is found to be 10.07 m/s²
Explanation:
We have the following data in this question:
Vf = Final Velocity of the Cheetah = 50 miles/hr
Vf = (50 miles/hr)(1 hr/ 3600 s)(1609.34 m/1 mile)
Vf = 22.35 m/s
Vi = Initial Speed of Cheetah = 0 m/s (Since, Cheetah starts from rest)
t = time interval passed = 2.22 s
So, in order to find the acceleration, we simply use the formula for acceleration, that is:
Acceleration = a = (Vf - Vi)/t
a = (22.35 m/s - 0 m/s)/2.22 s
a = 10.07 m/s²
The acceleration of the Cheetah is found to be 10.07 m/s²
How much heat is necessary to increase temperature of 4 kg water from 20 degree Celsius to 80 degree Celsius?
Answer:
2.4 × 10⁵ cal
Explanation:
Step 1: Given data
Mass of water (m): 4 kgInitial temperature: 20°CFinal temperature: 80°CSpecific heat capacity of water (c): 1 cal/g.°CStep 2: Convert the mass to grams
We will use the relationship 1 kg = 1,000 g.
[tex]4kg \times \frac{1,000g}{1kg} = 4,000 g[/tex]
Step 3: Calculate the change in the temperature
[tex]\Delta T = T_f - T_i = 80 \° C - 20 \° C = 60 \° C[/tex]
Step 4: Calculate the heat required (Q)
We will use the following expression.
[tex]Q = c \times m \times \Delta T = \frac{1cal}{g. \° C} \times 4,000 g \times 60 \° C = 2.4 \times 10^{5} cal[/tex]
A solenoid is 2.50 cm in diameter and 30.0 cm long. It has 300 turns and
carries 12 A.
(a) Calculate the magnetic field inside the solenoid.
(b) Calculate the magnetic flux through the surface of a circle of radius 1.00 cm, which is
positioned perpendicular to and centered on the axis of the solenoid.
(c) The perimeter of this circle is made of conducting material. The current in the solenoid
uniformly goes from 12 A to 10 A in 0.001 seconds. What e.m.f. is generated in the
conducting material?
(d) Now, inside the solenoid we put a bar of steel, a ferromagnetic material. Steel has got a
magnetic permeability μm = 4000μ0 . If the current in the solenoid is 12 A, what’s the
total magnetic field in the steel bar?
Answer:
a) B = 0.015 T
b) ФB = 4.71*10⁻6 W
c) emf = 7.89*10^-4 V
d) B = 60.31 T
Explanation:
(a) To find the magnitude of the magnetic field inside the solenoid you use the following formula:
[tex]B=\frac{\mu_oNI}{L}[/tex] (1)
B: magnitude of the magnetic field
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
N: turns of the solenoid = 300
L: length = 2.50cm = 0.025 m
I: current = 12 A
You replace the values of the variables in the equation (1):
[tex]B=\frac{(4\pi *10^{-7}T/A)(300)(12A)}{0.3m}=0.015T[/tex]
(b) The magnetic flux is given by:
[tex]\Phi_B=BA[/tex]
A: area = π(0.01m)^2 = 3.1415*10^-4 m^2
[tex]\Phi_B=(0.015T)(3.1415*10^{-4}m^2)=4.71*10^{-6}W[/tex]
(c) The induced emf is given by the following formula:
[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=\frac{A(B_f-B_i)}{\Delta t}[/tex] (2)
Bf and Bi are the final and initial magnetic field. You use the equation (1) into the equation (2):
[tex]emf=-A\mu_o\frac{N}{L}\frac{(I_2-I_1)}{\Delta t}\\\\emf=-\pi(0.01m)^2(4\pi*10^{-7}T/A)\frac{300}{0.3m}\frac{(10A-12A)}{0.001s}\\\\emf=7.89*10^{-4}V[/tex]
