Answer:
100j
Explanation:
10 A turning pork creates sound cares
with
Frequency of 170Hz: To the
speed of sound in is in 340mls
calculate the wave
wave length
of
in air is
the sound wales.
Answer:
2m
Explanation:
wavelength=speed/frequency
=340/170
=2m
Please show steps as to how to solve this problem
Thank you!
Explanation:
Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:
[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]
[tex] -m_1gx + m_2gd_2 = 0[/tex]
[tex]m_1x = m_2d_2[/tex]
Solving for x,
[tex]x = \dfrac{m_2}{m_1}d_2[/tex]
[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]
When should a line graph be used?
A. When the independent variable is continuous and does not show a relationship to the dependent variable
B. When the independent variable is composed of categories and does not show a relationship
C. When the independent variable is continuous and shows a casual link to the dependent variable
D. When there is no independent variable
A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
Explanation:
Given
Acceleration of the pebble is
At t=0, velocity is
considering horizontal motion
[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]
Velocity acquired during this time
[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]
Consider vertical motion
[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]
Net velocity is
[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]
Angle made is
[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]
7. If a load of 300N is pulled along the inclined plane shown in the figure, answer the following. B 200 N 0.5m 2m 300 N А i. Calculate the VR and MA of the inclined plane. Calculate the input work and output work. ii.what efficiency of inclined plane?iv.what should be the length of inclined plane if same load has to be pulled with a50N effort for the same efficiency as above
Explanation:
700n I think friend .. if worng
A small ball of uniform density equal to 1/2 the density of water is dropped into a pool from a height of 5m above the surface. Calculate the maximum depth the ball reaches before it is returned due to its bouyancy. (Omit the air and water drag forces).
Answer:
1.67 m
Explanation:
The potential energy change of the small ball ΔU equals the work done by the buoyant force, W
ΔU = -W
Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5
ΔU = mgΔh
ΔU = ρVgΔh
Now, W = ρ'VgΔh' where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h
So, ΔU = -W
ρVgΔh = -ρ'VgΔh'
ρVg(h - 5) = -ρ'Vgh
ρ(h - 5) = -ρ'h
Since the density of the small ball equals 1/2 the density of water,
ρ = ρ'/2
ρ(h - 5) = -ρ'h
(ρ'/2)(h - 5) = -ρ'h
ρ'(h - 5)/2 = -ρ'h
(h - 5)/2 = -h
h - 5 = -2h
h + 2h = 5
3h = 5
h = 5/3
h = 1.67 m
So, the maximum depth the ball reaches is 1.67 m.
a tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?
Answer: 2.4 km/hr
Explanation:
Distance = 600m
Time= 15 minutes = 15 x 60 second/minute = 900 seconds
Speed = [tex]\frac{Distance}{Time}[/tex] = [tex]\frac{600}{900}[/tex] = [tex]\frac{2}{3}[/tex] m/sec
⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)
What best describes a societal law
Answer:
Societal laws are based on the behavior and conduct made by society or government.hope it helps.stay safe healthy and happy.A pump lifts 400 kg of water per hour a height of 4.5 m .
Part A
What is the minimum necessary power output rating of the water pump in watts?
Express your answer using two significant figures.
Part B
What is the minimum necessary power output rating of the water pump in horsepower?
Express your answer using two significant figures.
Answer:
Power = Work / Time
P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts
Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp
Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb
Answer:
i₀ = V / R_i
Explanation:
For this exercise we use Ohm's law
V = i R
i = V / R
the equivalent resistance for
[tex]\frac{1}{R_{eq}}[/tex] = ∑ [tex]\frac{1}{R_i}[/tex]
if all the bulbs have the same resistance, there are N bulbs
[tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]
R_{eq} = R_i / N
we substitute
i = N V / Ri
where i is the total current that passes through the parallel, the current in a branch is
i₀ = i / N
i₀ = V / R_i
A 5 kg object is moving in a straight-line with an initial speed of v m/s. It takes 13 s for the speed of the object to increase to 13 m/s and it kinetic energy increases at a rate of 15 J/s. What is the initial speed v (in m/s)?
