Answer:
Given the potential, [tex] V = Ax^l+By^m+Cz^n+D [/tex]
The components of the electric field are:
[tex]E_x = \frac{-dV}{dx} = -Alx^l^-^1[/tex]
[tex]E_y = \frac{-dV}{dy} = - Bmy^m^-^1[/tex]
[tex]E_z = \frac{-dV}{dz} = - nCzn^n^-^1[/tex]
Let's calculate the potential difference for all given points.
[tex] V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10 [/tex]
[tex]=> D = 10[/tex]
[tex] V(1, 0, 0) = 4V => A + 10 = 4 [/tex]
Solving for A, we have:
[tex] A = 4 - 10 [/tex]
[tex] A = -6 [/tex]
[tex] V(0, 1, 0) = 6V => B + 10 = 6 [/tex]
Solving for B, we have:
[tex] B = 6 - 10[/tex]
[tex] B = -4 [/tex]
[tex] V(0, 0, 1) = 8V => C + 10 = 4 [/tex]
Solving for C, we have:
[tex] C = 8 - 10 [/tex]
[tex] C = -2 [/tex]
For all given points, let's calculate the magnitude of electric field as follow:
[tex]E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16[/tex]
[tex]Al = -16[/tex]
Solving for l, we have:
[tex]l = \frac{-16}{A}[/tex]
From above, A = -6
[tex]l = \frac{-16}{-6}[/tex]
[tex]l = \frac{8}{3}[/tex]
[tex] E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16 [/tex]
[tex]Bm = -16[/tex]
Solving for m, we have:
[tex]m = \frac{-16}{A}[/tex]
From above, B = -4
[tex]m = \frac{-16}{-4}[/tex]
[tex]m = 4[/tex]
[tex] E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16 [/tex]
[tex]nC = - 16[/tex]
Solving for n, we have:
[tex]n = \frac{-16}{C}[/tex]
From above, C = -2
[tex]n = \frac{-16}{-2}[/tex]
[tex]n = 8[/tex]
a small sphere of
mass 0.25 g that carries a charge of 9.0 × 10−10 C.
Two parallel vertical infinite charged sheets of
charge densities σ1= -30 × 10−6 C/m2 and σ2= ab ×
10−6 C/m2 respectively. The sphere is attached to
one end of a very thin silk string 5.0 cm long. The
other end of the string is attached to the 2nd sheet as
shown in the figure. At equilibrium, the string will
make an angle (ϴ) with the vertical. Calculate the
angle that the string makes with the vertical?
Answer:
θ = 39.7º
Explanation:
In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left
Y Axis
[tex]T_{y}[/tex] -W = 0
[tex]T_{y}[/tex] = W
X axis
-[tex]F_{e1}[/tex] - F_{e2} + Tₓ = 0
let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical
sin θ = Tₓ / T
cos θ = T_{y} / t
Tₓ = T sin θ
T_{y} = T cos θ
let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.
F = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate
F = 2E A = q_{int} / ε₀
let's use the concept of surface charge density
σ = q_{int} / A
we substitute
2E A = σ A /ε₀
E = σ / 2ε₀
this is the field created by each plate. The electric force is
[tex]F_{e}[/tex] = q E
for plate 1 with σ₁ = -30 10⁻⁶ C / m²
F_{e1} = q σ₁ /2ε₀
for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²
F_{e2} = q σ₂ /2ε₀
we substitute and write the system of equations
T cos θ = mg
- q σ₁ / 2ε₀ - q σ₂ /2ε₀ + T sinθ = 0
we introduce t in the second equations
- q /2 ε₀ (σ₁ + σ₂) + (mg / cos θ) sin θ = 0
mg tan θ = q /2ε₀ (σ₁ + σ₂)
θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)
data indicates the mass of 0.25 g = 0.25 10⁻³ kg
give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²
let's calculate
θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2 8.85 10⁻¹² 0.25 10⁻³ 9.8))
θ = tan⁻¹ 8.3 10⁻¹)
θ = 39.7º
What is the momentum of a 2 kg ball traveling at 2m/s
Answer:
4
Explanation:
p=m×v
m=2kg
v=2m/s
2×2=4
Answer:
4kgms/1
Explanation:
p=m×v
=2kg×2m/s
=4kgms/1
When a voltage difference is applied to a piece of metal wire, a 8-mA current flows through it. If this metal wire is now replaced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire
Answer:
Explanation:
The resistance of a wire can be given by the following expression
[tex]R=\frac{\rho\times L}{A }[/tex]
where R is resistance , ρ is specific resistance , L is length of wire and A is cross sectional area
specific resistance of metals are almost the same . So in the present case ρ and l are same . Hence the formula becomes
R = k / A where k is a constant .
