Answer:
the cubic meter (m3)
Explanation:
Volume is the measure of the 3-dimensional space occupied by matter, or enclosed by a surface, measured in cubic units. The SI unit of volume is the cubic meter (m3),
Answer:
please answer all component's
Explanation:
please answer all component's
Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work is needed to stretch the spring from 36 cm to 41 cm
Answer:
0.83 J of work
Explanation:
2 J of work is required to stretch a spring from 34cm to 46cm
So that is 12cm stretched with 2 J of work
We can make that 6cm for 1 J of work
So, we need the find the work for stretching 36cm to 41cm
Which is 5cm
So, What is the work required to stretch 5cm?
1 J of work for 6cm
x work for 5cm
So, by proportion method
1 : 6 :: x : 5
6 * x = 1 * 5
6x = 5
x = 5/6
= 0.83
So to stretch 36cm to 41cm we need 0.83 J of work
220V a.c. is more dangerous than 220V d.c why?
Answer:
220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.
(a) State Hook's law. [2]
(b) The walls of the tyres on a car are made of a rubber
compound. The variation with stress of the strain of a
specimen of this rubber compound is shown in Fig. 1.2.
As the car moves, the walls of the tyres end and straighten
continuously. Use Fig. 1.2 to explain why the walls of the
tyres become warm
[3]
hin
Fig. 1.2.
Hooke's law gives the relationship between the force applied and observed compression and expansion
(a) Hooke's law states that force applied is proportional to the deformation of an object
(b) The tube becomes warm from the excess of the energy absorbed but not given off back as the straightening of the tyre
The reason for the above explanation are as follows;
(a) Hooke's law states that for little deformations, the change in the dimension, Δx, of an extended or compressed object is directly proportional to the applied force, F
Mathematically, we get;
F = k × ΔL
[tex]k = \mathbf{\dfrac{F}{\Delta L}}[/tex]
Given that the material cross sectional area = A, and the original length of the material = L, we get;
F/A = Stress = σ, ΔL/L = strain = ε
[tex]\mathbf{Young's \ Modulus, \ E} = \mathbf{\dfrac{\sigma}{\varepsilon}} = \dfrac{\left(\dfrac{F}{A} \right) }{\left(\dfrac{\Delta L}{L} \right) } = \mathbf{\dfrac{F}{\Delta L } \times \dfrac{L}{A}}[/tex]
Therefore, Hooke's law can be expressed as a form of Young's Modulus for a given length to area ratio of a material
(b) The given graph of stress to strain curve, is attached, from which area under the curve gives the energy absorbed by the the material during deformation
Therefore during bending, the stress is increasing as shown in the top of the two curve, and the energy absorbed is given by the area under the curve
As the tyre straightens, the path of the stress change curve is given by the lower curve
The area under the top (bending) curve is larger than the area under the lower (straightening) curve, therefore, the energy absorbed during bending is larger than the energy given off during straightening
Energy absorbed < Energy released
The balance energy is transformed into other forms of energy, including heat energy, which is observed by the raising in temperature, warming, of the tyre
Energy absorbed = Energy released + Heat energy
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Here we will use a simple example to demonstrate the difference between the rms speed and the average speed. Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s. Find the rms speed for this collection. Is it the same as the average speed of these molecules
Explanation:
Given that,
Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.
The rms speed for this collection is as follows :
[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]
The average speed of these molecules is :
[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]
So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.
Required information
You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including
passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 m/s, the maximum descent speed
to be 10.0 m/s, and the maximum acceleration magnitude to be 1.80 m/s2. Ignore friction.
What is the minimum upward force that the supporting cables exert on the elevator car?
KN
Answer:
19,224 N
Explanation:
The given parameters are;
The mass limit of the elevator = 2,400 kg
The maximum ascent speed = 18.0 m/s
The maximum descent speed = 10.0 m/s
The maximum acceleration = 1.80 m/s²
Given that the acceleration due to gravity, g ≈ 9.81 m/s²
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;
[tex]F_{min}[/tex] = m·g - m·a
∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N
What would you expect to happen to the velocity of the bobber if the mass of the washers in the cylinder remained the same and the radius was doubled?
Answer:
The velocity becomes [tex]v\sqrt 2[/tex].