(d) With the ferromagnetic material inside the solenoid the magnetic field inside is modified as following:
[tex]B=\frac{\mu NI}{L}\\\\\mu=4000\mu_o=4000(4\pi*10^{-7}T/A)=5.026*10^{-3}T/A\\\\B=\frac{(5.026*10^{-3}T/A)(300)(12A)}{0.3m}=60.31T[/tex]
Following are the calculation to the given points:
Given:
[tex]r = 1.25\ cm \\\\l=30\ cm\\\\ N= 300 \\\\I=12\ A\\\\[/tex]
To find:
[tex]B=?\\\\\Phi_B=?\\\\[/tex]
Solution:
For point a:
Using formula:
[tex]\to \mu_o= 4 \pi 10^{-7} \ \frac{T}{A}[/tex]
[tex]\to N= 300\\\\ \to L= 2.50\ cm = 0.025\ m\\\\ \to I= 12 \ A\\\\[/tex]
[tex]\to B = \frac{\mu_o \ N\ I}{L}\\[/tex]
[tex]=\frac{4 \pi \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{4 \times 3.14 \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{48 \times 3.14\ T \times 300 }{0.3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 3000 }{3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 1000 }{10^{7}}\\\\= 0.015\ T\\\\[/tex]
For point b:
[tex]\to A : area = \pi (0.01m)^2 = 3.14 \times 10^{-4}\ m^2\\\\[/tex]
Using formula:
[tex]\to \Phi_B = BA\\\\[/tex]
[tex]= (0.015\ T)(3.14 \times 10^{-4} \ m^2)\\\\ = 4.71 \times 10^{-6}\ W[/tex]
For point c:
Using formula:
[tex]\to emf = -A \mu_o \frac{N}{L} \frac{(I_2-I_1)}{\Delta t}[/tex]
[tex]= -\pi(0.01 \ m)^2(4\pi \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3\ m} \frac{(10A-12A)}{0.001\ s} \\\\= -3.14 (0.01 \times 0.01 \ m)(4\times 3.14 \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3} \frac{(-2A)}{0.001\ s} \\\\= -3.14 (1 \ m)(12.56\times 10^{-7}\ \frac{T}{A}) 10^6 \frac{(-2A)}{ 10^3\ s} \\\\= -3.14 (1 \ m)(12.56 \frac{T}{A}) \frac{(-2A)}{ 10^4\ s} \\\\= 7.89 \times 10^{-4}\ V\\\\[/tex]
For point d:
[tex]\to B = \frac{\mu NI}{L}\\\\\to \mu = 4000 \mu_o = 4000(4\pi \times 10^{-7}\ \frac{T}{A}) = 5.026 \times 10^{-3} \frac{T}{A}\\\\ \to B = \frac{(5.026\times 10^{-3} \ \frac{T}{A})(300)(12\ A)}{0.3\ m} = 60.31\ T\\\\[/tex]
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A transverse sinusoidal wave is traveling rightward along a string stretched along the X-direction. The wavelength is 70.0 cm and 8.00 wave peaks per second pass any fixed point on the string. The maximum speed of the string particles in the direction perpendicular to the string is 3.20 m/s. At time t 0, the element of string at x = 20.0 cm is 4.00 cm above its resting position. (a) Construct two distinct wave functions yfx,t) consistent with the given information. (If the only difference between two wavefunctions is that their phase constants differ by an integer multiple of 2n, then they are not distinct.) (b) Sketch both of these wavefunctions for time t 0 between x 0 and x A har 2 20 /s A= 0.0Ch 560 Ch a)yo) Acos CRxt wt +o) w- 2TF : 50 265 S A Vwa 3 20 5a.2655 =0.0656n 0.7854
Answer:
1111
1Explanation:
111111
1
Dispositivo que muestra tanto la dirección como la magnitud de la corriente eléctrica.
Answer:
Galvanómetro
Explanation:
Un galvanómetro es un dispositivo eléctrico utilizado para detectar la presencia de corriente y voltaje pequeños o para medir su magnitud. Los galvanómetros se utilizan principalmente en puentes y potenciómetros.
