The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass [tex]1.67 \times 10^(-27)[/tex]kg, charge +e = [tex]+1.60 \times 10^(-19)[/tex] C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed [tex]2.50 \times 10^6[/tex] m/s. The proton comes momentarily to rest at a distance [tex]5.31 \times 10^(-13)[/tex] m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are [tex]5.31 \times 10^{-13}[/tex] m apart?
Explanation:
The given data is as follows.
Mass of proton = [tex]1.67 \times 10^{-27}[/tex] kg
Charge of proton = [tex]1.6 \times 10^{-19} C[/tex]
Speed of proton = [tex]2.50 \times 10^{6} m/s[/tex]
Distance traveled = [tex]5.31 \times 10^{-13} m[/tex]
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
[tex](K.E + P.E)_{initial}[/tex] = [tex](K.E + P.E)_{final}[/tex]
[tex](\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)[/tex]
where, [tex]\frac{kq_{p}q_{t}}{r} = U = Electric potential energy[/tex]
U = [tex](\frac{1}{2}m_{p}v^{2}_{p})[/tex]
Putting the given values into the above formula as follows.
U = [tex](\frac{1}{2}m_{p}v^{2}_{p})[/tex]
= [tex](\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})[/tex]
= [tex]5.218 \times 10^{-15} J[/tex]
Therefore, we can conclude that the electric potential energy of the proton and nucleus is [tex]5.218 \times 10^{-15} J[/tex].
A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Complete question:
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.
(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Answer:
The net force on the person as the air bad deploys is -6750 N backwards
Explanation:
Given;
mass of the passenger, m = 60 kg
velocity of the car at impact, u = 15 m/s
final velocity of the car after impact, v = 0
distance moved as the front of the car crumples, s = 1 m
First, calculate the acceleration of the car at impact;
v² = u² + 2as
0² = 15² + (2 x 1)a
0 = 225 + 2a
2a = -225
a = -225 / 2
a = -112.5 m/s²
The net force on the person;
F = ma
F = 60 (-112.5)
F = -6750 N backwards
Therefore, the net force on the person as the air bad deploys is -6750 N backwards
"A plane has an airspeed of 142 m/s. A 16.0 m/s wind is blowing southward at the same time as the plane is flying. If the velocity of the plane relative to Earth is due east, what is the magnitude of that velocity
Answer:
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
Explanation:
Given;
A plane has an airspeed of 142 m/s (eastward)
vi = 142 m/s
16.0 m/s wind is blowing southward at the same time as the plane is flying
vb = 16.0m/s
Writing the relative velocity vector, we have;
Taking north and south as positive and negative y axis respectively, east and west as positive and negative x axis respectively.
v = 142i - 16j
The magnitude of the velocity is;
vr = √(vi^2 + vb^2)
vr = √(142^2 + 16^2)
vr = √(20420)
vr = 142.8985654231 m/s
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A, how many turns must the solenoid have?
Answer:
16,931 turnsExplanation:
The magnetic field produced is expressed using the formula
[tex]B = \frac{\mu_0NI}{L}[/tex]
B is the magnetic field = 0.30T
I is the current produced in the coil = 4.5A
[tex]\mu_0[/tex] is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A
L is the length of the solenoid = 32 cm = 0.32 m
N is the number of turns in the solenoid.
Making N the subject of the formula from the equation above;
[tex]B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\[/tex]
[tex]N = \frac{BL}{\mu_0I}[/tex]
Substituting the give values to get N;
[tex]N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21[/tex]
The number of turns the solenoid must have is approximately 16,931 turns
A horizontal force of 480 n is applied to a stationary wooden box in one direction and a 600 n horizontal force is applied in the opposite direction. What is the additional force is needed for the box to remain stationary
Answer:
The additional force is [tex]F_3 = 120 \ N[/tex]
Explanation:
From the question we are told that
The horizontal force in one direction is [tex]F_i = 480 \ N[/tex]
The horizontal force in the opposite direction is [tex]F_f = -600 \ N[/tex]
The negative sign shows that it is acting in the opposite direction
Generally for the box to remain stationary the net force on it must be equal to zero that is
[tex]F_1 + F_2 +F_3 = 0[/tex]
Where [tex]F_3[/tex] is the additional force required
So
[tex]F_3 = -F_1 - F_2[/tex]
substituting values
[tex]F_3 = -480 - [-600][/tex]
[tex]F_3 = -480 + 600[/tex]
[tex]F_3 = 120 \ N[/tex]
HELP ILL MARK BRAINLIEST PLS!!!!