The object's kinetic energy changes according to
dK/dt = 15 J/s
If v is the object's initial speed, then its initial kinetic energy is
K (0) = 1/2 (5 kg) v ²
Use the fundamental theorem of calculus to solve for K as a function of time t :
[tex]K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t[/tex]
After t = 13 s, the object's kinetic energy is
K (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J
Put this as the left side in the equation above for K(t) and solve for v :
[tex]422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)[/tex]
==> v ≈ 9.5 m/s
Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.
The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explanation:
The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Initial: a 0 0
Equilibrium: (a - x) 0 2x
Total pressure = (a - x) + 2x = a + x
As it is given that the total pressure is 1.1 atm.
So, a + x = 1.1
a = 1.1 - x
Now, expression for equilibrium constant for this equation is as follows.
[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]
Hence, the value of 'a' is calculated as follows.
a + x = 1.1 atm
a = 1.1 atm - x
= 1.1 atm - 0.31 atm
= 0.79 atm
Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explain the following observations:
a) A balloon filled with hydrogen gas floats in air;
B) A ship made of steel floats on water.
Answer and Explanation:
a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.
b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x-axis.
Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?
Answer:
7.9 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:
[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]
Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:
[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]
Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]
Plug everything in:
7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]
(1590kg the initial weight plus the weight of the added passenger)
An elevator with its occupants weighs 2400 N and is supported by a vertical cable. What is the tension in the cable if the elevator is moving up with its speed decreasing at a rate of 1.7
Answer:
Hope you find it useful. please correct me if I am wrong
The tension in the cable if the elevator is moving upward with its speed decreasing at a rate of 1.7 m/s² is equal to 1983.67 N.
What is tension?Tension can be described as a force acting along the length of a medium such as a rope, mainly a force carried by a flexible medium.
Tension can be defined as an action-reaction pair of forces acting at each end of the elements. The tension force is in every section of the rope in both directions, apart from the endpoints. Each endpoint of the rope experience tension and force from the weight attached.
Given the force due to the weight of the elevator = mg = 2400N
m = 2400/9.8 Kg
The elevator deaccelerating while moving upward, a = -1.7 m/s²
According to Newton's 3rd law: T - mg = ma
T - 2400 = (2400/9.8) × (-1.7)
T = 2400 - 416.32
T = 1983.67 N
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I need help with this physics question.
Answer:
5.04 m
Explanation:
You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.
Answer:
5.0384m
Explanation:
% increase = 100 x (Final - Initial / | initial | )
( |~~| Bars indicate absolute value since you can't have a negative height)
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.
What is the minimum angular spread (in rad) of a 534 nm wavelength manganese vapor laser beam that is originally 1.19 mm in diameter
Answer:
Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad
Explanation:
GIven;
Wavelength of manganese vapor laser beam = 534 nm = 534 x 10⁻⁹ m
Diameter = 1.19 mm = 1.19 x 10⁻³ m
Find:
Minimum angular spread (in rad)
Computation:
Minimum angular spread (in rad) = 1.22[Wavelength / Diameter]
Minimum angular spread (in rad) = 1.222[(534 x 10⁻⁹) / (1.19 x 10⁻³)]
Minimum angular spread (in rad) = 2[448.73 x 10⁻⁶]
Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
Answer:
[tex]\sigma=0.014\ C/m^2[/tex]
Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,
[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]
So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].
Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that
is 6.0mm in diameter and lift a 1700kg object. Then, the wire elongates
0.17m. Calculate the Young’s modulus for the rope material.
Answer:
Young's modulus for the rope material is 20.8 MPa.