Using the following information to determine the young's modulus for the unknown material, the radius of the material is 4 cm. while gravitational
acceleration g=10m/s?
initial length (cm) 25
final length (cm)25.7
mass (gm) 200
Answer:
14285.71N/m2
Explanation:
Young modulus known as modulus of elasticity is defined as;
E = σ/ε
Where E is young modulus
ε is strain defined as extention / length
σ is stress defined as force /area
Let's calculate σ;
Force = mass× gravity =200/1000. × 10 =0.2kg× 10 = 2N
The area impacted by the force is raduis surface of the wire and it's calculated thus;
πr^2 = 22/7 × (0.04)^2 ; 4cm in m = 0.04m=0.005m2
Hence σ = 2/0.005=400N/M2
Let's calculate ε;
ε= extension/ original length
extension = final length - original length
extension=25.7-25=0.7cm= 0.007m
ε= 0.7/25
Hence E = 400/ 0.7/25
E = 400 × 25/ 0.7= 14285.71N/m2
2 pts.) An electron is placed in an electric field of intensity 700 N/Cj. What are themagnitude and direction of the acceleration of this electron due to this field? (melectron=9.1 × 10-31 kg, qe= 1.60 × 10-19 C)
Explanation:
We have,
Electric field intensity is 700 N/Cj
It is required to find the magnitude and direction of the acceleration of this electron due to this field.
The electric force is balanced by the force due to its motion i.e.
qE =ma
a is acceleration of this electron
[tex]a=\dfrac{qE}{m}\\\\a=\dfrac{1.6\times 10^{-19}\times 700}{9.1\times 10^{-31}}\\\\a=1.23\times 10^{14}\ m/s^2[/tex]
So, the acceleration of the electron is [tex]1.23\times 10^{14}\ m/s^2[/tex] and it is moving in +y direction.
An isotope has 46 electrons, 60 neutrons, and 46 protons. Name the isotope.
Answer:
Palladium
Explanation:
Answer:
palladium-106
Explanation:
46 protons -46 electrons=no charge
46 electrons +60neutron = 106
Thus this is called palladium -106
Consider five charged particles: A,B,C,D,and E.
(A attracts B),(C attracts D ), (B repels C), and (D repels E). If C is Positive,what is the charge of the other particles?
Answer:
A_negative
B_positive
C_positive
D_positive
E_negative
Explanation:
according to the law of electrostatic which is
like charge repels and unlike charge attract.
Answer:A=negative, B=positive, C=positive, D=negative, E=negative
Explanation:
Like poles or charges repels and unlike poles or charges attract each other
A plasma is a gas of ionized (charged) particles. When plasma is in motion, magnetic effects "squeeze" its volume, inducing inward pressure known as a pinch. Consider a cylindrical tube of plasma with radius R and length L moving with velocity v along its axis. If there are n ions per unit volume and each ion has charge q , we can determine the pressure felt by the walls of the cylinder.
Required:
a. What is the volume charge density p in terms of n and q?
b. The thickness of the cylinder surface is n^1/3. What is the surface charge density σ in terms of n and q?
Answer:
a
The volume charge density is [tex]\rho = nq[/tex]
b
The surface charge density is [tex]\sigma = n^{\frac{2}{3} } q[/tex]
Explanation:
From the question we are told that
The radius is R
The length is L
The velocity is v
The number of ions per unit volume is n
The charge is q
The thickness of the cylinder surface is [tex]n^{\frac{1}{3} }[/tex]
The volume charge density is mathematically represented as
[tex]\rho = nq[/tex]
The surface charge density is mathematically represented as
[tex]\sigma = \rho n^{\frac{1}{3} }[/tex]
substituting for [tex]\rho[/tex]
[tex]\sigma = n * n^{\frac{1}{3} } q[/tex]
[tex]\sigma = n^{\frac{2}{3} } q[/tex]
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behaves like a simple spring. (a) Calculate how much it would stretch if the same person were lying in it. (b) How much would it stretch if the person jumped from 38 m?
Answer:
a) x = 0.098
b) x = 2.72 m
Explanation:
(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.
