Answer:
103 g
Explanation:
given that
Mass of the attached mass, m = 40 g
Initial distance of balance, x1 = 49.7 cm = 0.497 m
Final distance of balance, x2 = 39.2 cm = 0.392 m
Equilibrium point of attachment, x3 = 14 cm = 0.14 m
To get the mass of the meter stick, we use the relation
M = m.[(x2 - x3) / (x1 - x2)]
And on solving in full, we gave
M = 40.[(0.392 - 0.14) / [0.497 - 0.392)]
M = 40(0.258 / 0.105)
M = 40 * 2.457
M = 102.8 g
Therefore the needed work good for mass of the stick j
Is 103 g
What is the potential difference across one wire of a 70 m extension cord made of 16 gauge copper wire carrying a current of 4 A at room temperature (20degrees Celsius)
Answer:
V = 3.6 volts
Explanation:
From Ohm's Law, we know that:
V = IR
but,
R = ρL/A
Therefore,
V = IρL/A
where,
V = Potential Difference = ?
I = Current = 4 A
ρ = resistivity of copper = 1.68 x 10⁻⁸ Ω.m
L = Length = 70 m
A = Cross-sectional Area = πd²/4 = π(1.29 x 10⁻³ m)²/4 [16 gauge wire has a diameter of 1.29 mm]
A = 1.31 x 10⁻⁶ m²
V = (4 A)(1.68 x 10⁻⁸ Ω.m)(70 m)/(1.31 x 10⁻⁶ m²)
V = 3.6 volts
The potential difference across the given length of copper wire is obtained to be 3.6 V.
Ohm's LawThe resistance of the wire depends on the length, area of cross-section of the wire and resistivity of the material.
Given the length of the wire, [tex]L = 70\,m[/tex].
The diameter of a 16 gauge copper wire is given by, [tex]d = 1.29\times 10^{-3}\,m[/tex].
Therefore, the radius of the cross-section is;
[tex]r=\frac{d}{2} = \frac{1.29\times 10^{-3}\,m}{2} =6.45\times 10^{-4}\,m[/tex]
So, the resistance of the given length of wire can be written as,
[tex]R= \rho\, \frac{L}{A} =(1.68 \times 10^{-8} \, \Omega .m)\times \frac{70\,m}{\pi \times (6.45\times 10^{-4}m)^2} = 0.9\, \Omega[/tex]
According to Ohm's Law, we can write;
[tex]V=IR[/tex]
Substituting the known values, we get the voltage drop is;
[tex]V = 4A \times 0.9\Omega = 3.6\,V[/tex]
Learn more about Ohm's law here:
https://brainly.com/question/3656434
what is the difference between a good conductor and a good insulator?
Answer:
Explanation:
In a conductor, electric current can flow freely, in an insulator it cannot.
Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.
Most atoms hold on to their electrons tightly and are insulators.
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated
Answer:
It is calculated as Force × perpendicular distance.
Explanation:
Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.
The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.
But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the junction. How many electrons per second move past a point in wire 3?
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
[tex]i_1+i_3=i_2[/tex] (1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):
[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]
Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:
[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]
The number of electrons that trough the wire 3 is 2.18*10^18 e/s
Calculate the ideal banking angle in degrees for a gentle turn of 1.88 km radius on a highway with a 136.3 km/hr speed limit, assuming everyone travels at the speed limit.
Answer:
Ф = 4.4°Explanation:
given:
radius (r) = 1.88 km
velocity (v) = 136.3 km/hr
required:
banking angle ∡ ?
first:
convert 1.88 km to m = 1.88km * 1000m / 1km
r = 1880 m
convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)
v = 37.86 m/s
now.. calculate the angle
Ф = inv tan (v² / r * g) we know that gravity = 9.8 m/s²
Ф = inv tan (37.86² / (1880 * 9.8))
Ф = 4.4°
6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.
