Answer:
E = 2.5 x 10⁶ N/C
Explanation:
V = Potential difference between the plates of a TV set = 25 kV = 25000 Volts
d = Distance between the plates of TV set = 1.0 cm = 0.01 m
E = Electric field in the region between the plates
Electric field between the plates is given as
[tex]E= \frac{V}{d}[/tex]
Inserting the values
[tex]E= \frac{25000}{0.01}[/tex]
E = 2.5 x 10⁶ N/C
The uniform electric field is the field in which the value of the electric field strength remain same at each point.
The magnitude of the uniform electric field in the region between the plates [tex]2.5\times10^6 \rm N/C[/tex].
What is uniform electric field?The uniform electric field is the field in which the value of the electric field strength remain same at each point.
It can be given as,
[tex]E=\dfrac{V}{d}[/tex]
Here, [tex]V[/tex] is the potential difference between two points and [tex]d[/tex] is the distance between two points.
Given information-
The potential difference between the accelerating plates of a TV set is about 25 k-V or 25000 V.
The distance between the plates is 1.0 cm or 0.01 meters.
Use the above formula to find the magnitude of the uniform electric field in the region between the plates as,
[tex]E=\dfrac{25000}{0.01}\\E=2.5\times10^6 \rm N/C[/tex]
Thus the magnitude of the uniform electric field in the region between the plates [tex]2.5\times10^6 \rm N/C[/tex].
Learn more about uniform electric field here;
https://brainly.com/question/14632877
The amplitude of a wave decreases gradually as the wave travels down a long, stretched string. What happens to the energy of the wave when this happens? (The frequency is the same).
Answer:
energy decreases
Explanation:
The energy of a wave is directly proportional to the square of amplitude of a wave.
Amplitude is defined as the maximum displacement of the wave particle from its mean position.
So, as the amplitude goes on decreasing, the energy of the wave is also decreasing.
If the amplitude of a wave decreases gradually as the wave travels down, the energy of the wave increases.
What is a wave?A wave can be defined as a disturbance in a medium that progressively transport energy from a source to another location, especially without the transportation of matter.
The characteristics of a wave.In Science, some of the characteristics of a wave include the following:
FrequencyWavelengthAmplitudeSpeedGenerally, the energy of a wave is directly proportional to the square of wave amplitude and the square of its frequency. Thus, the energy of the wave increases when the amplitude of a wave decreases gradually as the wave travels down.
Read more on waves here: https://brainly.com/question/19250621
Two parallel wires are separated by 6.65 cm, each carrying 3.25 A of current in the same direction. (a) What is the magnitude of the force per unit length between the wires?
Answer:
The magnitude of the force per unit of lenght between the wires are of F/L= 3.17 * 10⁻⁵ N/m.
Explanation:
d=0.0665m
I1=I2= 3.25A
μo= 4π * 10⁻⁷ N/A²
F/L= (μo * I1 * I2) / (2π * d)
F/L= 3.17 * 10⁻⁵ N/m
A toroid of mean radius 26 cm and circular cross section of radius 1.9 cm is wound with a superconducting wire of length 1000 m that carries a current of 410 A. (a) What is the number of turns on the coil?
Answer:
8377 turns
Explanation:
Given:
Mean radius of toroid = 26 cm
Circular cross-section radius, r = 1.9 cm = 0.019
Length of the wire, l = 1000 m
The number of turns (n) on the coil is given by
[tex]n = \frac{l}{s}[/tex]
where,
l is the length of the wire
s is the circumference of the circular wire
⇒[tex]n = \frac{1000}{2\pi r}[/tex]
or
[tex]n = \frac{1000}{2\pi \times 0.019}[/tex]
or
n = 8376.57 or 8377 turns
hence, the number of turns on the coils is 8377
The number of turns of the coil is determined as 8377 turns.
Number of turnsThe number of turns of the superconducting wire is calculated as follows;
n = L/2πr
where;
L is the length of superconducting wire = 1000 mr is the radius of the circular cross section = 1.9 cm = 0.019 mSubstitute the given parameters and solve for the number of turns of the coil,
n = 1000/(2π x 0.019)
n = 8377 turns
Thus, the number of turns of the coil is determined as 8377 turns.