When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.
Then, you have:
[tex]K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex] (1)
m: mass of the person = 62kg
k: spring constant = ?
v: velocity of the person just when he touches the fire net = ?
x: elongation of the fire net = 1.4 m
Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:
[tex]v^2=v_o^2+2gy[/tex]
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
y: height from the person jumps = 20.0m
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}[/tex]
With this value you can find the spring constant k from the equation (1):
[tex]mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}[/tex]
When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:
[tex]W=F_e\\\\mg=kx[/tex]
you solve the last expression for x:
[tex]x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m[/tex]
When the person is lying on the fire net the elongation of the fire net is 0.098m
b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}[/tex]
Next, you calculate x from the equation (1):
[tex]x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m[/tex]
The net fire is stretched 2.72 m
A) If the person is lying on the net, the net would stretch by : 0.0679 m
B) If the person jumped from 38m, the net would stretch by : 1.61 m
Given data :
Mass of person (m) = 62 kg
Initial height ( H₁ ) = 20 m
Net stretch ( H[tex]_{f}[/tex] ) = -1.2 m
Initial potential energy before jump ( PE₁ ) = mgH₁ = 62 * 9.81 * 20 = 12164.4J
Potential energy after the jump ( PE₂ ) = mgH[tex]_{f}[/tex] = 62 * 9.81 * ( -1.2 ) = -729.86J
The potential spring constant ( SE ) = 1/2 kx²
= 1/2 * k * ( 1.2 )²
Applying the principle of energy conservation
PE₁ = PE₂ + SE
12164.4 = - 729.86 + 0.5 * k * 1.44
∴ k = ( 12164.4 + 729.86 ) / ( 0.5 * 1.44 )
= 17908.69
A) Calculate the amount of stretch if same person lies on the net1/2 kx² = mgx ---- ( 1 )
Where x = stretch
equation ( 1 ) becomes equation ( 2 )
1/2 kx = mg ----- ( 2 )
x ( amount of stretch ) = ( 2mg ) / k
= ( 2 * 62 * 9.81 ) / 17908.69
= 0.0679 m
B ) Calculate How much the net would stretch if Height = 38 mHi = 38 m
MgHi = mgH[tex]_{f}[/tex] + 1/2 kx² ----- ( 3 )
where ; H[tex]_{f}[/tex] = -x
Back to equation ( 3 )
62 * 9.81 * 38 = 62 * 9.81 * (-x) + 1/2 ( 17908.69 x² )
23112.36 + 608.22x - [tex]\frac{17908.69 x^{2} }{2}[/tex] = 0
8954.345 x² - 608.22x - 23112.36 = 0 ----- ( 4 )
Resolving the quadratic equation ( 4 )
x = ± 1.60623 ≈ 1.61 m
Therefore we can conclude that the net would stretch 1.61 m if the person jumped from 38 m
Hence we can conclude that If the person is lying on the net, the net would stretch by : 0.0679 m, and If the person jumped from 38m the net would stretch by : 1.61 m
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A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m/s at an angle of 35.0 ∘ with the horizontal, as shown in the figure (Figure 1) . Determine the time taken by the projectile to hit point P at ground level.
Answer:
9.96 s
Explanation:
Given in the y direction:
Δy = -115 m
v₀ = 65.0 m/s sin 35.0° = 37.28 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-115 = 37.28 t − 4.9 t²
4.9 t² − 37.28 t − 115 = 0
t = [ 37.28 ± √(37.28² − 4(4.9)(-115)) ] / 9.8
t = (37.28 ± 60.37) / 9.8
t = 9.96
A 200-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s
Answer:
F = 187.5N.
Explanation:
So, from the question above we are given the following parameters or data or information which is going to assist us in solving the question/problem;
=> Mass= 200-kg , => radius = 1.50 m, => angular speed of 0.400 rev/s, and time = 2.0 seconds.
Step one: the first step is to calculate or determine the angular speed. Here, the angular speed is Calculated in rad/sec.