Explanation:
The Young's modulus is given by:
[tex] E = \frac{FL_{0}}{A\Delta L} [/tex]
Where:
F: is the force applied on the wire
L₀: is the initial length of the wire = 3.1 m
A: is the cross-section area of the wire
ΔL: is the change in the length = 0.17 m
The cross-section area of the wire is given by the area of a circle:
[tex] A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2} [/tex]
Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:
[tex] F = T_{w} = W_{o} [/tex]
Where:
[tex] T_{w} [/tex]: is the tension of the wire
[tex]W_{o} [/tex]: is the weigh of the object = mg
m: is the mass of the object = 1700 kg
g: is the acceleration due to gravity = 9.81 m/s²
[tex] F = mg = 1700 kg*9.81 m/s^{2} = 16677 N [/tex]
Hence, the Young's modulus is:
[tex] E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa [/tex]
Therefore, Young's modulus for the rope material is 20.8 MPa.
I hope it helps you!
If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is half of her average speed when she swam to the other side.a) trueb) false
Answer:
false.
Explanation:
Ok, we define average velocity as the sum of the initial and final velocity divided by two.
Remember that the velocity is a vector, so it has a direction.
Then when she goes from the 1st end to the other, the velocity is positive
When she goes back, the velocity is negative
if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:
AV = (V + (-V))/2 = 0
While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.
So we already can see that the average velocity will not be equal to half of the average speed.
The statement is false
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p
Answer:
a. 8p
Explanation:
We are given that
Radius of hollow sphere , R1=R
Density of hollow sphere=[tex]\rho[/tex]
After compress
Radius of hollow sphere, R2=R/2
We have to find density of the compressed sphere.
We know that
[tex]Density=\frac{mass}{volume}[/tex]
[tex]Mass=Density\times volume=Constant[/tex]
Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]
Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]
Using the formula
[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]
[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]
[tex]\rho_2=8\rho[/tex]
Hence, the density of the compressed sphere=[tex]8\rho[/tex]
Option a is correct.
Which of the following statements is correct about the magnitude of the static friction force between an object and a surface?
a. Static friction depends on the mass of the object.
b. Static friction depends on the shape of the object.
c. Static friction depends on what the object is made of but not what the surface is made of.
d. None of the above is correct.
Answer:
Static friction depends on the mass of the object.
Explanation:
Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;
1) nature of the object and the surface(how rough or smooth the surfaces are)
2)surface area of the object and the surface
3) mass of the object
Since;
F=μmg
Where;
μ= coefficient of static friction
m= mass of the object
g= acceleration due to gravity
Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.
The following contribute to accidents when a teen driver has other teens as passengers
Answer:
When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
What are the factors contribute to accidents when a teen driver has other teens as passengers?When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.
Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.
Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
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MCQ
................
Answer:
I think it would be (-7 C )..
NEED HELP ASAP. Please show all work.
A point on a rotating wheel (thin hoop) having a constant angular velocity of 200 rev/min, the wheel has a radius of 1.2 m and a mass of 30 kg. ( I = mr2 ).
(a) (5 points) Determine the linear acceleration.
(b) (4 points) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Look at work
Explanation:
a) I am not sure if you want tangential or centripetal but I will give both
Centripetal acceleration = r*α
Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2
Tangential acceleration = ω^2*r
convert 200rev/min into rev/s
200/60= 10/3 rev/s
a= 100/9*1.2= 120/9= 40/3 m/s^2
b) Rotational Kinetic Energy = 1/2Iω^2
I= mr^2
Plug in givens
I= 43.2kgm^2
K= 1/2*43.2*100/9=2160/9=240J
A small plane tows a glider at constant speed and altitude. If the plane does 2.00 * 105 J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the tow rope and the horizontal
Answer:
θ = 57.4°
Explanation:
The complete formula to find out the work done by the plane is as follows:
[tex]W = FdCos\theta[/tex]
where,
W = Work = 200000 J
F = Force = Tension = 2560 N
d = distance = 145 m
θ = angle between rope and horizontal = ?
Therefore,
[tex]200000\ J = (2560\ N)(145\ m)Cos\theta\\\\Cos\theta = \frac{200000\ J}{371200\ J}\\\\\theta = Cos^{-1}(0.539)[/tex]
θ = 57.4°
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]
So, the new work is more than 130 J.