An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke's Law) located at the end of the track. The spring constant is 52.1 N/m, the mass of the object 0.250 kg and the speed of the object is 1.70 m/s immediately before the collision.
(a) Determine the spring's maximum compression if the track is frictionless.
?? m
(b) If the track is not frictionless and has a coefficient of kinetic friction of 0.120, determine the spring's maximum compression.
??m
Answers
(a) As it gets compressed by a distance x, the spring does
W = - 1/2 (52.1 N/m) x ²
of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so
W = ∆K
- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.118 m
(b) Taking friction into account, the only difference is that more work is done on the object.
By Newton's second law, the net vertical force on the object is
∑ F = n - mg = 0
where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives
n = mg = 2.45 N
and from this we get the magnitude of kinetic friction,
f = µn = 0.120 (2.45 N) = 0.294 N
Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:
W (friction) = - (0.294 N) x
W (spring) = - 1/2 (52.1 N/m) x ²
==> W (total) = W (friction) + W (spring)
Solve for x :
- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.112 m
For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:
a) The spring's maximum compression when the track is frictionless is 0.118 m.
b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.
a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex] (1)
Where:
[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg
[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s
k: is the spring constant = 52.1 N/m
x: is the distance of compression
After solving equation (1) for x, we have:
[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]
Hence, the spring's maximum compression is 0.118 m.
b) When the track is not frictionless, we can calculate the spring's compression by work definition:
[tex] W = \Delta E = E_{f} - E_{i} [/tex]
[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex] (2)
Work is also equal to:
[tex] W = F*d = F*x [/tex] (3)
Where:
F: is the force
d: is the displacement = x (distance of spring's compression)
The force acting on the object is given by the friction force:
[tex] F = -\mu N = -\mu m_{o}g [/tex] (4)
Where:
N: is the normal force = m₀g
μ: is the coefficient of kinetic friction = 0.120
g: is the acceleration due to gravity = 9.81 m/s²
The minus sign is because the friction force is in the opposite direction of motion.
After entering equations (3) and (4) into (2), we have:
[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]
[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]
[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]
Solving the above quadratic equation for x
[tex] x = 0.112 m [/tex]
Therefore, the spring's compression is 0.112 m when the track is not frictionless.
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Related Questions
Part B
What is the approximate amount of thrust you need to apply to the lander to keep its velocity roughly constant? Explain why, using Newton's first
law of motion.
Answers
Answer:
Force is zero.
Explanation:
According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.
F = m a
if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.
Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.
Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 59.4 m/s and returns the shot with the ball traveling horizontally at 37.2 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction).
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball?
Answers
5 9 . 4
- 3 7 . 2
2 2 . 2
Explanation:
Use the algorithm method.
5 9 . 4
- 3 7 . 2
2 2 . 2
2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.
22.2
22.2
Determine the absolute pressure on the bottom of a swimming pool 27.0 m by 8.9 m whose uniform depth is 1.8 m . Express your answer using two significant figures.
Answers
Answer:
[tex]P=17658Pa[/tex]
Explanation:
From the question we are told that:
Dimension
[tex]L*B=27.0*8.9[/tex]
Depth [tex]d=1.8m[/tex]
Generally the equation for Volume of water is mathematically given by
[tex]V=L*B*D[/tex]
[tex]V=27.0*8.9*1.8[/tex]
[tex]V=432.54m^3[/tex]
Therefore
Force at the bottom of the Pool
[tex]F=\rho Vg[/tex]
Where
[tex]\rho \ density\ of \ water(1000kg/m^3)[/tex]
[tex]F=1000*432.54m^3*9.81[/tex]
[tex]F=4.2*10^{6}N[/tex]
Generally the equation for Pressure at the bottom is mathematically given by
[tex]P=\frac{Forece }{Area}[/tex]
[tex]P=\frac{4.2*10^{6}N}{27.0*8.9}[/tex]
[tex]P=17658Pa[/tex]
20. How much charge will flow through a 2002 galvanometer
connected to a 40092 circular coil of 1000 turns on a wooden
stick 2 cm in diameter? If a magnetic field B=0.011 T parallel to
the axis of the stick is decreased suddenly to zero?