A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
A solid non-conducting sphere of radius R carries a charge Q1 distributed uniformly. The sphere is surrounded by a concentric spherical shell of inner radius Ra and outer radius Rb . The shell carries a total charge Q2 distributed uniformly in its volume. What is the net electric field at a radial distance r such that R < r < Ra
Answer:
E = k Q₁ / r²
Explanation:
For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law
Ф = ∫ E .dA = [tex]q_{int}[/tex] / ε₀
where Ф the electric flow, qint is the charge inside the surface
To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone
R <r <R_a
for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.
E A = q_{int} /ε₀
The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field
q_{int} = Q₁
The surface area is
A = 4π r²
we substitute
E 4π r² = Q₁ /ε₀
E = 1 / 4πε₀ Q₁ / r²
k = 1/4πε₀
E = k Q₁ / r²
A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
I2 = 3.076 A
Explanation:
In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:
[tex]B=\frac{\mu_oI}{2R}[/tex] (1)
I: current in the wire
R: radius of the wire
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:
[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex] (2)
I1: current of the first ring = 8A
R1: radius of the first ring = 0.078m
I2: current of the second ring = ?
R2: radius of the first second = 0.03m
To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:
[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]
The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.
During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.
1) = what work do they do?
2) = what power do they develop if the work is done in 5.5 minutes?
Answer:
work is =7590joules
power = 23watts
Answer:
1) 7590 Joules
2) 23 Watts
Explanation:
1) Work = force × distance
W = mgh
W = (115 kg) (10 m/s²) (6.6 m)
W = 7590 J
2) Power = work / time
P = W / t
P = (7590 J) / (330 s)
P = 23 W
"water is circulating in pipes in a two floor house. on the first floor the pressure is 340 kpa and the speed 2.8 m/s. the second floor is 4m higher. calculate the pressure on the second floor of the areas are A1 1m2 and A2 2m2"
Answer:
Explanation:
We shall apply Bernoulli's equation of flow of liquid
1 / 2 ρ v² + ρ gh + P = constant
For calculating velocity in second floor
A₁ V₁ = A₂ V₂
1 x 2.8 = 2 x V₂
V₂ = 1.4 m /s
1 / 2 ρ v₁² + ρ gh₁ + P₁ = 1 / 2 ρ v₂² + ρ gh₂ + P₂
.5 x 10³ x 2.8² + 10³ x 9.8h₁ + 340 x 10³ = .5 x 10³ x 1.4² + 10³ x 9.8 x h₂ + P₂
P₂ = 3.92 x 10³ + 9.8 x 10³ ( h₁ - h₂ ) + 340 x 10³ - .98 x 10³
= 3.92 x 10³ - 9.8 x 10³ x 4 + 340 x 10³ - .98 x 10³
= 303.74 x 10³ Pa
= 303.74 kPa .
Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/kg. The final specific volume is 0.027 m^3/kg. Find the specific work in the process.
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
What is the equivalent resistance between the points A and B of the network?
Explanation:
First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.
Applying Gaussâs Law
When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of +3.0 μC charge put on a 5.0-cm aluminum spherical ball at the following two points in space: (a) a point 1.0 cm from the center of the ball (an inside point) and (b) a point 10 cm from the center of the ball (an outside point).
Answer:
a) E = 0
b) E = 2.697 MN/C
Explanation:
Solution:-
- The Gauss Law makes life simpler by allowing us to determine the Electric Field strength ( E ) of symmetrically charged objects. By choosing an appropriate Gaussian surface and determine the flux ( Φ ) that passes through an imaginary closed surface.
