Answers
Answer:
W = F • ∆x
so for work to be done, a force and displacement has to be in the same direction. (Ex: a box is being pushed forward and it's also moving forward.)
Related Questions
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
Answers
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.
Learn more here:https://brainly.com/question/2411455
The speed of light is the fastest in which medium
Answers
In vacuum, going at 2.99×10^8 m/s.
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?
Answers
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.
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
Answers
Answer:
Where is the figure ?????
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
Answers
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
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
Answers
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.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
Answers
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
4. What is constant displacement
Answers
Answer:
The displacement is constant when the rate of change in velocity is constant, for example, if a car’s speed increases by 5 km every 5 minutes it is called constant displacement
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
Answers
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?
Answers
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]
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
Answers
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]
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
Answers
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 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
Answers
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 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
Answers
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]
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
Answers
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
Two bodies with masses m1 and m2 are both moving east with velocities of magnitudes v1 and v2, where v1 is less than v2. The magnitude of the velocity of the center of mass of this system of two bodies is:_________
a. less than v1.
b. equal to v1.
c. equal to theaverage of v1 and v2.
d. greater than v1 and less than v2.
e. greater than v2.
Answers
Answer:
the correct one is D
Explanation:
The velocity of the center of mass of a body is
[tex]v_{cm} = \frac{1}{M } \ \sum m_i v_i[/tex]vcm = 1 / M sum mi vi
M = [tex]\sum m_i[/tex] sum my
in this case we have two bodies
v_{cm} = [tex]\frac{1}{m1+m2} \ ( m_1v_1 +m_2v_2)[/tex]1 / m1 + m2 (m1 v1 + m² v2)
v_{cm} = [tex]\frac{m_1}{m1+m2} \ v_1 + \frac{m2}{m_1+m_2} \ v_2[/tex]
we can see that the velocity is an average between v₁ and v₂, but the exact value depends on the ratio of the masses of the bodies
when checking the answers the correct one is D
If a thin rod of platinum is placed in an external magnetic field of 1.2000 T, with its axis parallel to that field, what will be the magnetic field inside the rod?
Answers
Answer:
[tex]B_1=1.29967T[/tex]
Explanation:
From the question we are told that:
Relative permeability Platinum [tex]\mu= 1.00026[/tex]
Magnetic field [tex]B_e= 1.3000 T[/tex]
Generally the equation for Magnetic field is mathematically given by
[tex]B_i=\frac{B_e}{\mu}[/tex]
[tex]B_i=\frac{1.3}{1.00026}[/tex]
[tex]B_1=1.29967T[/tex]
Simple physics question
Answers
Answer:
option (b) is the correct answer
What is the expression of thousand joules
Answers
1000 watts
Joules are a measure of energy; Watts are a measure of the rate of energy generated or consumed. A Watt is defined as one joule per second. So if your 1000 Joules flowed in one second, the power would be 1000 Watts or 1 kW. But if it took 1000 seconds for the 1000Joules to flow, the power would be 1 Watt.
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
Answers
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
the change of potential with respect to distance is equal to electric_____?
Answers
Answer:
energy
Explanation:
I am not sure but I think it's electric energy
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.
Answers
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.
Answers
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
47. Calculate the final steady temperature obtained when
6.0g of water at 50C is mixed with 3.0g of water at
40C. [specific heat capacity of water is 4200jkg K-]
Answers
Answer:
the final steady temperature of the mixture is 46.67 ⁰C
Explanation:
Given;
mass of first water,m₁ = 6 g = 0.006 kg
initial temperature, t₁ = 50 ⁰C
mass of the second water, m₂ = 3.0 g = 0.003 kg
initial temperature of the second water, t₂ = 40 ⁰C
specific heat capacity of water, C = 4,200 J/kg.K
Let the final temperature of the mixture = T
Based on Law of conservation of energy, the temperature of the final mixture is calculated as follows;
m₁c(t₁ - T) = m₂c(T - t₂)
0.006 x 4200 x (50 - T) = 0.003 x 4200 x (T - 40)
1,260 - 25.2T = 12.6T - 504
1,260 + 504 = 12.6T + 25.2T
1764 = 37.8T
T = 1764/37.8
T = 46.67 ⁰C
Therefore, the final steady temperature of the mixture is 46.67 ⁰C
220V a.c. is more dangerous than 220V d.c why?
Answers
Answer:
220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.
the product 17.10 ✕
Answers
Explanation:
pls write full question
Need help! Need help! Need help! Need help! Need help! Need help!
Answers
Answer:
i can help you i know this answer
Answer: the side two are 50 then the other two are 140 i thank
Explanation:
Why did the Prince go down on one knee?
Answers
Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 180 A and a height of 80 m above the ground, what magnetic field does the line produce at ground level?
Answers
Answer:
B = 4.5 x 10⁻⁷ T = 0.45 μT
Explanation:
From Ampere's law the magnetic field at a certain radius from straight current-carrying wire is given by the following formula:
[tex]B = \frac{\mu_oI}{2\pi r}[/tex]
where,
B = magnetic field = ?
[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ H/m
I = current = 180 A
r = radius = 80 m
Therefore,
[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(180\ A)}{2\pi (80\ m)}\\\\[/tex]
B = 4.5 x 10⁻⁷ T = 0.45 μT
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.
Answers
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