Answers
Answer:
5.76 μC
Explanation:
The induce emf, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = NAΔB where N = number of turns of coil = 1000, A = cross-sectional area of coil = πd²/4 where d = diameter of coil = 2 cm = 2 × 10⁻² m and ΔB = change in magnetic field strength = B' - B where B' = final magnetic field = 0 T and B = initial magnetic field strength = 0.011 T. So, ΔB = 0 T - 0.011 T = -0.011 T
So, ε = -ΔΦ/Δt
ε = -NAΔB/Δt
ε = -NAΔB/Δt
Also ε = iR where i = current and R = combined resistance of circular coil and galvanometer = 200 Ω + 400 Ω = 600 Ω (since they are in series)
So, iR = -NAΔB/Δt
iΔt = -NAΔB/R
Δq = -NAΔB/R where Δq = charge = iΔt
substituting the values of the variables into the equation, we have
Δq = -1000 × π(2 × 10⁻² m)²/4 × -0.011 T/600 Ω
Δq = -1000 × 4π × 10⁻⁴ m²/4 × -0.011 T/600 Ω
Δq = 0.011π × 10⁻¹ m²T/600 Ω
Δq = 0.03456 × 10⁻¹ m²T/600 Ω
Δq = 5.76 × 10⁻⁶ C
Δq = 5.76 μC
Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
Required:
What is the magnitude of the dipole moment of the arrangement?
Answers
Answer:
The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".
Explanation:
Given that,
q = 0.50 nC
d = 900 mm
As we know,
⇒ [tex]P=qd[/tex]
By putting the values, we get
⇒ [tex]=0.50\times 900[/tex]
⇒ [tex]=(0.50\times 10^{-9})\times 0.9[/tex]
⇒ [tex]=4.5\times 10^{-10} \ Cm[/tex]
Answer:
The dipole moment is 4.5 x 10^-10 Cm.
Explanation:
Charge on each ball, q = 0.5 nC
Length, L = 900 mm = 0.9 m
The dipole moment is defined as the product of either charge and the distance between them.
It is a vector quantity and the direction is from negative charge to the positive charge.
The dipole moment is
[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]
How do the magnitudes of the currents through the full circuits compare for Parts I-III of this exercise, in which resistors are combined in series, in parallel, and in combination
Answers
Answer: hello tables and data related to your question is missing attached below are the missing data
answer:
a) I = I₁ = I₂ = I₃ = 0.484 mA
b) I₁ = 0.016 amps
I₂ = 0.0016 amps
I₃ = 7.27 * 10^-4 amps
c) I₁ = 1.43 * 10^-3 amp
I₂ = 0.65 * 10^-3 amps
Explanation:
A) magnitude of current for Part 1
Resistors are connected in series
Req = r1 + r2 + r3
= 3300 Ω ( value gotten from table 1 ) ,
V = 1.6 V ( value gotten from table )
hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA
The magnitude of current is the same in the circuit
Vi = I * Ri
B) magnitude of current for part 2
Resistors are connected in parallel
V = 1.6 volts
Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) + R3 ]
= [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]
= 87.30 Ω
For a parallel circuit the current flow through each resistor is different
hence the magnitude of the currents are
I₁ = V / R1 = 1.6 / 100 = 0.016 amps
I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps
I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps
C) magnitude of current for part 3
Resistors are connected in combination
V = 1.6 volts
Req = R1 + ( R2 * R3 / R2 + R3 )
= 766.66 Ω
Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps
magnitude of currents
I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps
I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps
Give an example of a substance with an amorphous structure.
Answers
Answer:
Tempered glass
Explanation:
When warmed, an amorphous substance has a non-crystalline architecture that differentiates from its isochemical liquid, but this does not go through structural breakdown or the glass transition.
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answers
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm
Stars have different colors. What causes stars to have colors?