- The Law states that the net flux ( Φ ) that passes through a Gaussian surface is proportional to the net charged ( Q ) stored within that surface. We can mathematically express the flux ( Φ ) as follows:
Φ = Q / εo
Where, 1 / εo : The proportionality constant
εo: The permittivity of free space = 8.85*10^-12
- The flux produced by a charged object is also given in form of a surface integral of Electric Field ( E ) over the entire surface area ( A ) of the Gaussian surface as follows:
Φ = [tex]_S\int\int [ E ] . dA[/tex]
- We can combine the two relations as follows:
[tex]_S\int\int [ E ] . dA[/tex] = Q / εo
- Now we will consider a charged metal sphere. The important part to note is that the charge on a conducting sphere ( Q ) uniformly distributed on the outside surface of the charged sphere.
- Lets consider a case, where we set up our Gaussian surface ( spherical ) with radius ( r ) < radius of the charged metal surface ( a ). We will use the combined relation and determine the Electric Field ( E ) within a charged metal sphere as follows:
[tex]E. ( 4\pi*r^2 ) = \frac{Q_e_n_c}{e_o} \\\\E = \frac{Q_e_n_c}{e_o4\pi*r^2}[/tex]
- However, the amount of charge enclosed in our Gaussian surface is null or zero. As all the charge is on the surface r = a. Hence (Q_enc = 0 ),
[tex]E = 0[/tex] ..... ( r < a )
- For the case when we set up our gaussian surface with radius ( r ) > radius of the charged metal surface ( a ). We placed a charge of Q = +3.0uC on the surface of the metal sphere. Therefore, the electric field strength at a distance ( r ) from the center of metal sphere is:
[tex]E = \frac{Q_e_n_c}{e_o*4*\pi*r^2 } = k\frac{Q_e_n_c}{r^2 }[/tex] .... ( r > a )
- The above relation turns out to be the Electric Field strength ( E ) produced by a point charge at distance ( r ) from the center. Where, k = 8.99*10^9 is the Coulomb's constant.
a) The radius of the charged metal sphere is given to be a = 5.0 cm. The first point r = 1.0 cm lies within the metal sphere. We looked at the first case where, ( r < a ) the enclosed charge is zero. Hence, the magnitudue of Electric Field Strength ( E ) is zero. ( E = 0 )
b) The second point lies at 10 cm from the center. For this we will use the second case where, ( r > a ). The Electric Field Strength due to a point charge with an enclosed charge of Q = +3.0 uC is:
[tex]E = ( 8.99*10^9 ) * \frac{3.0*10^-^6}{0.1^2} \\\\E = 2697000 N / C[/tex]
Answer: The electric field strength at point 10 cm away from the center is 2.697 MN/C
At what minimum angle will you get total internal reflection of light traveling in flint glass and reflected from water?
Answer:
θ_c = 53.65°
Explanation:
The point after which the light ray had started reflecting internally will be when the reflecting angle is at 90°. The incident angle at this point is called the critical angle and this can be calculated through Snell's law as;
n1 sin θ_c = n2 sin 90
Where;
n1 is the refractive index of the medium through which the incident rays will pass through.
n2 is the Refractive index of the medium through which the refracted rays will pass through.
θ_c is the critical angle at which the incident ray will reflect totally internally.
Now, since sin 90 = 1
Thus;
n1 sin θ_c = n2
θ_c = sin^(-1) (n2/n1)
Now, we are told that the reflection travels in flint glass and reflected from water.
Thus, the first medium is flint glass and the second medium is water.
So, from tables,
Refractive index of flint glass; n1 = 1.655
Refractive index of water; n2 = 1.333
Thus;
θ_c = sin^(-1) (1.333/1.655)
θ_c = 53.65°
Suppose that the voltage of the battery in the circuit is 3.9 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the paper) is 1.1 T, and the length of the rod between the rails is 0.22 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.
Answer:
v = 16.11 m/s
Explanation:
In order to calculate the maximum speed of the rod, you use the following formula:
[tex]\epsilon=vBLsin\theta[/tex] (1)
ε = voltage of the circuit = 3.9V
v: maximum speed of the rod = ?
