6.07 m
Explanation:
Given:
[tex]v_0=24.5\:\text{m/s}[/tex]
[tex]\theta_0 = 35.5°[/tex]
First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that
[tex]x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}[/tex]
or
[tex]t = 1.29\:\text{s}[/tex]
To find the vertical height where the ball hit the wall, we use
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
[tex]\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2[/tex]
[tex]\:\:\:\:=6.07\:\text{m}[/tex]
How are Newton’s 1 and 2 law related?
find out the odd one and give reason (length, volume, time, mass
Answer:
Time
Explanation:
The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.
g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
The desk is in equilbrium, so Newton's second law gives
∑ F (horizontal) = p - f = 0
∑ F (vertical) = n - mg = 0
==> n = mg
==> p = f = µn = µmg = 0.400 (80.0 kg) g = 313.6 N
The student pushes the desk 3.00 m, so she performs
W = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J
of work.
In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.517 ss for the value of the period.
Trial 1 Spring constant is 117N/m, period of oscillations .37s, mass of the block is .400kg .
Trial 2 oscillation period is .52s
Answer:
[tex]M_2=0.79kg[/tex]
Explanation:
From the question we are told that:
Period [tex]T=0.517s[/tex]
Trial 1
Spring constant [tex]\mu=117N/m[/tex]
Period [tex]T_1=0.37[/tex]
Mass [tex]m=0.400kg[/tex]
Trial 2
Period [tex]T_2=0.52[/tex]
Generally the equation for Spring Constant is mathematically given by
\mu=\frac{4 \pi^2 M}{T^2}
Since
[tex]\mu _1=\mu_2[/tex]
Therefore
[tex]\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}[/tex]
[tex]M_2=M_1*(\frac{T_2}{T_1})^2[/tex]
[tex]M_2=0.400*(\frac{0.52}{0.37}})^2[/tex]
[tex]M_2=0.79kg[/tex]
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 40.0 mph and half the distance at 60.0 mph . On her return trip, she drives half the time at 40.0 mph and half the time at 60.0 mph.
Required:
a. What is Julie's average speed on the way to grandmother's house?
b. What is her average speed in the return trip?
Answer:
a. The average speed on her way to Grandmother's house is 48.08 mph
b. The average speed in the return trip is 50 mph.
Explanation:
The average speed (S) can be calculated as follows:
[tex] S = \frac{D}{T} [/tex]
Where:
D: is the total distance
T: is the total time
a. To find the total distance in her way to Grandmother's house, we need to find the total time:
[tex]T_{i} = t_{1_{i}} + t_{2_{i}} = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}}[/tex]
Where v is for velocity
[tex] T = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}} = \frac{(100/2) mi}{40.0 mph} + \frac{(100/2) mi}{60.0 mph} = 1.25 h + 0.83 h = 2.08 [/tex]
Hence, the average speed on her way to Grandmother's house is:
[tex]S_{i} = \frac{D}{T_{i}} = \frac{100 mi}{2.08 h} = 48.08 mph[/tex]
b. Now, to calculate the average speed of the return trip we need to calculate the total time:
[tex]D = v_{1_{f}}\frac{T_{f}}{2} + v_{2_{f}}\frac{T_{f}}{2} = \frac{T_{f}}{2}(v_{1_{f}} + v_{2_{f}})[/tex]
[tex]100 mi = \frac{T_{f}}{2}(40 mph + 60 mph)[/tex]
[tex] T_{f} = \frac{200 mi}{40 mph + 60 mph} = 2 h [/tex]
Therefore, the average speed of the return trip is:
[tex]S_{f} = \frac{D}{T_{f}} = \frac{100 mi}{2 h} = 50 mph[/tex]
I hope it helps you!
What is the strength of the magnetic field a distance 4.4 mm above the center of a circular loop of radius 0.8 mm and current 474.1 A
Answer:
B = 0.118 T
Explanation:
From Biot-Savart Law:
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where,
B = strength of magnetic field = ?
μ₀ = 4π x 10⁻⁷ Tm/A
I = current enclosed = 474.1 A
r = radius = 0.8 mm = 8 x 10⁻⁴ m
Therefore,
[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]
B = 0.118 T
A long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point
The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
If you and a friend are standing side-by-side watching a soccer game, would you both view the motion from the same reference frame?
a. Yes, we would both view the motion from the same reference point because both of us are at rest in Earth’s frame of reference.
b. Yes, we would both view the motion from the same reference point because both of us are observing the motion from two points on the same straight line.
c. No, we would both view the motion from different reference points because motion is viewed from two different points; the reference frames are similar but not the same.
d. No, we would both view the motion from different reference points because response times may be different; so, the motion observed by both of us would be different.
Answer:
the correct is C
Explanation:
The concept of a frame of reference is of crucial importance in physics, because it is the system from which measurements are made. Therefore, the relationships between the different reference frames must be clear so that the measurements made can be compared correctly.