Learn more about number of turns here: https://brainly.com/question/25886292
A thin conducing plate 2.3 m on a side is given a total charge of −20.0 µC. (Assume the upward direction is positive.) (a) What is the electric field (in N/C) 1.0 cm above the plate? (Indicate the direction with the sign of your answer.)
Answer:
The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Distance = 2.3 m
Charge [tex]q= -20.0\ \muC[/tex]
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{\sigma}{2\epsilon_{0}}[/tex]
Where, [tex]\sigma=\dfrac{Q}{A}[/tex]
Q = charge
A = area
Put the value into the formula
[tex]E=\dfrac{Q}{2A\epsilon_{0}}[/tex]
[tex]E=\dfrac{-20.0\times10^{-6}}{2\times2.3\times2.3\times8.85\times10^{-12}}[/tex]
[tex]E=-213599.91\ N/C[/tex]
[tex]E=-2.14\times10^{5}\ N/C[/tex]
Negative sign shows the direction of the electric field.
The direction of electric field is toward the plates.
Hence, The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Final answer:
The electric field 1.0 cm above a thin conducting plate with a side length of 2.3 m and a total charge of -20.0 µC is -213,559.32 N/C, indicating a downward direction.
Explanation:
A thin conducting plate 2.3 m on a side is given a total charge of -20.0 µC. To find the electric field 1.0 cm above the plate, we use the concept of surface charge density (σ) and the formula for the electric field near an infinite plane sheet of charge, E = σ / (2ε0), where ε0 is the vacuum permittivity (ε0 = 8.85 x [tex]10^-^1^0[/tex] C2/N·m²). The surface charge density is σ = Q / A, with Q being the charge and A the area of the plate.
The area of the plate, A, is 2.3 m × 2.3 m = 5.29 m2. Therefore, σ = (-20.0 x 10-6 C) / (5.29 m2) = -3.78 x [tex]10^-^6[/tex] C/m². Substituting σ into the electric field equation yields E = -3.78 x [tex]10^-^6[/tex] C/m² / (2 × 8.85 x [tex]10^-^1^2[/tex] C2/N·m2) = -213,559.32 N/C, indicating the direction is downward due to the negative sign.
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900c to 0.942c MeV
Answer:
The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
Explanation:
We know that,
Mass of electron [tex]m_{e}=9.11\times10^{-31}\ kg[/tex]
Rest mass energy for electron = 0.511 Mev
(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}[/tex]
[tex]E=0.582\ Mev[/tex]
(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}[/tex]
[tex]E=0.350\ Mev[/tex]
Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
Electric field intensity, E can be defined as the electric (Coulomb) force on a unit test (1C) charge. Find the electric field intensity at a distance, d from the center of a +Q charged proton. use k as Coulombe constant.
Explanation:
Electric field intensity E is defined as the electric (Coulomb) force on a unit test (1 C) charge. Mathematically, it is given by :
[tex]E=\dfrac{F}{q}[/tex]
The electric force is given by :
[tex]F=k\dfrac{qQ}{d^2}[/tex]
Where
Q and q are electric charges
d is the distance between charges
The electric field intensity at a distance d from the center is given by :
[tex]E=\dfrac{k\dfrac{qQ}{d^2}}{q}[/tex]
So,
[tex]E=\dfrac{kQ}{d^2}[/tex]
Hence, this is the required solution.
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subjected to an axial load of 200 N. Determine the deflection of the spring. C = 80 kN/mm2.
Answer:
The deflection of the spring is 34.56 mm.
Explanation:
Given that,
Diameter = 10 mm
Number of turns = 10
[tex]Radius_{mean} = 60\ mm[/tex]
[tex]Diameter_{mean} = 120\ mm[/tex]
Load = 200 N
We need to calculate the deflection
Using formula of deflection
[tex]\delta=\dfrac{8pD^3n}{Cd^4}[/tex]
Put the value into the formula
[tex]\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}[/tex]
[tex]\delta =34.56\ mm[/tex]
Hence, The deflection of the spring is 34.56 mm.