B: magnitude of the magnetic field = 1.1T
L: length of the rod = 0.22m
θ: angle between the direction of motion of the rod and the direction of the magnetic field = 90°
You solve the equation (1) for v and replace the values of the other parameters:
[tex]v=\frac{\epsilon}{BLsin\theta}=\frac{3.9V}{(1.1T)(0.22m)sin90\°}\\\\v=16.11\frac{m}{s}[/tex]
The maximum speed of the rod is 16.11 m/s
the time required for one cycle, a complete motion that returns to its starting point, it called the_____. period medium frequency periodic motion
Answer:
The time required for one cycle, a complete motion that returns to its starting point,it is called periodic motion
Explanation:
I hope this will help you:)
An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?
Answer:
The power used is 0.96 watts.
Explanation:
Recall the formula for electric power (P) as the product of the voltage applied times the circulating current:
[tex]P=V\,\,I[/tex]
and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:
[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]
Then we can re-write the power expression as:
[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]
which in our case becomes:
[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.00m/s. Determine the constants A and B with units.
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Answer:
Explanation:
F ×1 = 0.5×0.145×47×47
F = 160.15 N
g Suppose that you charge a 3 F capacitor in a circuit containing eight 3.0 V batteries, so the final potential difference across the plates is 24.0 V. How much charge is on each plate
The complex, highly technical formula for capacitors is
Q = C V
Charge = (capacitance) (voltage)
Charge = (3 F) (24 V)
Charge = 72 Coulombs
The positive plate of the capacitor is missing 72 coulombs worth of electrons. They were sucked into positive terminal of the battery stack.
The negative plate of the capacitor has 72 coulombs worth of extra electrons. They came from the negative terminal of the battery stack.
You should be aware that this is a humongous amount of charge ! An average lightning bolt, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around 15 coulombs of charge !
The scenario in the question involves a "supercapacitor". 3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.
Also, IF you can charge this animal to 24 volts, it will hold 864J of energy. You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.
An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.
Answer:
The change in gravitational potential energy of the climber-Earth system is [tex]\Delta PE = 396900 \ J[/tex]
Explanation:
From the question we are told that
The mass of the hiker is [tex]m = 75 \ kg[/tex]
The time taken is [tex]T = 2 \ hr = 2 * 3600 = 7200 \ s[/tex]
The vertical elevation after time T is [tex]H = 540 \ m[/tex]
The change in gravitational potential is mathematically represented as
[tex]\Delta PE = mgH[/tex]
here g is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
substituting values
[tex]\Delta PE = 75 * 9.8 * 540[/tex]
[tex]\Delta PE = 396900 \ J[/tex]
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the +x-axis does the third piece move?
(a) 39.8º from the +x-axis
(b) 36.9° from the +x-axis
(c) 39.9° from the +x-axis
(d) 216.9° from the +x-axis
(e) 219.8° from the +X-axis
Answer:
M1 Vx1 + M2 Vx2 + M3 Vx3 = 0 conservation of momentum in x direction
Vx3 = -(M1 Vx1 + M2 Vx2 ) / M3
Vx3 = - 320 * 2 / 100 = -6.4 m/s M2 has no x-component of momentum
Likewise:
Vy3 = -(M1 Vy1 + M2 Vy2 ) / M3
Vy3 = - 355 * 1.5 / 100 = -5.33 m/s
tan theta = -5.33 / -6.4 = .833 where theta is in the third quadrant and measured from the negative x-axis
theta = 39.8 deg
180 + 39.8 = 219.8 from the positive x-axis
Two children of mass 20.0 kg and 30.0 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3.00 m, at what distance from the pivot point is the small child sitting in order to maintain the balance
Answer:
The distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m
Explanation:
Given;
mass of the bigger child, M = 30 kg
mass of the smaller child, m = 20 kg
distance between the two children, d = 3 m
This information can be represented diagrammatically;
3m
|<------------------------------------------------>|
----------------------------------------------------------------------------
↓ x Δ 3-x ↓
20kg 30kg
x is the distance from the pivot point that the small child will sit in order to maintain the balance
Take moment about the pivot;
Clockwise moment = anticlockwise moment
30(3-x) = 20x
90 -30x = 20x
90 = 20x + 30x
90 = 50x
x = 90 / 50
x = 1.8 m
Therefore, the distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m
The distance from the pivot point which the small child must sit in order to maintain the balance is 1.8 meters.