The diameter of wire becomes two times hence area of cross section becomes 4 times or 4A .
Resistance becomes 1/4 times . Hence if resistance of metal wire is R , resistance of silver wire will be R / 4 .
current = voltage / resistance
In case of metal wire
8 x 10⁻³ = V / R
In case of silver wire
I = V / (R / 4 ) , I is current , V is potential difference .
I = 4 x V/R
= 4 x 8 x 10⁻³ A
= 32 mA.
Suppose that a uniform electric field exists in a certain region of space. Now consider a mathematical plane surface of area A. To maximize the flux through this surface, the face of the plane (not its normal)
Answer: The normal of the plane must be parallel to the electric field vector.
Explanation:
the normal to the surface is defined as a versor that is perpendicular to the plane.
Now, if the angle between this normal and the line of the field is θ, we have that the flux can be written as:
Φ = E*A*cos(θ)
Where E is the field, A is the area and θ is the angle already defined.
Now, this maximizes when θ = 0.
This means that the normal of the surface must be parallel to the electric field
Light requires 4.5 years to travel from the nearest star to earth. If we could travel there in a spaceship going 90% of the speed of light, the trip would require 5.0 years according to clocks on earth. How much time would pass for the passengers in the ship
Answer:
Time according to earth clock (T0) = 0.22 years (Approx)
Explanation:
Given:
Time taken by light = 4.5 years
Time taken by ship = 5 years
Speed of light = c
Speed of ship (v) = 90% of c = 0.9c
Find:
Time according to earth clock (T0) = ?
Computation:
Time dilation is ,
[tex]T(Difference) = \frac{T0}{\sqrt{1-\frac{v^2}{c^2} } }\\\\ (5-4.5)= \frac{T0}{\sqrt{1-\frac{(0.9c)^2}{c^2} } }\\\\ 0.5=\frac{T0}{\sqrt{1-0.81} }\\\\T0 =0.2179[/tex]
Time according to earth clock (T0) = 0.22 years (Approx)
Popping popcorn is a thermodynamic process. Assume the pot remains covered while popcorn is being popped and the contents of the pot are the system. Which of the following correctly describes some feature of the system or what happens to the system undergoing this thermodynamic process?A. W > 0.B. Q>0.C. Tincreases.D. AU < 0.E. Pincreases.F. V-constant.
Answer:
the answer is C
Explanation:
Mark me brainlest
Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI), a measure of the likelihood of injury, is defined as SI = a5/2t, where a is the acceleration in multiples of g and t is the time the acceleration lasts (in seconds). In one set of studies of rear end collisions, a person's velocity increases by 12 km/h with an acceleration of 35 m/s2.(a) What is the severity index for the collision? (s)(b) How far does the person travel during the collision if the car was initially moving forward at 7.0 km/h? (m)
Answer:
a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.
Explanation:
a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:
[tex]a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]a' = 3.569[/tex]
The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:
[tex]v_{f} = v_{o} + a \cdot t[/tex]
Where:
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v_{f}[/tex] - Final speed, measured in meter per second.
[tex]a[/tex] - Acceleration, measured in [tex]\frac{m}{s^{2}}[/tex].
[tex]t[/tex] - Time, measured in seconds.
[tex]t = \frac{v_{f}-v_{o}}{a}[/tex]
[tex]t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }[/tex]
[tex]t = 0.095\,s[/tex]
Lastly, the severity index is now determined:
[tex]SI = \frac{a'^{5}}{2\cdot t}[/tex]
[tex]SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}[/tex]
[tex]SI = 3047.749[/tex]
b) The initial and final speed of the injured are [tex]1.944\,\frac{m}{s}[/tex] and [tex]5.278\,\frac{m}{s}[/tex], respectively. The travelled distance can be determined from this equation of motion:
[tex]v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s[/tex]
Where [tex]\Delta s[/tex] is the travelled distance, measured in meters.
[tex]\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s = 0.345\,m[/tex].