In this case, the first observed sees the movement of the ball, suppose it moves a distance r, the second observed is next to me, separated by a distance x, therefore a frame of reference located in the movement of the ball. ball r '.
Consequently, the measurement carried out is related by
r = r’ + x
where the bold letters indicate wind blowers.
With these explanations we review the different answers, the correct one is C
A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?
A) 3.57 m.
B) 4.28 m.
C) 3.14 m.
D) 2.68 m.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.052 T, how many turns of wire are needed
Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns
[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]
[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]
Therefore, the number of turns of wire needed is 573.8 turns
Receptor elétrico 5 pontos Dispositivo que converte energia elétrica em outra forma de energia, não exclusivamente térmica. Exemplos: motores elétricos, ventiladores, liquidificadores, geladeiras, aparelhos de sons, vídeos, celulares, computadores?
Answer:
Electromechanical transducer and Electrical receiver.
Explanation:
Electromechanical transducer is the part of a communication system which converted electrical waves or electrical energy into sound waves. The most common example loudspeaker while on the other hand, Electrical receiver is a device that converts electrical energy into another form of energy, except thermal. Examples are cell phones, computers and television.
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
The gravitational field strength due to its planet is 5N/kg What does it mean?
Answer:
The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.
Weight can be calculated using the equation:
weight = mass × gravitational field strength
This is when:
weight (W) is measured in newtons (N)
mass (m) is measured in kilograms (kg)
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
15.1.2 Exam: Semester Exam
ŽA
This graph shows the energy of a reaction over time. Which statement is
true?
Potential energy
N
Reaction progress
O A. G represents the activation energy
OB. H represents the energy of the products,
O C. G represents the energy of the products.
< PREVIOUS
Answer: D. F represents the activation energy
Explanation:
The activation energy is the energy required to get the reactants to begin reacting with one another such that products are created. This energy ranges from the minimum to the maximum energy required.
F is therefore the activation energy because it shows the range between the minimum energy it took for the reaction to start and the maximum energy that was required to continue the reaction.
Two guitar strings, of equal length and linear density, are tuned such that the second harmonic of the first string has the same frequency as the third harmonic of the second string. The tension of the first string is 510 N. Calculate the tension of the second string.
Answer:
The tension in the second string is 226.7 N.
Explanation:
Length is L, mass per unit length = m
T = 510 N
Let the tension in the second string is T'.
second harmonic of the first string = third harmonic of the second string
[tex]2 f = 3 f'\\\\2\sqrt{\frac{T}{m}} = 3 \sqrt {\frac{T'}{m}}\\\\4 T = 9 T'\\\\4\times 510 = 9 T'\\\\T' = 226.7 N[/tex]
If an object travels on a circular path is an acceleration? What is changing to cause an acceleration?
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
The thrust F of a screw propeller is known to depend upon the diameter d, Speed of advance v, fluid density e, revolution per second N, and the coefficient of viscosity M, of the fluid. Find the expression for F, in terms of the quantities
Answer:
[tex]{ \bf{F = { \tt{ \frac{4}{3} \pi {r}^{3}v gM}}}}[/tex]
A car is moving north at 5.2 m/s2. Which type of motion do the SI units in this
value express?
Answer:
the SI unit (meter per second square) indicates a linear type of motion.
Explanation:
Given;
acceleration of the car, a = 5.2 m/s² North
the SI unit of the car, = m/s²
The SI unit of the given value (acceleration), indicates a linear type of motion.
Linear acceleration is the change in linear velocity with time. Also, the northwards direction indicates linear displacement of the car.
Therefore, the SI unit (meter per second square) indicates a linear type of motion.
Answer:
displacement
Explanation:
Hi can someon help me how to answer this?
Btw I'm from Philippines
Answer:
Test 1
1.True
2.True
3.True
4.False
5.True
6.True
7.False
8.True
9.True
10.True
yung iba nasa pic
Which is a mixture?
'a' sodium metal
'b' chlorine gas
'c' sodium metal and chlorine gas
'd' sodium chloride (salt) and water
Answer:
d. Sodium chloride (salt) + water
Explanation:
A mixture is made up of two or more substance combined together (combined chemically).NaCl (salt) can completely dissolve in water and sodium chlorine (aqueous) is a homogeneous mixture.sodium metal when extracted is a soft, silvery white solid.chlorine gas is a pure gas.sodium metal and chlorine gas are at pure state hence they are not mixture.learn more: https://brainly.com/question/2331419
Answer:D. Sodium chloride (salt) and water
Explanation:
I got it right on edge 2023
hope this is helpful!
Part C – RC Circuits in AC Mode 1. Derive Equation 5-6 from Equation 5-5. 2. Using the τ’s you calculated and your measured resistance: a. Calculate the capacitances of the capacitors. b. Compare your calculated and measured values via percent error.