Let the first child be A.Let the second child be B.Given the following data:
Mass of A = 20.0 kgMass of B = 30.0 kgDistance = 3.00 mTo determine what distance from the pivot point is the small child sitting in order to maintain the balance, we would take moment about a pivot:
Let the distance from the pivot be n.Note: The distance of the child from the pivot is equal to [tex]3-n[/tex]
For moment:
Clockwise moment = anticlockwise moment
[tex]30(3-n) = 20n\\\\90-30n=20n\\\\90=20+30n\\\\90=50n\\\\n=\frac{90}{50}[/tex]
n = 1.8 meters
Read more: https://brainly.com/question/2400211
Suppose that 300 keV X-ray photons are aimed at a zinc cube (Zinc, Z = 30). According to the chart below, what effect will predominate when the X-rays hit the metal?
a) Photoelectric Effect 3
b) Compton Effect 3
c) Pair Production
Answer:
the answer is option A = photoelectric effect
Explanation:
If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.
How much work is needed to move an object from one position to another when both positions are located the same distance from the center of the earth
Answer:
The product of the object's weight and the horizontal distance between the two positions.
Explanation:
Work is the product of force and the distance through which this force is moved. The distance moved can be vertical, or horizontal. For two bodies located the same distance from the center of the earth, the work done will be the product of the weight of the product and the horizontal distance between the two positions. If the vertical work is needed, then the work is zero, since there is no height gradient between them.
Explain the purpose of hot gravity filtration. Why is it good to use the stemless funnel for this experiment
Answer: It is done to prevent the necessary compound from solidifying along with the debasements. It expels any insoluble pollutions from the appropriate response (as opposed to separating the predetermined item). With since quite a while ago stemmed channels, the gems kick off inside the progression because the arrangement cools, obstructing the pipe. utilizing a stemless channel keeps this from occurring.
Explanation:
it is good to use the stemless funnel for hot gravity filtration experiment, to prevent the necessary compound from solidifying, expels any insoluble pollutions from the appropriate response.
what is hot gravity filtration ?Recrystallization is the process of getting pure crystals from an impure compound in a solvent and Hot gravity filtration remove the impurities from a solution prior to recrystallization.
In this technique the filtration equipment and the sample are heated and the filtration is needed for recrystallization which requires a hot solution as it need to be supersaturated for crystals to form on cooling.
Hot solutions hold more solute in a suspension than a cold solution as the solubility of solids increases with a increase in temperature, that means saturated solution contain more dissolved solute.
When the hot solution cool down, it will be supersaturated and hold more dissolved solute than its cold. The main objective to choose a solvent is that it dissolves the compound when heated, but that doesn’t dissolve the impurity at high temperatures.
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In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum
Explanation:
We have
A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.
The frequency of a pendulum is equal to the no of oscillation per unit time. so,
[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]
Tim period is reciprocal of frequency. So,
[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]
So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.
Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble
Answer:
Johannes KeplerExplanation:
He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.Hope this helped,
Kavitha
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
The combustion of propane (C3H8) in the presence of excess oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) When 2.0 mol of O2 are consumed in this reaction, ________ mol of CO2 are produced.
Answer:
1.2
Explanation:
2.0 mol O₂ × (3 mol CO₂ / 5 mol O₂) = 1.2 mol CO₂
5. Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car
Answer:
The work done by both Joel and Jerry is equal to 0 J.
Explanation:
The work done on a body by an external agency is the product of the force applied on the body and the distance through which the body moves. Therefore,
W = F.d
where,
W = Work Done on the Body
F = Force Applied on the Body
d = displacement covered by the body
In the given case of both Joel and Jerry, they are unable to move the car. It means that the displacement covered by the car is zero. Hence,
W = F(0)
W = 0 J (For both Joel and Jerry)