Cart A, with a mass of 0.4 kg, travels on a horizontal air track at 6 m/s and hits cart B, which has a mass of 0.8 kg and is initially at rest. After the collision, cart A has a velocity of 2 m/s in the opposite direction. What is the final kinetic energy of cart B
Answer:
6.4 J
Explanation:
For cart A
mA = 0.4 kg, uA = 6 m/s, vA = - 2 m/s
For cart B
mB = 0.8 kg, uB = 0, vB = ?
use the law of conservation of momentum
momentum of system before coliision = momentum of system after collision
mA uA + mB uB = mA vA + mB vB
0.4 x 6 + 0 = 0.4 x (- 2) + 0.8 x vB
vB = 4 m/s
Kinetic energy of Cart B after collision = 1/2 mB vB^2
= 1/2 x 0.8 x 4 x 4 = 6.4 J
A ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, what is the refractive index of the glass?
Answer:
[tex]n_{glass}[/tex] = 1.6
Explanation:
[tex]\theta _{i}[/tex] = Angle of incidence = 65.9°
[tex]\theta _{r}[/tex] = Angle of refraction = 34.8°
[tex]n_{air}[/tex] = Index of refraction of air = 1
[tex]n_{glass}[/tex] = Index of refraction of glass = ?
Using Snell's law
[tex]n_{air}[/tex] Sin[tex]\theta _{i}[/tex] = [tex]n_{glass}[/tex] [tex]\theta _{r}[/tex]
(1) Sin65.9 = [tex]n_{glass}[/tex] Sin34.8
[tex]n_{glass}[/tex] = 1.6
Since the index of refraction of air is 1, the refractive index of the glass is 1.6 approximately
Refractive IndexRefractive Index is the measure of refraction or bending when light passes from one medium to another.
Given that a ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, the refractive index of the glass can be calculated with the formula below
n = sin i / sin r
Where
n = refractive indexi = angle of incidencer = angle of refractionSubstitute all the parameters
n = sin 65.9 / sin 34.8
n = 0.913 / 0.5707
n = 1.599
n = 1.6 approximately
Therefore, the refractive index of the glass is 1.6 approximately.
Learn more about Refractive Index here: https://brainly.com/question/10729741
Determine the position in which a solid cylindrical block of wood of diameter 0.3 m and length 0.4 m will float in water. Take specific gravity of wood as 0.5
Answer:
0.2 m
Explanation:
Diameter = 0.3 m
radius, r = 0.15 m
Length, H = 0.4 m
density of wood, d = 0.5 g/cm^3 = 500 kg/m^3
density of water, d = 1000 kg/m^3
Let h be the depth of cylinder immersed in water.
By the principle of floatation.
Buoyant force = Weight of cylinder
Volume immeresed x density of water x g = Volume of cylinder x density of wood x g
A x h x 1000 x g = A x H x 500 x g
1000 h = 500 x 0.4
h = 0.2 m
The cheetah is the fastest land animal, reaching speeds as high as 33 m/s, or roughly 75 mph. What is the kinetic energy of a 65 kg cheetah running at top speed? Express your answer in SI units.
Answer:
Ek= 35.392 kJ or 35.392 10³ J
Explanation:
Formula for Kinetic energy:
Ek= [tex]\frac{1}{2}[/tex] × m × [tex]v^{2}[/tex]
since m= 65kg and v= 33m/s
Ek= [tex]\frac{1}{2}[/tex]× 65×[tex]33^{2}[/tex]
Ek= 35.392 kJ or 35.392 10³ J
Final answer:
The kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is calculated using the formula KE = 1/2 mv², resulting in a kinetic energy of 35442.5 Joules.
Explanation:
The question asks for the kinetic energy of a 65 kg cheetah running at its top speed, which is given to be 33 m/s. To calculate kinetic energy, we use the formula KE = 1/2 mv², where m is the mass of the object (in this case, the cheetah) and v is the velocity of the object. Substituting the given values, we get:
KE = 1/2 × 65 kg × (33 m/s)²
KE = 1/2 × 65 × 1089
KE = 1/2 × 70885
KE = 35442.5 Joules
Therefore, the kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is 35442.5 Joules.