Explanation:
The force acting on the bobber is centripetal force.
The centripetal force is given by
[tex]F =\frac{mv^2}{r}[/tex]
when mass remains same, radius is doubled and the force is same, so the velocity is v'.
[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
On the average, in a ferromagnetic domain, permanent atomic magnetic moments are aligned ____ to one another.a. antiparallel b. parallel c. perpendicular d. alternately parallel and antiparallel e. randomly relative
Answer:
b. parallel
Explanation:
Ferromagnetism is a magnetism that is associated with iron and cobalt and nickel. Ferromagnetisms material are magnetics easily and in strong magnetic fields are magnetized by a defined limit called a situation. The force keeps magnetic moments of many atoms parallel to each other.Example 2.13 The acceleration a of a particle in a time t is given by the equation a = 2+ 5t^2. Find the instantaneous velocity after 3s. Solution
Answer:
the instantaneous velocity is 51 m/s
Explanation:
Given;
acceleration, a = 2 + 5t²
Acceleration is the change in velocity with time.
[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]
Therefore, the instantaneous velocity is 51 m/s
A 0.4 m long solenoid has a total of 356 turns of wire and carries a current of 79 A. What is the magnitude of the magnetic field at the center of the solenoid?
Answer:
B = 0.088 T
Explanation:
Given that,
The length of a solenoid, l = 0.4 m
No. of turns of wire, N = 356
Current, I = 79 A
We need to find the magnitude of the magnetic field at the center of the solenoid. It is given by the formula.
[tex]B=\mu_onI\\\\B=\mu_o \dfrac{N}{l}\times I\\\\B=4\pi \times 10^{-7}\times \dfrac{356}{0.4}\times 79\\\\B=0.088\ T[/tex]
So, the magnitude of the magnetic field at the center of the solenoid is 0.088 T.
once the object is seen clearly (Figure 5).
Cliary muscles
Nea
Obec
image
25 cm
FIGURE 3
dusion : Thus, we observe that the focal length of the eye
ically by the action of cilin
Answer:
Where is the figure ?????
You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 690.0 kg and was traveling eastward. Car B weighs 520.0 kg and was traveling westward at 74.0 km/h. The cars locked bumpers and slid eastward with their wheels locked for 6.00 m before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. 1) How fast (in kilometer per hour) was car A traveling just before the collision
Answer:
The speed of car A before collision is 3.5 km/h.
Explanation:
Mass of car A = 690 kg eastwards
Mass of car B = 520 kg at 74 km/h west wards
Distance, s = 6 m
coefficient of friction = 0.75
Let the speed after collision is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]
Let the initial speed of car A is v'.
Use conservation of momentum
690 x v' - 520 x 74 = (690 + 520) x 33.8
690 v' + 38480 = 40898
v' = 3.5 km/h
A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant
Answer:
[tex]T=8.1N[/tex]
Explanation:
From the question we are told that:
Mass m=0.40
Radius r=1.8m
Angle Beneath the Horizontal \theta =40 \textdegree
Speed v=5.0m/s
The Tension Angle
[tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]
[tex]\alpha=50 \textdegree[/tex]
Generally the equation for Tension is is mathematically given by
[tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]
[tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]
[tex]T=8.1N[/tex]
A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing
Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children plus sled is 55 kg. The mother has a mass of 61 kg. Find the static friction acting on the mother.
Answer:
f = 106.3 N
Explanation:
The force applied on the sled must be equal to the static frictional force to move the sled:
Tension Force Horizontal Component = Static Frictional Force
[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]
where,
T = Tension = 120 N
θ = angle of rope = 37°
μ = coefficient of static friction = ?
m = mass of children plus sled = 55 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]
Now, the static friction acting on the mother will be:
[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 106.3 N
If you double the current in a long straight wire, the magnetic field at a fixed point will... be cut in half. triple. double. quadruple.
Answer:
the magnetic field must double
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
Where the bold indicate vectors
With this expression we can see that if we double the current, keeping the same trajectory, the magnetic field must double
The pan flute is a musical instrument consisting of a number of closed-end tubes of different lengths. When the musician blows over the open ends, each tube plays a different note. The longest pipe is 0.31 m long.