Angular speed, w = 0.400 × 2π
= 2.51 rad/s.
Step two: determine the value of a.
Using the formula below;
W = Wo + a × time,t.
2.51 = 0 + a(2.0).
a= 1.25 rad/s^2.
Step three: determine constant force from the Torque.
Torque = I × a.
F = 1/2 × (200 kg) × 1.50 × 1.25.
F = 187.5N
A 16.0 g bullet is moving to the right with speed 250 m/s when it hits a target and travels an additional 22.6 cm into the target. What are the magnitude (in N) and direction of the stopping force acting on the bullet
Answer:
The magnitude will be "2212 N".
Explanation:
The given values are:
Initial speed, u = 250 m/s
Final speed, v = 0
Mass, m = 16.0 = 0.016 kg
Distance, d = 22.6 cm
On converting cm into m, we get
d = 0.226 m
As we know,
⇒ [tex]v^2-u^2=-2ad[/tex]
⇒ [tex]a=\frac{v^2-u^2}{-2d}[/tex]
On putting the estimated values, we get
⇒ [tex]=\frac{(0)^2-(250)^2}{-2\times 0.226}[/tex]
⇒ [tex]=\frac{0-62500}{-0.452}[/tex]
⇒ [tex]=13,8276 \ m/s^2[/tex]
Now,
Force, [tex]F=ma[/tex]
On putting values, we get
⇒ [tex]=0.016\times 138276[/tex]
⇒ [tex]=2212 \ N[/tex] (magnitude)
The magnitude of the stopping force on the bullet is 2,212.4 N in opposite direction to the bullet.
The given parameters;
mass of the bullet, m = 16 gspeed of the bullet, u = 250 m/sdistance traveled by the bullet and the target, d = 22.6 cm = 0.226 mThe acceleration of the bullet is calculated as follows;
v² = u² - 2ad
where;
v is the final velocity when the bullet stops = 00 = 250² - a(2 x 0.226)
0.452a = 62500
[tex]a = \frac{62500}{0.452} \\\\a = 138,274.34 \ m/s^2\\\\[/tex]
The magnitude of the stopping force on the bullet is calculated as follows;
F = ma
F = 0.016 x 138,274.34
F = 2,212.4 N
Thus, the magnitude of the stopping force on the bullet is 2,212.4 N in opposite direction to the bullet.
Learn more here: https://brainly.com/question/19887955
A long metallic wire is stretched along the x direction. The applied potential difference along
the wire is 12 volts. The resistance of the wire is estimated to be 8 Ohm.
a. Calculate the current in the wire and the passing charge in 2 seconds.
b. Calculate the electric power dissipated in the wire.
This wire has a length of 100 m and the material of the wire has a resistivity of 1.6 x 10 -8 Ωm.
c. Calculate the cross-section area of the wire.
d. Calculate the conductivity of the wire.
e. What is the capacitance of the capacitor, that is connected to a series resistor of 8 MΩ in an RC
circuit that has a time constant of one second.
Answer:
a) I = 1.5 A , b) P = 18 W , c) A = 8 10⁻⁷ m², d) σ = 0.625 10⁸ (Ω m)⁻¹,
e) C = 0.125 10⁻⁶ F
Explanation:
a) for this exercise let's use Ohm's Law
V = I R
I = V / R
I = 12/8
I = 1.5 A
b) the power is given by the expression
P = V I
P = V V / R = V² / R
P = 12²/8
P = 18 W
c) the resistance of the wire is given by
R = ρ l / A
A = ρ l / R
A = 1.6 10⁻⁸ 100/8
A = 8 10⁻⁷ m²
d) conductivity is the inverse of resistivity
σ = 1 / ρ
σ = 1 / 1,610⁻⁸
σ = 0.625 10⁸ (Ω m)⁻¹
e) In an RC circuit the response time is
τ = RC
C = τ / R
C = 1/8 10⁶
C = 0.125 10⁻⁶ F
An AC voltage is applied to a purely capacitive circuit. Just as the applied voltage is crossing the zero axis going negative, what is the value of capacitor current? Ic is at its positive peak Ic is zero Ic is at its negative peak
Answer:
The Ic will be zero.