The carbon isotope C-14 is used for carbon dating. C-14 (mass 2.34 x 10-26 kg) decays by beta-decay, in which the nucleus emits an electron (the beta particle) and a subatomic particle called a neutrino. In one such decay, the electron and neutrino are emitted at right angles to each other. The electron (mass 9.11 x 10-31 kg) has a speed of 5.0 x 107 m/s and the neutrino has a momentum of 8.0 x 10-24 kg m/s. What is the recoil speed of the daughter nucleus?
Answer:
Explanation:
We shall apply the law of conservation of momentum to calculate the momentum and the speed of daughter nucleus .
Since the velocity of electron is very high we shall apply relativistic formula to calculate its momentum .
= [tex]\frac{mv}{\sqrt{1-(\frac{v}{c})^2 } }[/tex]
[tex]= \frac{9.11 \times 10^{-31}\times 5\times 10^7}{\sqrt{1-(\frac{5\times10^7}{3\times 10^8})^2 } }[/tex]
45.55 x 10⁻²⁴ x 1.176
= 53.58 x 10⁻²⁴ .
momentum of neutrino = 8 x 10⁻²⁴ . They are perpendicular to each other so total momentum
= √ [( 53.58 x 10⁻²⁴ )²+(8 x 10⁻²⁴)²]
= 54.17 x 10⁻²⁴
Hence the momentum of recoiled daughter nucleus will be same but in opposite direction
velocity of recoil = momentum / mass
= 54.17 x 10⁻²⁴ / 2.34 x 10⁻²⁶
= 23.15 x 10² m /s
Which angle is the angle of reflection? 1, 2, 0R 3?
Answer:
The answer is angle 2.
Explanation:
Angle 2 is the Reflected Ray of angle 1.
Angle 3 is the Refracted angle.
Angle 2 is the angle of reflection. the light incident will bounce to the opposite end after striking is called the reflection.
What is the law of reflection?The law of reflection specifies that upon reflection from a downy surface, the slope of the reflected ray is similar to the slope of the incident ray.
The reflected ray is consistently in the plane determined by the incident ray and perpendicular to the surface at the point of reference of the incident ray.
When the light rays descend on the smooth surface, the angle of reflection is similar to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in a similar plane.
In the reflection, the light incident will bounce to the opposite end after striking. Angle 2 satisfies the properties of the reflection.
Hence angle 2 is the angle of reflection.
To learn more about the law of reflection refer to the link;
brainly.com/question/12029226
On the moon, what would be the force of gravity acting on an object that has a mass of 7kg?
Answer:
Force of gravity, F = 70 N
Explanation:
It is required to find the force of gravity acting on an object that has a mass of 7 kg. Force of gravity always acts in downward direction.
The force of gravity is equal to the weight of an object. It is given by :
[tex]F=mg[/tex]
g = acceleration due to gravity, for Earth, g = 10 m/s²
So,
[tex]F=7\times 10\\\\F=70\ N[/tex]
So, 70 N of force of acting on an object.
An antiproton is identical to a proton except it has the opposite charge, −e. To study antiprotons, they must be confined in an ultrahigh vacuum because they will annihilate−producing gamma rays−if they come into contact with the protons of ordinary matter. One way of confining antiprotons is to keep them in a magnetic field. Suppose that antiprotons are created with a speed of 1.5 × 10^4 m/s and then trapped in a 4.5 mT magnetic field.What minimum diameter must the vacuum chamber have to allow these antiprotons to circulate without touching the walls?Express your answer with the appropriate units.d = _________
Answer:
Explanation:
We know that, the force responsible for circulating in circular path is the centripetal force given by the force on charged particle due to magnetic field.
Here the charge is antiproton is
p = -1.6 * 10⁻¹⁹C
the speed of proton is given by 1.5 * 10 ⁴ m/s
the magnetic field is B = 4.5 * 10⁻³T
we have force due to magnetic field is equal to centripetal force
Bqv = mv² / r
Bq = mv / r
[tex]r = \frac{mv}{Bq} \\\\r=\frac{mv}{Bq} \\\\r=\frac{1.67 \times 10^-^2^7\times 1.5 \times 10^4}{4.5 \times 10^-^3\times 11.6\times 10^-^1^9} \\\\r=347.9\times 10^-^4\\\\r=3.479cm[/tex]
The diameter d of the vacuum chamber have to allow these antiprotons to circulate without touching the walls is
d = 2r
d = 2 * 3.479
d = 6.958
d ≅ 7cm
R1=3 ohms
R2=6 ohms
R4=18 ohms
R5= 15 ohms
R5=9 ohms
90 volts
What is the current running through the entire circuit?