A. location
B. temperature
C. oxygen
D. carbon dioxide
Answers
Answer:
temperature
Explanation:
temperature change forms different elements and different element sustain different colour
A wheel has a diameter of 10m and weight 360N what minimum horizontal force is necessary to pull the wheel over a brick 0.1m when a force is applied at the wheel
Answers
Cold air rises because it is denser than water, is this true?
Answers
Answer:
true
Explanation:
im not sure please dont attack me
A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at
t = 0
from the position
x = 0.0480 m.
Determine the location of the mass at
t = 5.85 s?
Answers
Answer:
[tex]X=0.0389m[/tex]
Explanation:
From the question we are told that:
Period of spring [tex]T_s=2.25s[/tex]
Initial Position of Mass [tex]x=0.0480m[/tex]
Final Mass period [tex]T_f=5.85s[/tex]
Generally the equation for the Mass location is mathematically given by
[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]
[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]
[tex]X=0.0389m[/tex]
A ball drops from a height, bounces three times, and then rolls to a stop when it reaches the ground the fourth time.
At what point is its potential energy greatest?
At what points does it have zero kinetic energy?
At what point did it have maximum kinetic energy?
Answers
Answer:
Greatest potential: moment before being dropped
Zero Kinetic: when it comes to rest
Greatest Kinetic: moment before first bounce
Explanation:
A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance d by a rope supplying a force F at an angle of elevation q. The surface has a frictional force acting during this motion. How much work was done by friction during this motion? The student calculates the value to be –Fd sinq. How does this value compare to the correct value?
a. It is the correct value.
b. It is too high.
c. It is too low.
d. The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Answers
Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resistor that makes up part of the parallel circuit?
Answers
Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesn't skid, what is the force exerted on it by static friction?
Answers
Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
Oxygen is obtained through various methods. Which of the following methods involves a chemical
change?
1. Electrolysis of water
2. Distillation of liquid air
3. Heating of KCIO,
02
1 and 2
1 and 3
Answers
Answer:
1
Explanation:
Electrolysis is the passing of an current through a conducting solution, when the occurs, a chemical reaction takes place.
Heating a chemical will always cause a chemical reaction, which is why 3 is also correct
Some information as to why 2 is NOT correct.
2 is NOT a chemical reaction, but rather a process of physical separation. It uses selective boiling and condensation, but is not considered a chemical reaction.
as with 3, heating is not considered a chemical reaction, but rather a physical temperature change. This is always what it is considered to be (e.g boiling water is a physical temperature change, not a chemical reaction)
Hope this helps.
Hope this helps.
In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a 120-V receptacle on a wall. The picture tube of the television set uses 76 W, and there is 5.5 mA of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio Ns/Np of the transformer.
Ns/Np = ______.
Answers
Answer:
c) N_s / N_p = 115.15
Explanation:
Let's look for the voltage in the secondary, they do not indicate the power dissipated
P = V_s i
V_s = P / i
V_s = 76 / 5.5 10⁻³
V_s = 13.818 10³ V
the relationship between the primary and secondary of a transformer is
[tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]
[tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]
Ns / Np = 13,818 10³ /120
N_s / N_p = 115.15
An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine
Answers
Answer:
If efficiency is .22 then W = .22 * Q where Q is the heat input
Heat Input Q = 2510 / .22 = 11,400 J
Heat rejected = 11.400 - 2510 = 8900 J of heat wasted
Also, 8900 J / (4.19 J / cal) = 2120 cal
An efficiency is the measure of productivity of an engine. The heat rejected by the engine is 8900 Joules.
What is efficiency?An efficiency of a heat engine is the ratio of the work done and heat supplied.
Given is the automobile engine has the efficiency 22% and Work done is 2510 Joules.
The efficiency is written as,
η= W / Qs.
The work done is W= Qs - Qr, where Qr is the rejected heat.
The heat rejected can be represented as
Qr = W ( 1/η -1)
Substituting the value into the equation, we get the rejected heat.
Qr = 2510 (1/0.22 -1)
Qr = 8900 Joules.
Thus, the heat rejected by the engine is 8900 Joules.
Learn more about efficiency.
#SPJ2
A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18 m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.
Required:
a. Find the motional emf induced between the ends of the segment.
b. What would the motional emf be if the wire segment was parallel to the y-axis?