Consider two objects on the Moon's surface that can just be resolved by one of the 10.0 m telescopes at the Keck Observatory. What is the separation of the two objects? Assume the resolving power is only limited by diffraction effects. The distance from the Earth to the Moon is 3.8 x 105km, assume the wavelength of the light is 550 nm.
Answer: 25 m
Explanation:
The resolving power of a telescope is given by:
[tex] \theta = 1.22\frac{\lambda}{diameter}[/tex]
Diameter of the telescope = 10.0 m
Wavelength of the telescope = 550 nm = 550 ×10⁻⁹m
Thus, the resolving power of one of telescopes at the Keck Observatory is:
[tex] \theta = 1.22 \frac{550\times 10^{-9}}{10.0} =671 \times 10^{-10} rad = 3.85 \times 10^{-6} deg[/tex]
Also, [tex] tan\theta = \frac {\text {separation between two objects}}{\text{distance to the objects}} [/tex]
separation between objects = tan (3.85×10⁻⁶)° × 3.8×10⁵km = 2.5 × 10⁻²km =25 m
What is the magnitude of a the vertical electric field that will balance the weight of a plastic sphere of mass that has been charged to -3.0 nC? g
Answer:
Explanation:
Let E be the strength of electric field and m be the mass of plastic sphere.
charge on plastic sphere, q = - 3 n C = - 3 x 10^-9 C
The force due to electric field on the charged plastic sphere = q E
The weight of the plastic sphere = m g
where, g is the acceleration due to gravity.
Now the force due to electrostatic field is balanced by teh weight of the plastic sphere.
q E = m g
E = m g / q
E = m x 9.8 / (3 x 10^-9) = 3.26 x 10^9 x m
Substitute the value of m and get the value E.
The magnitude of the vertical electric field required to balance the weight of the charged plastic sphere with a mass of 2.1 g and a charge of -3.0 nC is [tex]6.86 * 10^6 N/C[/tex].
To find the magnitude of the vertical electric field that will balance the weight of a charged plastic sphere, we need to equate the electrostatic force on the sphere due to the electric field with the weight of the sphere.
Given information:
- Mass of the plastic sphere, m = 2.1 g = [tex]2.1 * 10^-3 kg[/tex]
- Charge on the plastic sphere, q = -3.0 nC = [tex]-3.0 * 10^-9 C[/tex]
- Acceleration due to gravity, g = [tex]9.8 m/s^2[/tex]
- Coulomb's constant, k = 1/(4πε₀) =[tex]9.0 * 10^9 N.m^2/C^2[/tex]
Step 1: Calculate the weight of the plastic sphere.
Weight of the sphere = mg
Weight = [tex](2.1 * 10^-3 kg)[/tex] × [tex](9.8 m/s^2)[/tex]
Weight = 2.058 × [tex]10^{-2}[/tex] N
Step 2: Equate the electrostatic force on the sphere to its weight.
Electrostatic force = qE
Weight = qE
[tex]2.058 * 10^{-2} N = (-3.0 * 10^{-9} C) * E[/tex]
Step 3: Rearrange the equation to find the electric field magnitude E.
Equation to find the electric field magnitude E.
E = (2.058 × [tex]10^{-2}[/tex] N) / (-3.0 × [tex]10^{-9}[/tex]C)
[tex]E = 6.86 * 10^6 N/C[/tex]
Complete question:
what's the magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2.1g that has been charged to -3.0nC? (k=[tex]1/4\pi[/tex], [tex]E_0[/tex]=[tex]9.0*10^9 N*m^2/C^2[/tex])
An angle has a value of 2.2 radians. Find its value in degrees.
Answer:
The value of 2.2 radians in degree is 126.11°.
Explanation:
Given that,
Angle = 2.2 radians
We need to calculate the value in degrees
We know that,
[tex]1\ radians = \dfrac{180}{\pi}[/tex]
[tex]2.2\ radians = \dfrac{180}{3.14}\times2.2[/tex]
[tex]2.2\ radians = 126.11^{\circ}[/tex]
Hence, The value of 2.2 radians in degree is 126.11°.
