Answer:
The electric field strength between the plates can be increased by decreasing the length of each side of the plates.
Explanation:
The electric field strength is given by;
[tex]E = \frac{V}{d}[/tex]
where;
V is the electric potential of the two opposite charges
d is the distance between the two parallel plates
[tex]E =\frac{V}{d} = \frac{\sigma}{\epsilon _o} \\\\(\sigma = \frac{Q}{A} )\\\\E = \frac{Q}{A\epsilon_o} \\\\E = \frac{Q}{L^2\epsilon_o}[/tex]
Where;
ε₀ is permittivity of free space
L is the length of each side of the plates
From the equation above, the electric field strength can be increased by decreasing the length of each side of the plates.
Therefore, decreasing the length of each side of the plates, could be made to increase the strength of the electric field between the plates
A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .
Answer:
Option 3 = both spheres are at the same potential.
Explanation:
So, let us complete or fill the missing gap in the question above;
" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"
The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:
=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.
To work on your car at night, you use an extension cord to connect your work light to a power outlet near the door. How would the illumination provided by the light be affected by the length of the extension cord
Answer:
The longer the cord, the lower the illumination
Explanation:
The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.
Long wires have more electrical resistance than shorter ones.
Let us consider this formula:
Resistance =[tex]\frac{\rho L}{A}[/tex]
From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.
Three point charges (some positive and some negative) are fixed to the corners of the same square in various ways, as the drawings show. Each charge, no matter what its algebraic sign, has the same magnitude. In which arrangement (if any) does the net electric field at the center of the square have the greatest magnitude?
Answer:
The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.
Two identical wooden barrels are fitted with long pipes extending out their tops. The pipe on the first barrel is 1 foot in diameter, and the pipe on the second barrel is only 1/2 inch in diameter. When the larger pipe is filled with water to a height of 20 feet, the barrel bursts. To burst the second barrel, will water have to be added to a height less than, equal to, or greater than 20 feet? Explain.
Answer:
The 1/2 inch barrel will burst at the same height of 20 ft
Explanation:
The pressure on a column of fluid increases with depth, and decreases with height. This means that if you increase the height of the fluid in the column, the pressure at the bottom will increase.
From the equation of fluid pressure,
P = ρgh
where
P is the pressure at the bottom of the fluid due to its height
ρ is the density of the fluid in question
h is the height to which the water stand.
You notice how apart from the height 'h' in the equation, all the other parts of the right hand side of the equation cannot be varied; they are a fixed property of the fluid and gravity. And there is no consideration for the horizontal diameter of the water's cross section area.
We can also think of the pressure at the bottom of the fluid to be as a result of an incremental weight of an infinitesimally small vertical section of the water down.
That been said, we can then say that if the barrel with the 1 ft diameter dimension bursts when filled with water up to 20 ft, then, the barrel with the reduced diameter will still burst at the same height as the former pipe.
NB: The only way to stop the pipe from bursting is to increase the thickness of the barrel wall to counteract the pressure forces due to the height.
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.
Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑ [tex]q_{i} / r_{i}[/tex]
where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm. The interference pattern on the screen 4.2 m away shows the first maximum 5.5 mm from the center of the pattern. What is the wavelength of the light in nm
Answer:
Explanation:
The double slit interference phonemene is described for the case of constructive interference
d sin θ= m λ (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
The double slit interference phonemene is described for the case of constructive interference
d sin θ = m lam (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in equation 1
d y / L = m λ
λ = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 mm = 0.44 10⁻³ m
y = 5.5 mm = 5.5 10⁻³ m
L = 4.2m
m = 1
let's calculate
λ = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
λ = 5.76190 10-7 m
let's reduce to num
lam = 5.56190 10-7 m (109 nm / 1m)
lam = 556,190 nmtea
we substitute
without tea = y / L
we substitute in equation 1
d y / L = m lam
lam = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 me = 0.44 10-3 m
y = 5.5 mm = 5.5 10-3
L = 4.2m
m = 1
let's calculate
lam = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
lam = 5.76190 10⁻⁷ m
let's reduce to num
lam = 5.56190 10⁻⁷ m (109 nm / 1m)
lam = 556,190 nm
A computer has a mass of 3 kg. What is the weight of the computer?
A. 288 N.
B. 77.2 N
C. 3N
D. 29.4 N
Answer:
29.4 NOption D is the correct option.
Explanation:
Given,
Mass ( m ) = 3 kg
Acceleration due to gravity ( g ) = 9.8 m/s²
Weight ( w ) = ?