What is the frequency of the note it plays? Assume room temperature of 20∘C.
Answer:
f = 276.6 Hz
Explanation:
This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.
In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is
L = λ/ 4
speed is related to wavelength and frequency
v = λ f
λ = v / f
we substitute
L = v / 4f
f = v / 4L
the speed of sound at 20ºC is
v = 343 m / s
let's calculate
f = [tex]\frac{343 }{4 \ 0.31}[/tex]
f = 276.6 Hz
What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 m
Answer:
The gauge pressure is equal to 147 kPa.
Explanation:
The pressure exerted by fluid is given by :
[tex]P=\rho gh[/tex]
Where
[tex]\rho[/tex] is density of water
h is height
So, put all the values,
[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]
or
P = 147 kPa
So, the gauge pressure is equal to 147 kPa.
Answer:
The gauge pressure is 147000 Pa.
Explanation:
Height, h = 15 m
density of water, d= 1000 kg/m^3
gravity, g = 9.8 m/s^2
The gauge pressure is the pressure exerted by the fluid.
The pressure exerted by the fluid is given by
P = h d g
P = 15 x 1000 x 9.8 = 147000 Pa
A marble rolling with speed 20cm/s rolls off the edge of a table that is 80cm high. How far horizontally from the table edge does the marble strike the floor
Answer:
8 cm
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 cm/s
Height (h) = 80 cm
Horizontal distance (s) =?
Next, we shall determine the time taken for marble to get to the ground. This can be obtained as follow:
Height (h) = 80 cm
Acceleration due to gravity (g) = 1000 cm/s²
Time (t) =?
t = √(2h/g)
t = √[(2 × 80)/1000]
t = √(160/1000)
T = √0.16
t = 0.4 s
Finally, we shall determine the horizontal distance travelled by the marble. This can be obtained as illustrated below:
Initial velocity (u) = 20 cm/s
Time (t) = 0.4 s
Horizontal distance (s) =?
s = ut
s = 20 × 0.4
s = 8 cm
Thus, the horizontal distance travelled by the marble is 8 cm.
The horizontal distance traveled by the marble is 8 cm.
The given parameters;
speed of the marble, v = 20 cm/sheight of the table, h = 80 cmThe time of motion of the marble is calculated as follows;
[tex]h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.8}{9.8} }\\\\t = 0.4 \ s[/tex]
The horizontal distance traveled by the marble is calculated as follows;
[tex]X = v_0_x t\\\\X = (20 \times 0.4)\\\\X = 8 \ cm[/tex]
Thus, the horizontal distance traveled by the marble is 8 cm.
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If 10 W of power is supplied to 1 kg of water at 100℃, how long will it take to for the water to completely boil away? The time calculated is a little less than actual time of boiling in practice. Why?
Answer:
t = 2.26 x 10⁵ s
Explanation:
The energy supplied to the water will be equal to the heat required for the boiling of water:
E = ΔQ
Pt = mL
where,
P = Power = 10 W
t = time = ?
m = mass of water = 1 kg
L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]
t = 2.26 x 10⁵ s
This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.
A single-turn circular loop of wire of radius 55 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10 s time interval, the magnitude of the field increases uniformly from 350 to 450 mT.
Required:
a. Determine the emf induced in the loop (in V). (Enter the magnitude.) V
b. If the magnetic field is directed out of the page, what is the direction of the current induced in the loop?
Answer:
Explanation:
Area of the loop = π x ( 55 x 10⁻³ )²
= 9.5 x 10⁻³ m²
Change in Magnetic flux dφ = 450 x 10⁻³ - 350 x 10⁻³ = 150x 10⁻³ Weber.
time dt =.10 s
emf induced = dφ / dt = 150x 10⁻³ Weber / .10 s
= 1.5 V .
b )
Magnetic field is directed outwards and it is increasing so according to Lenz's law , direction of induced current will be clockwise in the loop.
Answer:
(a) 9.5 mV
(b) clockwise
Explanation:
Radius, r = 55 mm
Time, t = 0.1 s
Change in magnetic field, B = 450 - 350 = 100 mT =0.1 T
(a) induced emf is given by
[tex]e = A \frac{dB}{dt}[/tex]
[tex]e = A \frac{dB}{dt}\\\\e=3.14\times 0.055\times0.055\times \frac{0.1}{0.1}\\\\e= 9.5 \times 10^{-3} V = 9.5 mV[/tex]
(b) According to the Lenz law, the direction of current is clockwise.