Calculate the force (in pounds) ona 4.0 ft by 7.0 ft door due to a 0.4 psi pressure difference.
Answer:
7174.09 N
Explanation:
Area = 7 ft x 4 x ft = 28 ft^2 = 2.60129 m^2
Pressure = 0.4 psi = 2757.9 Pa
Force = Pressure x Area
Force = 2757.9 x 2.60129 = 7174.09 N
As you move away from a positive charge distribution, the electric field:
Always decreases
Always increases
Must change
None of the above
please explain this thoroughly!
Answer:
The electric field always decreases.
Explanation:
The electric field due to a point charge is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
Where
k = electric constant
q = charge
r = distance from the charge
It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".
If the gravitational force between two masses located a distance 100 m apart is F, what would be the force between these objects if the distance between them increases to 800 m?
Answer:
[tex] \frac{F}{64}[/tex]
Explanation:
m = product of masses of two objects
r = distance between the two objects = 100 m
F = initial force between the two object
r' = new distance between the two objects = 800 m
F' = new force between the two objects = ?
k = constant of proportionality
initial force between the two object is given as
[tex]F=\frac{km}{r^{2}}[/tex] eq-1
new force between the two objects is given as
[tex]F'=\frac{km}{r'^{2}}[/tex] eq-2
Dividing eq-2 by eq-1
[tex]\frac{F'}{F}=\frac{r^{2}}{r'^{2}}[/tex]
Inserting the values
[tex]\frac{F'}{F}=\frac{100^{2}}{800^{2}}[/tex]
[tex]F' = \frac{F}{64}[/tex]
An electron and a proton are each placed at rest in an electric field of 500 N/C. Calculate the speed (and indicate the direction) of each particle 54.0 ns after being released. electron m/s O in the same direction as the field O in a direction opposite to the field proton m/s O in a direction opposite to the field O in the same direction as the field
Answer:
For proton: 2592 m/s In the same direction of electric field.
For electron: 4752000 m/s In the opposite direction of electric field.
Explanation:
E = 500 N/C, t = 54 ns = 54 x 10^-9 s,
Acceleration = Force /mass
Acceleration of proton, ap = q E / mp
ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2
Acceleration of electron, ae = q E / me
ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2
For proton:
u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s
use first equation of motion
v = u + at
vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.
For electron:
u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s
use first equation of motion
v = u + at
vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.
A sound wave from a siren has an intensity of 100.0 point, what the sound intensity level of the sound?
Answer:
Sound intensity level of the sound = 140 dB
Explanation:
We have expression for sound intensity level
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]
We have sound intensity, I = 100 W/m²
I₀ = 10⁻¹² W/m²
We need to find sound intensity level
Substituting
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )=10log_{10}\left ( \frac{100}{10^{-12}}\right )=10x14=140dB[/tex]
Sound intensity level of the sound = 140 dB
When four people with a combined mass of 315 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) What is the effective force constant of the springs?
Answer:
3.43 x 10⁵ N/m
Explanation:
M = mass of the car alone = 2000 kg
m = mass of the four people combined = 315 kg
x = compression of the spring when there is car alone
x' = compression of the spring when there is car and four people = x + 0.009
k = spring constant
When there is car, the weight of the car is balanced by the spring force and the force equation is given as
k x = Mg
k x = (2000)(9.8)
k x = 19600 eq-1
When there is car and four people, the weight of the car and four people is balanced by the spring force and the force equation is given as
k x' = (M + m) g
k (x + 0.009) = (2000 + 315) (9.8)
k x + 0.009 k = (2000 + 315) (9.8)
using eq-1
19600 + 0.009 k = 22687
k = 3.43 x 10⁵ N/m
Final answer:
The effective force constant of the springs k = 343000 N/m
Explanation:
To solve for the effective force constant of the springs in a car's suspension system, we can apply Hooke's Law, which is stated as F = -kx. Here, F represents the restoring force supplied by the springs, k is the spring constant or force constant, and x is the displacement of the springs from their equilibrium position.