Now, let's find the weight :
[tex]w \: = \: m \times g[/tex]
plug the values
[tex] = 3 \times 9.8[/tex]
Multiply the numbers
[tex] = 29.4 \: [/tex] Newton
Hope this helps!!
best regards!!
Given that the mass of the Earth is 5.972 * 10^24 kg and the radius of the Earth is
6.371 * 10^6 m and the gravitational acceleration at the surface of the Earth is 9.81
m/s^2 what is the gravitational acceleration at the surface of an alien planet with
2.4 times the mass of the Earth and 1.9 times the radius of the Earth?
Although you do not necessarily need it the universal gravitational constant is G =
6.674 * 10^(-11) N*m^2/kg^2
9
Answer:
gₓ = 6.52 m/s²
Explanation:
The value of acceleration due to gravity on the surface of earth is given as:
g = GM/R² -------------------- equation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
M = Mass of Earth
R = Radius of Earth
Now, for the alien planet:
gₓ = GMₓ/Rₓ²
where,
gₓ = acceleration due to gravity at the surface of alien planet
Mₓ = Mass of Alien Planet = 2.4 M
Rₓ = Radius of Alien Planet = 1.9 R
Therefore,
gₓ = G(2.4 M)/(1.9 R)²
gₓ = 0.66 GM/R²
using equation 1
gₓ = 0.66 g
gₓ = (0.66)(9.81 m/s²)
gₓ = 6.52 m/s²
The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?
Answer:
1.03A
Explanation:
For computing the magnitude of the current in the circuit we need to do the following calculations
LCR circuit impedance
[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]
= 148.7Ω
Now the phase angle is
[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]
Now the rms current flowing in the circuit is
[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]
= 0.98 A
The current flowing in the circuit is
[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]
= 1.39 A
And, finally, the current across the generator is
[tex]I'= I cos \phi[/tex]
[tex]= (1.39) cos 42.3^{\circ}[/tex]
= 1.03A
Hence, the magnitude of the circuit current is 1.03A
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
The number of neutrons in the nucleus of zinc 65 Zn 30 is:
35
Need more data to answer
65
30
Explanation:
proton number + neutron number = atomic mass
30 + 35 = 65
Zack is driving past his house. He wants to toss his physics book out the window and have it land in his driveway. If he lets go of the book exactly as he passes the end of the driveway. Should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)? Explain.
Answer:
Zack should direct his throw outward and toward the back of the car.
Explanation:
As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.
The solution is throw 3.
I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
Which statement best applies Newton’s laws of motion?The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.
When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.
The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.
Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
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dandre expands 120w of power in moving a couch 15 meters in 5 seconds how much force does he exert ?
Answer:
The answer is 40 N for APX
Explanation:
A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70
Answer:
The induced current is [tex]I = 6.25*10^{-4} \ A[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1[/tex]
The cross-sectional area is [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]
The initial magnetic field is [tex]B_i = 0.500 \ T[/tex]
The magnetic field at time = 1.02 s is [tex]B_t = 2.60 \ T[/tex]
The resistance is [tex]R = 2.70\ \Omega[/tex]
The induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]
The negative sign tells us that the induced emf is moving opposite to the change in magnetic flux
Here [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as
[tex]d \phi = dB * A[/tex]
Where dB is the change in magnetic field which is mathematically represented as
[tex]dB = B_t - B_i[/tex]
substituting values
[tex]dB = 2.60 - 0.500[/tex]
[tex]dB = 2.1 \ T[/tex]
Thus
[tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]
[tex]d \phi = 1.722*10^{-3} \ weber[/tex]
So
[tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]
[tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]
The induced current i mathematically represented as
[tex]I = \frac{\epsilon}{ R }[/tex]
substituting values
[tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]
[tex]I = 6.25*10^{-4} \ A[/tex]
What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?
Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑ [tex]q_{i} / r_{i}[/tex]
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m
The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.
Answer:
27°
Explanation:
The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)
So theta = arcsin(0.45)
=27°
The angle between the wire and the magnetic field is 27°.
Calculation of the angle:Since The magnetic force per meter on a wire is measured to be only 45 %
So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field
Therefore,
theta = arcsin(0.45)
=27°
Hence, The angle between the wire and the magnetic field is 27°.
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What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda = 622 nm[/tex]
Explanation:
From the question we are told that
The distance of the slit to the screen is [tex]D = 5 \ m[/tex]
The order of the fringe is m = 6
The distance between the slit is [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]
The fringe distance is [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]
Generally the for a dark fringe the fringe distance is mathematically represented as
[tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]
=> [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]
substituting values
=> [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]
=> [tex]\lambda = 6.22 *10^{-7} \ m[/tex]
[tex]\lambda = 622 nm[/tex]
A 285-kg object and a 585-kg object are separated by 4.30 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object placed midway between them.