Answer: current I = 1.875A
Explanation:
If the resistors are connected in series,
Then the equivalent resistance will be
R = 6 + 18 + 15 + 9
R = 48 ohms
Using ohms law
V = IR
Make current I the subject of formula
I = V/R
I = 90/48
I = 1.875A
And if the resistors are connected in parallel, the equivalent resistance will be
1/R = 1/6 + 1/18 + 1/15 + 1/9
1/R = 0.166 + 0.055 + 0.066 + 0.111
R = 1/0.3999
R = 2.5 ohms
Using ohms law
V = IR
I = 90/2.5
Current I = 35.99A
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current of 27.5 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 295 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?
Answer:
The y-value of the line in the xy-plane where the total magnetic field is zero [tex]U = 0.1355 \ m[/tex]
Explanation:
From the question we are told that
The distance of wire one from two along the y-axis is y = 0.340 m
The current on the first wire is [tex]I_1 = (27.5i) A[/tex]
The force per unit length on each wire is [tex]Z = 295 \mu N/m = 295*10^{-6} N/m[/tex]
Generally the force per unit length is mathematically represented as
[tex]Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}[/tex]
=> [tex]\frac{\mu_o I_1I_2}{2\pi y} = 295[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value of [tex]\mu_o = 4\pi *10^{-7} \ N/A2[/tex]
substituting values
[tex]\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}[/tex]
=> [tex]I_2 = 18.23 \ A[/tex]
Let U denote the line in the xy-plane where the total magnetic field is zero
So
So the force per unit length of wire 2 from line U is equal to the force per unit length of wire 1 from line (y - U)
So
[tex]\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }[/tex]
substituting values
[tex]\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }[/tex]
[tex]6.198 -18.23U = 27.5U[/tex]
[tex]6.198=45.73U[/tex]
[tex]U = 0.1355 \ m[/tex]
A tank holds a 2.38-m thick layer of oil that floats on a 1.24-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.27 and 1.81, respectively. A ray originating at O reaches the brine-oil interface at the critical angle. What is the distance of this point from the axis?
Answer:
1.22m
Explanation:
Since
sinθ = refraction-of-the-oil/refraction-of-the-brine =1.27/1.81 = 0.702
θ = [tex]sin^-^1[/tex](0.702)
Hence
Critical angle = θ = 44.58°
tan(θ) = d/1.24
tan(44.58°) = d/1.24
Hence, 0.98 = d/1.24
The distance d = 0.98 x 1.24 = 1.22m
Write the second law of motion’s formula and its unit.
please answer this question, for 13 points!!!!!!!!!
Answer:
The Formula Is F = m * a And It's Units Is (kg)(m/s2)
Explanation:
A segment of wire of length L is along the x axis centered at x=0. Which of the following is a correct integral expression for the magnetic field at point P (centered on the wire segment at y=b) due to the current I flowing left to right in the segment of length L? In all answers below the limits of integration are from -L/2 to L/2.
a. μ0I/4π∫ dx b/(b2 + x2)3/2 kb. μ0I/4π∫ dx b/(b2 + x2)3/2 j c. μ0I/4π∫ dx /(b2 + x2) kd. -μ0I/4π∫ dx /(b2 + x2)1/2 ke. none of 1-5
Answer:
b. μ0I/4π∫ dx b/(b2 + x2)³/² j
Explanation:
The wire of length L centered at origin ( x =0 and y=0 ) carries current of I . We have to find magnetic field at point ( x = 0 , y = b ) .
First of all we shall consider magnetic field due to current element idx which is at x distance away from origin . magnetic field
dB = [tex]\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }[/tex]
component of magnetic field along y- axis at point P
[tex]\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }cos\theta[/tex]
where θ is angle between y - axis and dE .
component of magnetic field along y- axis at point P
[tex]\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }\times \frac{b}{\sqrt{x^2+b^2} }[/tex]
[tex]\frac{\mu _0ibdx}{4\pi(x^2+b^2)^\frac{3}{2} }[/tex]
The same magnetic field will also exist due to current element dx at x distance away on negative x - axis
The perpendicular component will cancel out .