When the four people with a combined mass of 315 kg sit in the car, their weight causes an additional displacement of 0.90 cm, or 0.009 m.
Since weight acts as the restoring force F, we can calculate it as F = mg, where m is the total mass and g is the acceleration due to gravity (9.80 m/s²). So, F = (315 kg)(9.80 m/s²) = 3087 N.
Using Hooke's Law, we can rearrange the formula to solve for k: k = F/x. Plugging in the values gives us k = 3087 N / 0.009 m, which provides the spring constant for the suspension system.
The effective force constant of the springs k = 343000 N/m
A gas is equilibrium at T kelvin. If mass of one molecule is m and its component of velocity in x direction is v. Then mean of its v2 is 3kT 2kT 2 kT 3 (4) zero
Answer:
The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
Explanation:
A gas is equilibrium at T kelvin.
Mass = M
We know that,
The average square of the velocity in the x,y and z direction are equal.
[tex]\bar{v}_{x}^2=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}[/tex]
[tex]v_{rms}^2=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}[/tex]
[tex]v_{rms}^2=3v_{x}^2[/tex]
[tex]v_{x}^{2}=\dfrac{v_{rms}^2}{3}[/tex]
Equation of ideal gas
[tex]PV=RT[/tex]
[tex]P=\dfrac{nKT}{V}[/tex]
Here, R = nK
We know that,
[tex]\dfrac{nKT}{V}=\dfrac{nM}{3V}v_{rms}^2[/tex]
[tex]v_{rms}^2=\dfrac{3KT}{M}[/tex]....(I)
Put the value of [tex]v_{rms}^2[/tex] in the equation (I)
[tex]v_{x}^2=\dfrac{1}{3}\times\dfrac{3KT}{M}[/tex]
[tex]v_{x}^2=\dfrac{KT}{M}[/tex]
Hence, The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
A sled is pulled along a level path through snow by a rope. A 20-lb force acting at an angle of 40° above the horizontal moves the sled 90 ft. Find the work done by the force. (Round your answer to the nearest whole number.)
Answer:
Work one by the force, W = 1378.87 lb-ft
Explanation:
It is given that,
Force acting on the sled, F = 20 lb
Angle between the force and the horizontal, θ = 40°
Distance moved, d = 90 ft
We need to find the work done by the force. We know that the work done can be calculated as :
[tex]W=Fd\ cos\theta[/tex]
[tex]W=20\ lb\times 90\ ft\ cos(40)[/tex]
W = 1378.87 lb-ft
So, the work done by the force is 1378.87 lb-ft. Hence, this is the required solution.
You heat a circular steel sheet of diameter 50.0 cm in an oven. What will be the radius for 50 K increase in temperature?
Answer:
R = 50.016 cm
Explanation:
Coefficient of thermal expansion for steel is given as
[tex]\alpha = 13 \times 10^{-6} per ^0C[/tex]
now as we know that the change in the length due to thermal expansion depends of change in temperature and initial length of the object
so here we will have
[tex]\Delta R = R_o \alpha \Delta T[/tex]
here we know that
[tex]R_o = \frac{50}{2} cm = 25 cm[/tex]
[tex]\Delta T = 50 K[/tex]
now we will have
[tex]\Delta R = (25)(13 \times 10^{-6})(50)[/tex]
[tex]\Delta R = 0.016 cm[/tex]
so final radius of the disc will be R = 50.016 cm
A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40° above the horizontal. How far above or below its original level will the ball strike the opposite wall?
Answer:
Ball hit the tall building 50 m away below 10.20 m its original level
Explanation:
Horizontal speed = 20 cos40 = 15.32 m/s
Horizontal displacement = 50 m
Horizontal acceleration = 0 m/s²
Substituting in s = ut + 0.5at²
50 = 15.32 t + 0.5 x 0 x t²
t = 3.26 s
Now we need to find how much vertical distance ball travels in 3.26 s.
Initial vertical speed = 20 sin40 = 12.86 m/s
Time = 3.26 s
Vertical acceleration = -9.81 m/s²
Substituting in s = ut + 0.5at²
s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²
s = -10.20 m
So ball hit the tall building 50 m away below 10.20 m its original level
The distance the ball will strike the opposite wall is 32.79 m.