Answer:
The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Explanation:
Given;
first object with mass, m₁ = 285 kg
second object with mass, m₂ = 585 kg
distance between the two objects, r = 4.3 m
The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m
Gravitational force between the first object and the 42 kg object;
[tex]F = \frac{GMm}{r^2}[/tex]
where;
G = 6.67 x 10⁻¹¹ Nm²kg⁻²
[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]
Gravitational force between the second object and the 42 kg object
[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]
Magnitude of net gravitational force exerted on 42kg object;
F = 3.545x 10⁻⁷ N - 1.727 x 10⁻⁷ N
F = 1.818 x 10⁻⁷ N
Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
λ = 5.2 x 10⁻⁷ m = 520 nm
An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field
Answer:
Explanation:
For circular path in magnetic field
mv² / R = Bqv ,
m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .
a )
R = mv / Bq
If v is changed to 2v , keeping other factors unchanged , R will be doubled
b )
magnitude of acceleration inside field
= v² / R
= Bqv / m
As v is doubled , acceleration will also be doubled
c )
If T be the time inside the magnetic field
T = π R / v
= π / v x mv / Bq
= π m / Bq
As is does not contain v that means T remains unchanged .
d )
Net force acting on electron
= m v² / R = Bqv
Net force = Bqv
As v becomes twice force too becomes twice .
So a . b , d are correct answer.
Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
The object with the twice the area of the other object, will have the larger drag coefficient.
Explanation:
The equation for drag force is given as
[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]
where [tex]F_{D}[/tex] IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
[tex]C_{D}[/tex] = coefficient of drag
A = area of the object
Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A
where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A
Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]
which also clarifies that the drag force is approximately proportional to the abject's area.
In this case, the object with the twice the area of the other object, will have the larger drag coefficient.
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
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A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N
The velocity and force are required.
The speed of the racket is 8.7 m/s
The required force is 471.43 N.
[tex]m_1[/tex] = Mass of racket = 1000 g
[tex]m_2[/tex] = Mass of ball = 60 g
[tex]u_1[/tex] = Initial velocity of racket = 12 m/s
[tex]u_2[/tex] = Initial velocity of ball = -15 m/s
[tex]v_1[/tex] = Final velocity of racket
[tex]v_2[/tex] = Final velocity of ball = 40 m/s
[tex]\Delta t[/tex] = Time = 7 ms
The equation of the momentum will be
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]
Force is given by
[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]
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Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration
Answer:
I believe it's called rapid growth
Explanation:
that is my answer no matter what
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
An electron moves to the left along the plane of the page, while a uniform magnetic field points into the page. What direction does the force act on the moving electron
Answer:
acting force is the answer
The direction of the magnetic force on the moving electron is upward.
The direction of the magnetic force on the electron can be determined by applying right hand rule.
This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.
Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.
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gravity can be described as..?
A. an magnetic force found in nature
B.the force that moves electrical charges
C.the force that repels object with like chargers
D.the force of attraction between two objects
Answer:
D
Explanation:
Gravity is the force of attraction between two objects.
Each object creates a gravitational field in wich every other object is affected by it.
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m 3 kg/m3 , length 81.2 cmcm and diameter 2.60 cmcm from a storage room to a machinist. Calculate the weight of the rod, www. Assume the free-fall acceleration is ggg = 9.80 m/s2m/s2 .
Answer:
The weight of the rod is 32.87 N
Explanation:
Density of the rod = 7800 kg/m
length of the rod = 81.2 cm = 0.812 m
diameter of rod = 2.60 cm = 0.026 m
acceleration due to gravity = 9.80 m/s^2
The rod can be assumed to be a cylinder.
The volume of the rod can be calculated as that of a cylinder, and can be gotten as
V = [tex]\frac{\pi d^{2} l}{4}[/tex]
where d is the diameter of the rod
l is the length of the rod
V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3
We know that the mass of a substance is the density times the volume i.e
mass m = ρV
where ρ is the density of the rod
V is the volume of the rod
m = 4.3 x 10^-4 x 7800 = 3.354 kg
The weight of a substance is the mass times the acceleration due to gravity
W = mg
where g is the acceleration due to gravity g = 9.80 m/s^2
The weight of the rod W = 3.354 x 9.80 = 32.87 N