Answer:
same direction = 80 (n)
opposite direction = 20 (n) going one direction
Explanation:
same direction means they are added to each other
and opposite means acting on eachother
A soccer ball is kicked straight upwards with an initial vertical speed of 8.0\,\dfrac{\text m}{\text s}8.0 s m 8, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. How long does it take the ball to have a downwards speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
Answer:
1.2 s
Explanation:
Given:
v₀ = 8.0 m/s
v = -4.0 m/s
a = -10 m/s²
Find: t
v = at + v₀
(-4.0 m/s) = (-10 m/s²) t + (8.0 m/s)
t = 1.2 s
find the value of k for which the given pair of vectors are not equal
2ki +3j and 8i + 4kj
Answer:
5
Explanation:
An automobile accelerates from zero to 30 m/s in 6 s. The wheels have a diameter of 0.4 m. What is the angular acceleration of each wheel
Answer:
12.5 rad/s²
Explanation:
Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²
From the question,
a = αr................... Equation 1
Where a = linear acceleration, α = angular acceleration, r = radius.
But,
a = (v-u)/t.............. Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
(v-u)/t = αr
make α the subject of the equation
α = (v-u)/tr................. Equation 3
Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m
Substitute into equation 3
α = (30-0)/(0.4×6)
α = 30/2.4
α = 12.5 rad/s²
An object 9.00 mm tall is placed 12.0 cm to the left of the vertex of a convex mirror whose radius of curvature has a magnitude of 20.0 cm.(a) calculate the position, size, orientation (erect or inverted), and nature (real or virtual) of the image. (b) draw a principal-ray diagram showing the formation of the image (how would i do that.. i know you can't draw it here)..
Answer:
Explanation:
a )
for convex mirror, focal length
f = + 20 / 2 = 10 cm
object distance u = 12 cm ( negative )
Using mirror formula
[tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]
Putting the values
[tex]\frac{1}{v} - \frac{1}{12} = \frac{1}{10}[/tex]
[tex]\frac{1}{v} =\frac{1}{10} + \frac{1}{12}[/tex]
[tex]v=\frac{12\times 10}{12+10}[/tex]
v = 5.45 cm
Size of image
=[tex]\frac{v}{u} \times size of object[/tex]
[tex]=\frac{5.45}{12} \times 9[/tex]
= 4.08 mm .
It will be erect because only erect image is formed in convex mirror .
It will be virtual .
b )
Draw ray from tip of object parallel to principal axis and after reflection appears to be coming from focus . Second ray falls on the pole and reflects back making equal angle with principal axis . Draw the reflected ray backwards to cut the other ray . They cut each other where image is formed .
Briefly describe the relationship between an equipotential surface and an electric field, and use this to explain why we will plot equipotential lines.
Answer:
E = - dV/dx
Explanation:
Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.
El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.
De los antes expuesto las dos magnitudes están relacionadas
E = - dV/dx
por lo cual el potenical es el gradiente del potencial eléctrico.
Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial
g At 4 ft. from a point source of gamma radiation the intensity is 158 mRem/hr. How far away would you have to move to decrease the intensity to 17 mRem/hr.? [I1/I2 = (D2)2/(D1)2] I1(D1)2 = I2(D2)2
Answer:
You will have to move 12.2 ft to decrease the initial intensity to 17 mRem/h.
Explanation:
The distance can be calculated using the following equation:
[tex] \frac{I_{1}}{I_{2}} = \frac{D_{2}^{2}}{D_{1}^{2}} [/tex]
Where:
I₁ = 158 mRem/h
I₂ = 17 mRem/h
D₁ = 4 ft.
D₂=?
Hence, the distance D₂ is:
[tex] D_{2} =\sqrt{\frac{I_{1}}{I_{2}}*D_{1}^{2}} = \sqrt{\frac{158 mRem/h}{17 mRem/h}*(4 ft)^{2}} = 12.2 ft [/tex]
Therefore, you will have to move 12.2 ft to decrease the initial intensity to 17 mRem/h.
I hope it helps you!