Time of motion of the ball
The time of motion of the ball from the given height is calculated as follows;
h = vsinθ(t) + ¹/₂gt²
50 = 20 x sin(40)t + 0.5(9.8)t²
50 = 12.86t + 4.9t²
4.9t² + 12.86t - 50 = 0
solve the quadratic equation using formula method,
t = 2.14 s
Horizontal distance of the ballThe horizontal distance of the ball from the initial position is calculated as follows;
X = vcosθ(t)
X = 20 x cos(40) x 2.14
X = 32.79 m
Thus, the distance the ball will strike the opposite wall is 32.79 m.
Learn more about horizontal distance here: https://brainly.com/question/24784992
A soccer player swings his leg forward to kick the ball. If his leg has a moment of inertia of 0.9 kg m2, and is initially at rest, how much torque must he generate with his hip flexor muscle in order to reach an angular velocity of -8 rad/s in 0.5 seconds?
Answer:
14.4 Nm
Explanation:
Moment of Inertia, I = 0.9 kg m^2, w0 = 0, w = 8 rad/s, t = 0.5 second
Use first equation of motion for rotational motion
w = w0 + α t
where, α be the angular acceleration
8 = 0 + α x 0.5
α = 16 rad/s^2
Now Torque = Moment of inertia x angular acceleration
τ = I x α
τ = 0.9 x 16 = 14.4 Nm
A person pushes a 16.0 kg lawn mower at constant speed with a force of 87.5 N directed along the handle, which is at an angle of 45.0 to the horizontal. Part D friction force? What force must the person exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, assuming the same Express your answer to three significant figures and Include the appropriate units
The person must exert a force of 8.32 N to accelerate the lawn mower from rest to 1.3 m/s in 2.5 seconds.
When an object is moving at a constant speed, the net force acting on it is zero.
In this case, the force applied by the person along the handle is balanced by the frictional force opposing the motion.
So, the frictional force [tex](\(F_{\text{friction}}\))[/tex] is equal in magnitude but opposite in direction to the force applied by the person.
Given:
- Force applied by the person = 87.5 N
- The angle of the handle = 45.0 degrees to the horizontal
To find the horizontal component of the force applied by the person, use trigonometry:
[tex]\[F_{\text{person-horizontal}} = F_{\text{person}} \cdot \cos(\theta)\][/tex]
[tex]\[F_{\text{person-horizontal}} = 87.5 \, \text{N} \cdot \cos(45.0^\circ)\][/tex]
[tex]= 87.5 \, \text{N} \cdot 0.707 = 61.29 \, \text{N}[/tex]
So, the person exerts a horizontal force of 61.29 N.
Since the lawn mower moves at a constant speed, the frictional force must be equal in magnitude and opposite in direction to this horizontal force:
[tex]\[F_{\text{friction}} = -61.29 \, \text{N}\][/tex]
To calculate this force [tex](\(F_{\text{acceleration}}\))[/tex], we can use Newton's second law:
[tex]\[F = m \cdot a\][/tex]
Given:
Mass of the lawn mower = 16.0 kg
Acceleration = [tex]\(\frac{\Delta v}{\Delta t}\)[/tex],
Now, calculate [tex]\(\Delta v\):[/tex]
[tex]\[\Delta v = 1.3 \, \text{m/s} - 0 \, \text{m/s} = 1.3 \, \text{m/s}\][/tex]
Now, using Newton's second law:
[tex]\[F_{\text{acceleration}} = m \cdot a = (16.0 \, \text{kg}) \cdot \left(\frac{1.3 \, \text{m/s}}{2.5 \, \text{s}}\right)\][/tex]
[tex]\[F_{\text{acceleration}} = 8.32 \, \text{N}\][/tex]
Thus, the person must exert a force of 8.32 N.
Learn more about Force here:
https://brainly.com/question/13191643
#SPJ12
To accelerate a 16.0 kg lawn mower from rest to 1.3 m/s in 2.5 seconds, a person needs to exert a force of 136 N along the handle, assuming that the same friction force is present as when the mower is pushed at constant speed.