Answers
Answer:
Explanation:
From the information given:
The motional emf can be computed by using the formula:
[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]
[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]
[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]
[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]
[tex]E = 0.50*((18*0.800)[/tex]
E = 0.72 volts
According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.
As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .
Then the motional emf will be:
[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]
E = 0 (zero)
Solids diffuse because the particles cannot move.
A. Can
B. Not enough info
C. Cannot
D. Sometimes will
Answers
Solids cannot diffuse.
They don’t space to move.
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
Answers
[tex]E_0=1.5033×10^{-10}\:\text{J}[/tex]
Explanation:
The rest energy [tex]E_0[/tex] of a proton of mass [tex]m_p[/tex] is given by
[tex]E_0 = m_pc^2[/tex]
[tex]\:\:\:\:\:\:\:=(1.6726×10^{-27}\:\text{kg})(2.9979×10^8\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:=1.5033×10^{-10}\:\text{J}[/tex]
A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy
Answers
Answer:
elastic energy
Explanation:
When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.
During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.
The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:
elastic energy
Explain why the flow from the battery increases when the switch is closed. Give the label of the concept(s) that you use from the model of electricity. [
Answers
Answer:
Due to the applied filed the electrons move in a particular direction.
Explanation:
Initially when the switch is off, the free electrons move here and there in any random directions in the conductor with the random speeds called thermal velocity. So, tat the net flow is almost zero.
When the battery is connected is switch is ON, the random motion of the electrons aligned in a particular direction due to the force applied by the electric filed, so the net flow is not zero it increases and thus the current flow.
In the following calculations, be sure to express the answer in standard scientific notation with the appropriate number of
significant figures.
3.88 x 1079 - 4.701 x 1059
x 10
g
Answers
Answer:
-45,597.07
Explanation:
if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh
1. What is the total distance the car moves until it stops?
a. 250 m
b. 450 m
c. 300 m
d. 600 m.
Answers
Just took the quiz bro it was easy
A proton is held at rest in a uniform electric field. When it is released, the proton will gain:_________
a) electrical potential energy.
b) kinetic energy.
c) both kinetic energy and electric potential energy.
d) either kinetic energy or electric potential energy.
Answers
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Answers
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
An object with mass m is located halfway between an object of mass M and an object of mass 3M that are separated by a distance d. What is the magnitude of the force on the object with mass m?A) 8GMm/d^2B) GMm/(4d^2)C) 4GMm/d^2D) GMm/(2d^2)E) 3GMm/2d^2
Answers
Answer:
A) 8GMm/d^2
Explanation:
We are given that
[tex]m_1=M[/tex]
[tex]m_2=3M[/tex]
[tex]m_3=m[/tex]
Distance between m1 and m2=d
Distance of object of mass m from m1 and m2=d/2
Gravitational force formula
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
Using the formula
Force acting between m and M is given by
[tex]F_1=\frac{GmM}{d^2/4}[/tex]
Force acting between m and 3M is given by
[tex]F_2=\frac{Gm(3M)}{d^2/4}[/tex]
Now, net force acting on object of mass is given by
[tex]F=F_2-F_1[/tex]
[tex]F=\frac{Gm(3M)}{d^2/4}-\frac{GmM}{d^2/4}[/tex]
[tex]F=\frac{12GmM}{d^2}-\frac{4GmM}{d^2}[/tex]
[tex]F=\frac{12GmM-4GmM}{d^2}[/tex]
[tex]F=\frac{8GmM}{d^2}[/tex]
Hence, the magnitude of the force on the object with mass m=[tex]\frac{8GmM}{d^2}[/tex]
Option A is correct.
which is the correct formula for calculating the age of meteor right if using half life
Answers
Answer:
n × t_1/2
Exmplanation:
The age of meteorite is calculated by multiplying it's quantity n with the half life . This means that the formula is for age of this meteorite is;
Age of meteorite= n × t_1/2
where;
n = quantity of the meteorite
t_/2 = half life of the meteorite
Thus:
The correct formula is; n × t_1/2