A farmer with 600 ft of available fencing wants to enclose a rectangular are located adjacent to a farmhouse (with no fencing required along the farmhouse) and then divide it into four sections with fencing perpendicular to the side of the farmhouse. Show that the largest possible area of the four sections will be 18,000 square feet, and state the corresponding dimensions of the rectangular area.
Answer:
The corresponding dimensions will be "x = 300 & y = 60".
Explanation:
Available fencing = 600 ft
Fencing,
⇒ [tex]5y+x=600[/tex]
⇒ [tex]x=600-5y[/tex]
As we know,
⇒ [tex]Area =xy[/tex]
On substituting the given values, we get
[tex]=(600-5y)y[/tex]
[tex]=600y-5y^2[/tex] ...(equation 1)
On differentiating with respect to y, we get
⇒ [tex]\frac{dA}{dy} =600-10y[/tex]
[tex]0 = 600-10y[/tex]
[tex]600=10y[/tex]
[tex]y=\frac{600}{10}[/tex]
[tex]y=60[/tex]
On putting the values of y in equation 1, we get
⇒ [tex]600(60)-5(60)^2[/tex]
⇒ [tex]600(60)-5(3600)[/tex]
⇒ [tex]36000 - 18000[/tex]
⇒ [tex]18000[/tex]
Dimensions of the rectangular area:
x = 300
y = 60
An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos(t) + 9 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.
Answer:
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
Explanation:
Given that
s = 9 cos(t) + 9 sin(t), t ≥ 0
Then acceleration and velocity is
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
A body is sent out in space. Which of the following statements is true of this body as it moves away from Earth?
A. The body's mass and weight remain equal.
B. The body's mass remains constant, and its weight decreases.
C. The body's mass decreases, and its weight remains constant.
The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period
Answer:
Q = 23.49 C
Explanation:
We have,
Electric current drawn by the air conditioner is 16.2 A
Time, t = 1.45 s
It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,
[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]
So, the charge of 23.49 C is passing through the circuit during this period.
Consider a conducting rod of length 31 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails. At what speed (in m /s) must the sliding rod move to produce an emf of 0.75 V in a 1.75 T field?
Answer:
The speed of the rod is 1.383 m/s
Explanation:
Given;
length of the conducting rod, L = 31 cm = 0.31 m
induced emf on the rod, emf = 0.75V
magnetic field around the rod, B = 1.75 T
Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.
emef = BLv
where;
B is the magnetic field
L is length of the rod
v is the speed of the rod
v = emf / BL
v = (0.75) / (1.75 x 0.31)
v = 1.383 m/s
Therefore, the speed of the rod is 1.383 m/s
Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive force Fi. The particles are moved without altering their charges. Now they experience an attractive force with a magnitude of 36Fi. Find an expression for their new separation.
Answer:
[tex]r_f=\frac{1}{6}r_i[/tex]
Explanation:
To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.
You use the following expression:
[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex] (1)
k: Coulomb's constant
qA: charge of A particle
qB: charge of B particle
When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:
[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex] (2)
To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:
[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]
The new separation of the charges is 1/6 times of the initial separation
The speed of a sound wave in air is 343m/s. If the density of the air is 1.2kg/m3, find the bulk modulus.
Answer:
141178.8
Explanation:
use : density x velocity²
1.2 x 343² = 141178.8 pa
A bucket of water with total mass 23 kg is attached to a rope, which in turn is wound around a 0.050-m radius pulley at the top of a well. The bucket is raised to the top of the well and released from rest. The bucket is falling for 2 s and has a speed of 8.0 m/s upon hitting the water surface in the well. What is the moment of inertia of the pulley?
Answer:
[tex]I = 0.083 kg m^2[/tex]
Explanation:
Mass of the bucket, m = 23 kg
Radius of the pulley, r = 0.050 m
The bucket is released from rest, u = 0 m/s
The time taken to fall, t = 2 s
Speed, v = 8.0 m/s
Moment of Inertia of the pulley, I = ?