Explanation:
Capacitors have a working principal as follows:
As the current flows through the circuit, they store the electrical energy according to certain attributes they have such as the area of the plates and the material's capacitence in between the plates.An AC voltage increases and decreases between certain maximum and minimum points periodically. So while the AC voltage is on the positive side, the capacitor charges up and when the AC voltage crosses to the negative side, the capacitor takes over and it's current starts increasing as the current coming from the AC source decreases.
So in this case, as the AC voltage crosses zero, the capacitor current was decreasing because the AC voltage was on the positive side and it was charging. The capacitor current will be zero as well and it will start to increase when AC voltage is on the negative.
I hope this answer helps.
The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.95 × 103 N with an effective perpendicular lever arm of 3.1 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia of the boxer's forearm?
Answer:
I = 0.483 kgm^2
Explanation:
To know what is the moment of inertia I of the boxer's forearm you use the following formula:
[tex]\tau=I\alpha[/tex] (1)
τ: torque exerted by the forearm
I: moment of inertia
α: angular acceleration = 125 rad/s^2
You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)
[tex]\tau=Fr=(1.95*10^3N)(0.031m)=60.45J[/tex]
Next, you replace this value of τ in the equation (1) and solve for I:
[tex]I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2[/tex]
hence, the moment of inertia of the forearm is 0.483 kgm^2
At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's velocity at t = 4.1 s, given that the components of its average acceleration during this period are ax = 2 m/s2 and ay = -1 m/s2.
Answer:
vx' = 8.7 m/s
vy' = - 2.9 m/s
Explanation:
The average acceleration during a time period is given by the formula:
a = (v' - v)/(t' - t)
where,
a = average acceleration
v' = Final Velocity
v = Initial Velocity
t' = Ending Time
t = Starting Time
Now, for x-component of velocity:
a = ax = 2 m/s²
v' = vx' = ?
v = vx = 0.5 m/s
t' = 4.1 s
t = 0 s
Therefore,
2 m/s² = (vx' - 0.5 m/s)/(4.1s - 0s)
(2 m/s²)(4.1 s) = vx' - 0.5 m/s
8.2 m/s + 0.5 m/s = vx'
vx' = 8.7 m/s
For y- component of velocity:
a = ay = -1 m/s²
v' = vy' = ?
v = vy = 1.2 m/s
t' = 4.1 s
t = 0 s
Therefore,
- 1 m/s² = (vy' - 1.2 m/s)/(4.1s - 0s)
(- 1 m/s²)(4.1 s) = vy' - 1.2 m/s
- 4.1 m/s + 1.2 m/s = vy'
vy' = - 2.9 m/s
Name and draw the devices that can convert digital signal to analog.
Answer:
modem, digital music players, optical communication
Explanation:
Photos of devices are attached. An example of R2R logic circuit is also attached. DAC conversion is also done via pulse code modulation
a body of 2.0kg mass makes an elastic collision with another at rest and continues to move in the original direction but with 1/4 of it's original speed.what is the mass of the struck body?
2.0 kg into the bifactor of 1/4 so the mass struck would be 32
3. Calculate the work done to move an object of 50g in a circular path of radius 10cm.
Answer:
Work done in moving an object in circular path is always 0 J
Explanation:
Work doen on any moving object, in general, is given by the following formula:
W = F d Cos θ
where,
W = Work Done
F = Force applied
d = Distance covered
θ = Angle between the direction of motion and the force applied
When the object is moving in a circular path the direction of its motion is always tangential to the circular path. While, the force applied on the object is the centripetal force. And centripetal force is always directed towards the center of the circular path. Therefore, the force and direction of motion are always perpendicular to each other. This means θ = 90°
Therefore,
W = F d Cos 90° = F d(0)
W = 0 J
A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it is released and explain how you know. b. If work is done on the particle after its release, is the work positive or negative
Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.