This is magnetic field dE due to small current element
Magnetic field due to whole wire
[tex]\int\limits^\frac{L}{2} _\frac{-L }{ 2 } }\frac{\mu _0ibdx}{4\pi(x^2+b^2)^\frac{3}{2} } \, dx[/tex]
The sound of a tuba is very low. Why?
0
O When you blow into a tuba the air vibrates very strongly.
) O When you blow into a tuba the air vibrates very quickly.
O When you blow into a tuba the air vibrates very softly.
O When you blow into a tuba the air vibrates very slowly.
Answer:
When you blow into a tuba the air vibrates very slowly.
Explanation:
Tuba is a buzz instrument ie sound is produced in it with the help of lip vibration . It is the lowest pitched musical instrument in the brass family .
Due to absence of resonance in it , it produces music of lowest pitch , So when one blows into it the air column of the instrument vibrates very slowly producing low pitched sound.
Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0.034 m. You wish the mass to have a kinetic energy of 13.0 Joules when released. How strong will the string need to be
Answer:
T = 764.41 N
Explanation:
In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:
[tex]F_c=m\frac{v^2}{r}[/tex] (1)
m: mass object = 2.3 kg
r: radius of the circular orbit = 0.034 m
v: tangential speed of the object
However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:
[tex]K=\frac{1}{2}mv^2[/tex]
You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:
[tex]v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}[/tex]
you replace this value of v in the equation (1). Also, you replace the values of r and m:
[tex]F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N[/tex]
hence, the tension in the string must be T = Fc = 764.41 N
The brake in most cars makes use of a hydraulic system. This system consists of a fluid filled tube connected at each end to a piston. Assume that the piston attached to the brake pedal has a cross sectional area of 3 cm2 and the piston attached to the brake pad has a 2 cross section area of 15 cm . When you apply a force of 50 Newton to the piston attached to the brake pedal, how much will be the force at the brake pad
Answer:
The force at the brake pad = 250 N
Explanation:
The hydraulic brake system works on the Pascal's Principle for pressure transmission in fluids; the pressure applied to a fluid is transmitted undiminished in all directions.
For hydraulic systems, the pressure applied to the brake pedal is transmitted undiminished through the fluid filled tube, connected at each end to a piston, to the brake pad.
Hence, mathematically,
P(brake pedal) = P(break pad)
Pressure is given as the force applied divided by the cross sectional Area perpendicular to the direction of applied force.
P(brake pedal) = (Force applied on the brake pedal) ÷ (Cross Sectional Area of the brake pedal)
Force applied on the brake pedal = 50 N
Cross Sectional Area of the brake pedal = 3 cm²
P(brake pedal) = (50/3) = 16.67 N/cm²
P(brake pad) = P(brake pedal) = 16.67 N/cm²
P(brake pad) = (Force applied on the brake pad) ÷ (Cross Sectional Area of the brake pad)
Force applied on the brake pad = F = ?
Cross Sectional Area of the brake pad = 15 cm²
16.67 = (F/15)
F = 16.67 × 15 = 250 N
Hence, the force at the brake pad = 250 N
Hope this Helps!!!
2. If electrons are removed from an object, is the object positively or negatively
charged
positive
Because adding electrons makes an object to be negative and removing makes it to be positive
You are throwing a stone straight-up in the absence of air friction. If the stone is caught at the same height it was thrown from, which of the following is true of the stones motion?
1. the acceleration is zero at the top point of the motion.
2. the acceleration is minimum at the top point of the motion.
3. acceleration and velocity always point in opposite directions.
4. the time going up is equal to the time coming down the acceleration is directly proportional to the velocity.
Answer:
4. the time going up is equal to the time coming down the acceleration is directly proportional to the velocity.
Explanation:
1. FALSE
The acceleration at the top point is equal to the acceleration due to gravity (9.8 m/s²). It is the velocity of the stone that becomes zero for a moment, at the top point.
2. FALSE
The acceleration of the body in a free fall motion always remains constant, and its value is equal to the acceleration due to gravity (9.8 m/s²)
3. FALSE
The velocity and acceleration point in opposite direction, during upward motion only (Velocity points up and acceleration points down). But they point in same direction during the downward motion (Both velocity and acceleration point down).
4. TRUE
Since, there is no air friction. Therefore, the acceleration for both upward an downward motion will be constant, without any opposing force. Hence, the time taken by the stone to go up will be equal to time taken by the stone to come down.
dose sound travel faster in a warm room or a cold room? explain your answer
Answer:Sound travel faster in warm room.