Explanation:In order to find the force that the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, we can use the second law of motion that states that the net force acting on an object is equal to its mass times its acceleration: F = m*a.
In this case, the acceleration would be the change in speed over time, which is (1.3 m/s - 0 m/s) / 2.5 s = 0.52 m/s². Consequently, F = (16.0 kg) * (0.52 m/s²) = 8.3 N.
It's important to mention this force is in addition to the friction force and the force to move the mower horizontally. Asume that the 87.5 N force previously exerted was just overcoming friction, then the total horizontal force required would be 87.5 N (for friction) + 8.3 N (for acceleration) = 95.8 N. However, since this force makes an angle of 45.0 with the horizontal, the person must exert a larger force along the handle, about 135.5 N. Hence, the correct answer, expressed to three significant figures, is 136 N.
Learn more about Force and Acceleration here:https://brainly.com/question/30959077
#SPJ3
An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of 3.5× 104 kg, and the force causes its speed to change from 6600 to 5700m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.5× 106 m determine the magnitude of the force.
Try this option; answers are marked with red colour: a) 19.215 GJ; b) 12.81 kN.
All the details are in the attached picture.
Two resistors connected in series have an equivalent resistance of 724.5 Ω, when they are connected in parallel, their equivalent resistance is 135.2 Ω. Find the resistance of each resistor.
Answer:
544.65 ohm and 179.84 ohm
Explanation:
Hello
Let
Resistor 1 (R1) and Resistor 2 (R2)
in series [tex]R_{eq} =R1+R2[/tex]
in parallel
[tex]\frac{1}{R_{eqp} } =\frac{1}{R1} +\frac{1}{R2}\\\\R1=R_{eqs} -R2\\\\R1 =724.5-R2 (equation 1)\\ R_{eqp}}=\frac{R1*R2}{R1+R2}\\\\\ 135.2 = \frac{R1*R2}{R1+R2}(equation 2)\\\\\\\\replacing 1 in \ 2\\\\probema data[/tex]
[tex]135.2=\frac{(724.5-R2)(R2)}{724.5-R2+R2} }\\135.2=\frac{(-R_{2} ^{2}+724.5R2) }{724.5}\\ 135.2*724.5=-R_{2} ^{2}+724.5R_{2}\\\\R_{2} ^{2} -724.5R_{2} +97952.4 =0\\R_{2} =\frac{724.5 \±\sqrt{(-724.5^{2})-4(1)(97952.4) } }{2(1)}\\R_{2} =\frac{724.5\±\sqrt{(133092.25) } }{2(1)}\\\\R_{2} =\frac{724.5+364.81810}{2} \\R_{2} =544.65 ohm\\\\R_{2} =\frac{724.5-364.81810}{2} =179.84 ohm[/tex]\\\\
let R2=544.65 and replace in equation 1
R1=724.5-544.65
R1=179.85
so, the resistors are 544.65 ohm and 179.84 ohm
Have a great day
Is finding the distance between two points in the rectangular coordinate system related to the Pythagorean theorem? Why or why not? Provide an example in your answer.
Answer:
Yes it is related
Explanation:
Consider 2 points in a rectangular co-ordinate system
[tex](x_{1},y_{1}),(x_{2},y_{2})[/tex]
The distance between them can be found by
d = [tex]\sqrt{(x_{1}-x_{2})^{2}-(y_{2}-y_{1})^{^{2}}}[/tex]
This is clear from the figure attached below
Is the following statement true or false? The time constant of an RC circuit is the time it takes to completely charge or discharge the capacitor. O True False
Answer:
FALSE
Explanation:
As we know that equation for charging of a capacitor is given by
[tex]q = CV(1 - e^{-\frac{t}{\tau}})[/tex]
now at t = 0 we can say there is no charge on the capacitor
now as the time will increase the value of charge will increase on the plates of capacitor
now after t = 1 time constant
we will have
[tex]q = CV(1 - e^{-1})[/tex]
[tex]q = 0.63 CV[/tex]
so after one time constant the capacitor will charged upto 63% of its maximum value.