A 0.22LR caliber bullet has a mass of 1.90 g and a muzzle velocity of 500 m/s. The bullet is fired into a door made of a single thickness of pine boards, with a thickness of 0.75 in. The average stopping force exerted by the wood is 960 N. How fast (in m/s) would the bullet be traveling after it penetrated through the door
Answer:
The final speed of the bullet is 480.4 m/s.
Explanation:
mass of bullet, m = 1.9 g
initial speed, u = 500 m/s
thickness, d = 0.75 inch = 0.01905 m
Force, F = 960 N
Let the final speed is v.
According to the work energy theorem,
Work = change in kinetic energy
[tex]W = F d = 0.5 m{\left (v^2 - u^2 \right )}[/tex]
-960 x 0.01905 = 0.5 x 0.0019 x (v^2 - 500 x 500)
-18.288 = 0.00095 (v^2 - 250000)
v = 480.4 m/s
Why did the Prince go down on one knee?
Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction
Answer:
a) Light that passes through the floor to reveal yourself (not shadow).
b) 2 rays of light that bounce between 2 transparent media.
c) I don't know what is Diffraction?
The speed of light is the fastest in which medium
In vacuum, going at 2.99×10^8 m/s.
the product 17.10 ✕
Explanation:
pls write full question
A satellite measures a spectral radiance of 8 Watts/m2/um/ster at a wavelength of 10 microns. Assuming a surface emissivity of 0.90, what would be the estimated temperature
Charge q is 1 unit of distance away from the source charge S. Charge p is six times further away. The force exerted
between S and and q is __ the force exerted between S and p.
1/6
6 times
1/36
36 times
Answer:
Fq = k q Q / R^2
Fp = k q Q / (6 R)^2
The force exerted between S and p is 1 / 36 of that between S and q
or the force between S and q is 36 times that between S and p.
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B=+5\ m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]
So, the magnitude is 20 m/s and the direction is westward (negative sign).
What is the electric potential 15 cm above the center of a uniform charge density disk of total charge 10 nC and radius 20 cm?
a) 360 V
b) 450 V
c) 22.5 V
d) 0 V
Answer:
b) 450 V
Explanation:
We are given that
Total charge, q=10nC=[tex]10\times 10^{-9} C[/tex]
[tex]1nC=10^{-9}C[/tex]
Radius, r=20 cm=[tex]\frac{20}{100}=0.2m[/tex]
1 m=100 cm
x=15 cm=0.15 cm
We have to find the electrical potential 15 cm above the center of a uniform charge density disk .
We know that
[tex]\sigma=\frac{q}{A}=\frac{q}{\pi r^2}[/tex]
[tex]\sigma=\frac{10\times10^{-9}}{3.14\times (0.2)^2}[/tex]
Where [tex]\pi=3.14[/tex]
[tex]\sigma=7.96\times 10^{-8}C/m^2[/tex]
Electric potential,[tex]V=\frac{\sigma}{2\epsilon_0}(\sqrt{x^2+r^2}-x)[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]V=\frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}(\sqrt{(0.15)^2+(0.2)^2}-0.15)[/tex]
[tex]V=449.7 V\approx 450V[/tex]
Hence, option b is correct.
Answer:
The potential is given by 449.7 V.
Explanation:
radius of disc, R = 20 cm = 0.2 m
distance, x = 15 cm = 0.15 m
charge, q = 10 nC
surface charge density
[tex]\sigma = \frac{q}{\pi R^2}\\\\\sigma = \frac{10\times 10^{-9}}{3.14\times 0.2\times 0.2 }\\\\\sigma = 7.96\times 10^{-8} C/m^2[/tex]
The electric potential is given by
[tex]V=\frac{\sigma}{2\varepsilon 0}\left ( \sqrt{R^2 + x^2} - x \right )\\\\V = \frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}\left ( \sqrt{0.2^2 + 0.15^2} - 0.15 \right )\\\\V = 449.7 V[/tex]