Answer: hi your question is incomplete attached below is the complete question
1) attached below
2) a) 31 Ω, 302.9 Ω
b) 17.3 Ω , 26.4 Ω
Explanation:
1) Deriving Eqn 5-6 from Eqn 5-5
attached below
2) using τ’s calculated and measured resistance
use given data
Tm1 = Rm1 * Cm1
= 329.3 * 333 * 10^-9 = 109.65 μs
Tm2 = Rm2 * Cm2
= 329.3 * 200 * 10^-9 = 658.6 μs
a) Capacitance of capacitors
For Cm1
t₁₂ = 72 μs = R₁' Cm1 log²
∴ R₁' = ( 72 * 10^-6 ) / ( 333 * 10^-9 log² )
= 31 Ω
For Cm2
t₁₂ = 40 μs , ∴ R₂' = 302.9 Ω
b) comparing calculated and measured values via percent error
errors
Rm1 - R₁' = 17.3 Ω
Rm2 - R₂' = 26.4 Ω
Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 412 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 412 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
Required:
Find ÎEint, 1, Q1, and W1 for the process at constant volume.
Answer:
ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Explanation:
Given the data in the question;
T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol
since helium is monoatomic;
Cv = (3/2)R, Cp = (5/2)R
W₁ = 0 J [ at constant volume or ΔV = 0]
Now for the first cylinder; from the first law of thermodynamics;
Q₁ = ΔE[tex]_{int[/tex],₁ + W₁
Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT
we substitute
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112
Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J
Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m along the axis and above the center of the loop
Answer:
[tex]B=2.91\ \mu T[/tex]
Explanation:
Given that,
The current in the loop, I = 2 A
The radius of the loop, r = 0.4 m
We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :
[tex]B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}[/tex]
Put all the values,
[tex]B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T[/tex]
So, the required magnetic field is equal to [tex]2.91\ \mu T[/tex].
Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
In what direction is the centripetal force directed?
Answer:
towards the center
Explanation:
that is the solution above
Explain the following defects of a simple electric cell:
a.Polarization,
ß. Local action.
Answer:
Explanation:
The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.
a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.
b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.
two point charges two point charges are separated by 25 cm in the figure find The Net electric field these charges produced at point a and point b
The image is missing and so i have attached it.
Answer:
A) E = 8740 N/C
B) E = -6536 N/C
Explanation:
The formula for electric field is;
E = kq/r²
Where;
q is charge
k is a constant with value 8.99 x 10^(9) N•m²/C²
A) Now, to find the net electric field at point A, the formula would now be;
E = (kq1/(r1)²) - (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point A
r2 is distance from charge q2 to point A.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m
r2 = 10 cm = 0.1 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))
E = 8740 N/C
B) similarly, electric field at point B;
E = (kq1/(r1)²) + (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point B
r2 is distance from charge q2 to point B.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 10 cm = 0.1 m
r2 = 25cm + 10 cm = 35 cm = 0.35 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))
E = -6536 N/C
Two balls of known masses hang from the ceiling on massless strings of equal length. They barely touch when both hang at rest. One ball is pulled back until its string is at 45 ∘, then released. It swings down, collides with the second ball, and they stick together.The problem can be divided into three parts: (1) from when the first ball is released and to just before it hits the stationary ball, (2) the two balls collide, and (3) the two balls swing up together just after the collision to their highest point. ..............conserved in parts (1) and (3) as the balls swing like pendulums. During the collision in part (2) ................. conserved as the collision is ................. Explain.Match the words in the left column to the appropriate blanks in the sentences on the rightboth energy and momentum areonly energy is only momentum is.........both energy and momentum are only energy is only momentum iselasticinelastic
Answer:
In parts 1 and 3 the energy
In part 2 moment. inelastic
conserved
Explanation:
In this exercise, we are asked to describe the conservation processes for each part of the exercise.
In parts 1 and 3 the energy is conserved since the bodies do not change
In part 2 the bodies change since they are united therefore the moment is conserved and part of the kinetic energy is converted into potential energy.
Energy
moment .inelastic
conserved
The two balls swing up together just after the collision to their highest point. energy is conserved.
What is the law of conservation of momentum?According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
When the first ball is released and just before it hits the stationary ball, The two balls collide, The two balls swing up together just after the collision to their highest point. energy is conserved.
The balls swing like pendulums. During the collision in part (2) energy is conserved as the collision is inelastic.
We are requested to describe the conservation methods for each element of the activity in this exercise.
Because the bodies do not change in sections 1 and 3, energy is conserved.
Because the bodies change in part 2 is joined, the moment is conserved and some of the kinetic energy is transformed into potential energy.
Hence the two balls swing up together just after the collision to their highest point. energy is conserved.
To learn more about the law of conservation of momentum refer to;
https://brainly.com/question/1113396
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
Yes, when an apple falls towards the earth, the apple gets accelerated and comes down due to the gravitational force of attraction used by the earth. The apple also exerts an equal and opposite force on the earth but the earth does not move because the mass of the apple is very small, due to which the gravitational force produces a large acceleration in it (a = F/m) but the mass of the earth is very large, the same gravitational force produces very small acceleration in the earth and we don't see the earth rising towards the apple.