Using the equation of motion:
v = u + at
8 = 0 + 2a
a = 8/2
a = 4 m/s²
The relationship between the linear and angular accelerations is given by the equation:
[tex]a = \alpha r[/tex]
Angular acceleration, [tex]\alpha = a/r[/tex]
[tex]\alpha = 4/0.050\\\alpha = 80 rad/s^2[/tex]
Since the bucket is falling, it can be modeled by the equation:
mg - T = ma
T = mg - ma = m(g-a)
T = 23(9.8 - 4)
The tension, T = 133.4 N
The equation for the pulley can be modeled by:
[tex]T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2[/tex]
How much electricity is in human body?
Answer:
100 on average
Explanation:
The human body is a physical substance from where a human can breathe, drink, eat, walk, etc, and consists of various cells. A human being cannot live with their functions and when functions stop working that means a human being is dead.
Therefore, in the human body, there is also electricity that contains 100 watts on average.
So, According to the given situation, the correct answer is 100 on average.
This questic concerns a tablet that
has remained in storage for 2 years
since manufacture. From the drug
monograph the decomposition
rate constant (k) was found to be 6
x 10-2 year-1. The degradation
reaction is known to proceed via
first-order kinetics. Which one of
the following is the remaining
percentage of the original drug
? concentration
Answer:
The remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
Explanation:
Recall the formula for the decay of a substance at an initial [tex]N_0[/tex] concentration at manufacture:
[tex]N(t)=N_0\,e^{-k\,\,t}[/tex]
where k is the decay rate (in our case 0.06/year), and t is the elapsed time in years. Therefore, after 2 years since manufacture we have:
[tex]N=N_0\,e^{-0.06\,\,(2)}\\N=N_0\,e^{-0.12}\\N/N_0=e^{-0.12}\\N/N_0=0.8869[/tex]
This in percent form is 88.7 %. That is, the remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
A 12.5-µF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.
(a) How much energy is stored in the capacitor before and after the dielectric is inserted?
(b) By how much did the energy change during the insertion? Did it increase or decrease?
Answer:
a) 0.0036 J & 0.0135 J
b) 0.0099 J
Explanation:
Given that
Capacitance of a capacitor, C(i) = 12.5*10^-6 F
Potential difference across the capacitor, V = 24 V
Dielectric constant, k = 3.75
Energy in a capacitor is given by the formula
U = ½ CV²
Now, applying this to our solution, we have.
a, the energy stored before the dielectric was inserted is 0.0036. After inserting the dielectric, we then take cognizance into it, essentially multiplying by the constant of the dielectric, we got 0.0135 J
b) Now, Change in energy is gotten by subtracting the two energies, and we have
ΔU = U - U(1)
ΔU = 0.0036 - 0.0135
ΔU = 00099 J
See attachment for calculation
The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?
Answer:
The potential at the center of the sphere is -924 V
Explanation:
Given;
radius of the sphere, R = 0.22 m
electric field at the surface of the sphere, E = 4200 N/C
Since the electric field is directed towards the center of the sphere, the charge is negative.
The Potential is the same at every point in the sphere, and it is given as;
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)
The electric field on the sphere is also given as;
[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]
[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]
Substitute in the value of q in equation (1)
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]
Therefore, the potential at the center of the sphere is -924 V
A force of 30 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How much work is done in stretching the spring from 35 cm to 40 cm
Answer:
0.25 J
Explanation:
Given that
Force on the spring, F = 30 N
Natural length of the spring, l1 = 20 cm = 0.2 m
Final length of the spring, l2 = 35 cm = 0.35 m
Extension of the spring, x = 0.35 - 0.2 = 0.15 m
The other extension is 40 - 35 cm = 5 cm = 0.05 m
Work done = ?
Considering Hooke's Law of Elasticity.
We're told that the spring stretches through 15 cm, and thereafter asked to find the work done in stretching it through 5 cm
The question is solved in the attachment below
Gravitational potential energy is greatest at the highest point of a roller coaster and least at the lowest point.