Explanation:The speed of sound depends on the temperature of the medium. Mathematically, the relation between the speed of the sound and the temperature is give by:v=
is the ratio of the specific heats
R is the gas constant
T is the temperature of the medium
We know that the temperature of the warm room is more as compared to the cold room.
So, it is clear that the sound travel faster in a warm room. The particles move faster when the temperature is high.
A ladder 7.90 m long leans against the side of a building. If the ladder is inclined at an angle of 66.0° to the horizontal, what is the horizontal distance from the bottom of the ladder to the building? 10.3 Incorrect: Your answer is incorrect.
Answer:3.21 m
Explanation:
Given
Length of ladder [tex]L=7.9\ m[/tex]
inclination of ladder [tex]\theta =66^{\circ}[/tex]
If x is the horizontal distance between building and ladder then,
Using trigonometric relation
we get
[tex]\cos \theta =\frac{x}{L}[/tex]
[tex]x=L\cos \theta [/tex]
[tex]x=7.9\times \cos (66)[/tex]
[tex]x=3.21\ m[/tex]
Consider two sinusoidal waves traveling along a string, modeled as y1(x, t) = (0.4 m)sin[(3 m−1)x + (2 s−1)t] and y2(x, t) = (0.8 m)sin[(6 m−1)x − (5 s−1)t]. What is the vertical displacement (in m) of the resultant wave formed by the interference of the two waves at the position x = 0.9 m at time t = 0.4 s? (Indicate the direction with the sign of your answer.)
Answer:
y(x,t)=-0.395m
Explanation:
The wave function modeling waves are:
[tex]y_1(x,t)[/tex]= (0.4 m)sin[(3 [tex]m^-^1[/tex])x + (2 [tex]s^-^1[/tex])t]
[tex]y_2(x,t)[/tex]=(0.8 m)sin[(6 [tex]m^-^1[/tex])x − (5 [tex]s^-^1[/tex])t]
The principle of superposition can be defined as the resultant of [tex]y_1[/tex] and [tex]y_2[/tex] is equal to their algebraic sum
[tex]y(x,t)=y_1(x,t)+y_2(x,t)[/tex]
y(x,t)= (0.4 m)sin[(3 [tex]m^-^1[/tex])x + (2 [tex]s^-^1[/tex])t] + (0.8 m)sin[(6 [tex]m^-^1[/tex])x − (5 [tex]s^-^1[/tex])t]
Substitute 0.9m for x and 0.4s for t as required
y(x,t)= (0.4 )sin[(3 [tex]m^-^1[/tex])(0.9) + (2 [tex]s^-^1[/tex])(0.4)] + (0.8 m)sin[(6 [tex]m^-^1[/tex])(0.9)− (5 [tex]s^-^1[/tex])(0.4)]
y(x,t)= -0.14 - 0.255
y(x,t)=-0.395m
The vertical displacement of the resultant wave formed by the interference of the two waves is -0.395m
Interference of waves:Two sinusoidal waves are given such that:
y₁(x, t) = (0.4 m)sin[(3 m⁻¹)x + (2s⁻¹)t] and
y₂(x, t) = (0.8 m)sin[(6 m⁻¹)x − (5s⁻¹)t]
The superposition of the two waves gives the outcome of the interference at x = 0.9 m and t = 0.4 s.
y(x,t) = y₁(x, t) + y₂(x, t)
y(0.9,0.4) = y₁(0.9, 0.4) + y₂(0.9, 0.4)
y(0.9,0.4) = (0.4 m)sin[(3 m⁻¹)0.9 + (2s⁻¹)0.4] + (0.8 m)sin[(6 m⁻¹)0.9 − (5s⁻¹)0.4]
y(0.9,0.4) = -0.395m
Learn more about interference:
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Mike ran ten laps around his school's 1/4 mile track in 15 minutes. He
finishes his run at the exact same point that he started. What was his average
velocity in miles per hour?
Answer:
v = 0
Explanation:
It is given that,
Mike ran ten laps around his school's 1/4 mile track in 15 minutes. He finishes his run at the exact same point that he started. We need to find his average velocity. Average velocity is equal to net displacement divided by total time taken.
As he returns exactly at the same point that he started, it means his displacement or the shortest distance covered is equal to 0. As a result, its average velocity is equal to 0.
Give 2 examples for Newton’s first law of motion.
please answer this question!!!!!
A ball is kicked horizontally with a speed of 5.0 ms-1 from the roof of a house 3 m high. When will the ball hit the ground?
Answer:
the time taken for the ball to hit the ground is 0.424 s
Explanation:
Given;
velocity of the ball, u = 5 m/s
height of the house which the ball was kicked, h = 3m
Apply kinematic equation;
h = ut + ¹/₂gt²
where;
h is height above ground
u is velocity
g is acceleration due to gravity
t is the time taken for the ball to hit the ground
Substitute the given values and solve for t
3 = 5t + ¹/₂(9.8)t²
3 = 5t + 4.9t²
4.9t² + 5t -3 = 0
a = 4.9, b = 5, c = -3
Solve for t using formula method
[tex]t = \frac{-5 +/-\sqrt{5^2-4(4.9*-3)}}{2(4.9)} = \frac{-5+/-(9.154)}{9.8} \\\\t = \frac{-5+9.154}{9.8} \ or \ \frac{-5-9.154}{9.8} \\\\t = \frac{4.154}{9.8} \ or \ \frac{-14.154}{9.8} \\\\t = 0.424 \ sec \ or -1.444 \ sec\\\\Thus, t = 0.424 \ sec[/tex]
Two circular coils are concentric and lie in the same plane. The inner coil contains 110 turns of wire, has a radius of 0.014 m, and carries a current of 9.0 A. The outer coil contains 160 turns and has a radius of 0.022 m. What must be the magnitude of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero?
Answer:
The current flowing through the outer coils is
Explanation:
From the question we are told that
The number of turn of inner coil is [tex]N _i = 110 \ turns[/tex]
The radius of inner coil is [tex]r_i = 0.014 \ m[/tex]
The current flowing through the inner coil is [tex]I_i = 9.0 \ A[/tex]
The number of turn of outer coil is [tex]N_o = 160 \ turns[/tex]
The radius of outer coil is [tex]r_o = 0.022\ m[/tex]
For net magnetic field at the common center of the two coils to be zero the current flowing in the outer coil must be opposite to current flowing inner coil
The magnetic field due to inner coils is mathematically represented as
[tex]B_i = \frac{N_i \mu I}{2 r_i}[/tex]
The magnetic field due to inner coils is mathematically represented as
[tex]B_o = \frac{N_o \mu I_o}{2 r_o}[/tex]
Now for magnetic field at center to be zero
[tex]B_o = B_i[/tex]
So
[tex]\frac{N_i \mu I_i}{2 r_i} = \frac{N_o \mu I_o}{2 r_o}[/tex]
=> [tex]\frac{110 * 9}{2 * 0.014} = \frac{160 *I_o}{2 0.022}[/tex]
[tex]I_o = 9.72 \ A[/tex]
Name and draw the devices that can convert the analog signal to digital
Answer:
Analog to digital converters
Explanation:
An analog-to-digital converter (ADC) is a device that converts analog signals such as sound into digital signals. Analog information is transmitted by modulating the signal and then amplifying the signal's strength. The conversion from analog to digital involves quantization of the input thereby reducing error or noise. The ADC produces the output data as a series of digital values (0's and 1's) with fixed precision.
Answer:
Analog to digital converter.
Explanation:
Three important type that covert analog signal to digital
1 flash ADC
2 Digital Ramp ADC
3 successive Approximation
If the circuit in the motorcycle runs off a standard 12 V battery, and one of the headlights is switched on, current flowing through the headlight is measured at 3.75 A. What is the power usage of the headlamp? Report the answer to two significant digits
Answer:
45 W
Explanation:
Power: This can be defined as the rate at which electrical energy is consumed in a circuit. The unit of power is Watt (W).
From the question,
The expression for power is given as,
P = VI.................. Equation 1
Where V = Voltage, I = current.
Given: V = 12 V, I = 3.75 A.
Substitute into equation 1.
P = 12(3.75)
P = 45 W.
Answer:
P = 45 Watt
Explanation:
The electrical power used by an electrical device or the electrical circuit is given by the following formulae:
P = VI
P = I²R
P = V²/R
where,
P = Electrical Power Consumed by the Device
V = The Voltage applied to the circuit or device
R = Resistance of device or circuit
I = Current passing through the device or circuit
We have the following data for our circuit:
V = 12 volts
I = 3.75 A
Therefore, it is clear from the data that we will use the first formula:
P = VI
P = (12 volts)(3.75 A)
P = 45 Watt