Answer:
because gravitational potential enegry is directly proportional to the height so more the height more the gravitational potential enegry. therefore gravitational potential enegry is greatest at high point than lower points.
Using the ideas of electric field and force, explain what would happen to an electron if released from rest at r
Given that,
Distance = r
Electric field :
Electric field is equal to the multiplication of electric constant and charge divided by square of distance.
In mathematically form,
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}[/tex]
Electric force :
Electric force is equal to the multiplication of electric constant and both charges divided by square of distance.
In mathematically form,
[tex]E=\dfrac{kqQ}{r^2}[/tex]
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{qQ}{r^2}[/tex]
We know that,
The relation between electric field and electric force is
[tex]E=\dfrac{F}{q}[/tex]
According to question,
If electron is released from rest then electron move towards the source point due to attractive force.
Hence, The electron move towards the source point due to attractive force.
An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?
Answer:
(a) max. height = 3.641 m
(b) flight time = 1.723 s
(c) horizontal range = 31.235 m
(d) impact velocity = 20 m/s
Above values have been given to third decimal. Adjust significant figures to suit accuracy required.
Explanation:
This problem requires the use of kinematics equations
v1^2-v0^2=2aS .............(1)
v1.t + at^2/2 = S ............(2)
where
v0=initial velocity
v1=final velocity
a=acceleration
S=distance travelled
SI units and degrees will be used throughout
Let
theta = angle of elevation = 25 degrees above horizontal
v=initial velocity at 25 degrees elevation in m/s
a = g = -9.81 = acceleration due to gravity (downwards)
(a) Maximum height
Consider vertical direction,
v0 = v sin(theta) = 8.452 m/s
To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,
solve for S in equation (1)
v1^2 - v0^2 = 2aS
S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s
(b) total flight time
We solve for the time t when the vertical height of the object is AGAIN = 0.
Using equation (2) for vertical direction,
v0*t + at^2/2 = S substitute values
8.452*t + (-9.81)t^2 = 3.641
Solve for t in the above quadratic equation to get t=0, or t=1.723 s.
So time for the flight = 1.723 s
(c) Horiontal range
We know the horizontal velocity is constant (neglect air resistance) at
vh = v*cos(theta) = 25*cos(25) = 18.126 m/s
Time of flight = 1.723 s
Horizontal range = 18.126 m/s * 1.723 s = 31.235 m
(d) Magnitude of object on hitting ground, Vfinal
By symmetry of the trajectory, Vfinal = v = 20, or
Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s
What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)
Answer: Use this F=Ma.
Explanation: So your answer will be
F=1 Kg+9.8 ms-2
So the answer will be
F=9.8N
How'd I do this?
I just used Newton's second law of motion.
I'll also put the derivation just in case.
Applied force α (Not its alpha, proportionality symbol) change in momentum
Δp α p final- p initial
Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)
or then
F α m(v-u)/t
So, as we know v=final velocity & u= initial velocity and v-u/t =a.
So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.
So, F=kma (where k is the proportionality constant)
k is 1 so you can ignore it.
So, our equation becomes F=ma
The density of a sample of metal was measured to be 8.91 g/cm3. An X-ray diffraction experiment measures the edge of a face-centered cubic cell as 352.4 pm. Part APart complete What is the atomic weight of the metal
Answer:
The atomic weight of the metal is 58.7 g/mol
Explanation:
Given;
density of the metal sample, ρ = 8.91 g/cm³
edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm
Volume of the unit cell of the metal;
V = α³
V = (352.4 x 10⁻¹⁰ cm)³
V = 4.376 x 10⁻²³ cm³
Mass of the metal in unit cell
mass = density x volume
mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³
mass = 3.899 x 10⁻²² g
Atomic weight, based on 4 atoms per unit cell;
4 atoms = 3.899 x 10⁻²² g
6.022 x 10²³ atoms = ?
= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)
= 58.699 g/mol
= 58.7 g/mol (this metal is Nickel)
Theerefore, the atomic weight of the metal is 58.7 g/mol