Answer:
Explanation:
Given that:
length of side , a = 0.5 m
charge , q = 6.65 mC
length of diagonal , d = 0.5 * sqrt(2)
d = 0.707 m
F is the force due to adjacent particle ,
F1 is the force due to diagonal particle
Now , for the net charge on a particle
Fnet = 2 * F * cos(45) + F1
Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2
Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)
Fnet = 3.05 *10^6 N
the magnitude of net force acting on each particle is 3.05 *10^6 N
part B)
for the direction of particle
d) along the line between the charge and the center of the square outward of the center
An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the negative x direction and the magnetic field is pointing in the negative y direction. In which direction does the released electromagnetic wave travel
Answer: the magnetic wave will travel out of the screen.
Explanation:
Electric field direction is perpendicular to the magnetic field direction. Both are also perpendicular to the direction of the particles.
Using right hand rule to solve this problem,
This pointed finger depicts the electric field direction which the curly fingers depict the direction of the magnetic field. The pointed thumb will depict the direction in which the wave travel. Which is out of the screen.
A human expedition lands on an alien moon. One of the explorers is able to jump a maximum distance of 16.0 m with an initial speed of 2.90 m/s. Find the gravitational acceleration on the surface of the alien moon. Assume the planet has a negligible atmosphere. (Enter the magnitude in m/s2.)
Answer:
Gravitational acceleration (g) = 0.4205 m/s²
Explanation:
Given:
Distance (R) = 20 m
Initial speed (u) = 2.90 m/s
Find:
Gravitational acceleration (g)
Computation:
⇒ Distance (R) = [Initial speed (u)]²/ Gravitational acceleration (g)
⇒ Gravitational acceleration (g) = [Initial speed (u)]² / Distance (R)
⇒ Gravitational acceleration (g) = 2.90 ² / 20
⇒ Gravitational acceleration (g) = 8.41 / 20
⇒ Gravitational acceleration (g) = 0.4205 m/s²
A spring stretches by 0.0190 m when a 3.36-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Answer:
m = 4.87 kg
Explanation:
In order to find the required mass you first calculate the spring constant of the spring. When the system reaches the equilibrium you obtain the following equation:
[tex]Mg=kx[/tex] (1)
That is, the weight of the object is equal to the restoring force of the spring.
M: mass of the object = 3.36 kg
g: gravitational constant = 9.8m/s^2
k: spring constant = ?
x: elongation of the spring = 0.0190m
You solve the equation (1) for k:
[tex]k=\frac{Mg}{x}=\frac{(3.36kg)(9.8m/s^2)}{0.0190m}=1733.05\frac{N}{m}[/tex]
Next, to obtain a frequency of 3.0Hz you can use the following formula, in order to calculate the required mass:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex] (2)
You solve the equation (2) for m:
[tex]m=\frac{1}{4\pi^2}\frac{k}{f^2}\\\\m=\frac{1}{4\pi^2}\frac{1733.05N/m}{(3.0Hz)^2}=4.87kg[/tex]
The required mass to obtain a frequency of 3.0Hz is 4.87 kg
This problem concerns the properties of circular orbits for a satellite of mass m orbiting a planet of mass M in an almost circular orbit of radius r. In doing this problem, you are to assume that the planet has an atmosphere that causes a small drag due to air resistance. "Small" means that there is little change during each orbit so that the orbit remains nearly circular, but the radius can change slowly with time. The following questions will ask about the net effects of drag and gravity on the satellite's motion, under the assumption that the satellite's orbit stays nearly circular. Use G if necessary for the universal gravitational constant.
What is the potential energy U of the satellite?Express your answer in terms ofm, M, G, and r.What is the kinetic energy K of the satellite?Express the kinetic energy in termsof m, M, G, and r.
Answer:
A) U = - GMm/r
B) K = 0.5 mGM/r
Explanation:
A) The potential energy U of the satellite
U = - GMm/r
G = universal gravitational constant which is ( 6.67e-11 Nm^2/c^2 )
M = mass of the planet
m = mass
r = distance ( radius )
B) Kinetic energy
kinetic energy expressed as K = 0.5 m Vo^2
NOTE : Vo^2 = GM / r
hence kinetic energy will be expressed as
K = 0.5 mGM/r
Linear charge density 4.00×10−12 C/m surrounds an infinitely long line charge. A positively charged elementary particle (mass 1.67×10−27 kg, charge +1.60×10−19 C) is 15.0 cm from this line charge. Consider that this elementary particle is moving at speed 3.20×103 m/s directly toward the line charge.
Part A- Find the initial kinetic energy of this elementary particle.
Part B- Find the closest distance that the elementary particle get to the line charge?
Answer:
A)Kopya
B)YASAK
Explanation:
kopya yasak dostum adın da belli. Başın belaya girmesin
A transverse sinusoidal wave is traveling rightward along a string stretched along the X-direction. The wavelength is 70.0 cm and 8.00 wave peaks per second pass any fixed point on the string. The maximum speed of the string particles in the direction perpendicular to the string is 3.20 m/s. At time t 0, the element of string at x = 20.0 cm is 4.00 cm above its resting position. (a) Construct two distinct wave functions yfx,t) consistent with the given information. (If the only difference between two wavefunctions is that their phase constants differ by an integer multiple of 2n, then they are not distinct.) (b) Sketch both of these wavefunctions for time t 0 between x 0 and x A har 2 20 /s A= 0.0Ch 560 Ch a)yo) Acos CRxt wt +o) w- 2TF : 50 265 S A Vwa 3 20 5a.2655 =0.0656n 0.7854
Answer:
1111
1Explanation:
111111
1
Name and draw the devices that can convert digital signal to analog.
Answer:
modem, digital music players, optical communication
Explanation:
Photos of devices are attached. An example of R2R logic circuit is also attached. DAC conversion is also done via pulse code modulation
An isotope has 46 electrons, 60 neutrons, and 46 protons. Name the isotope.
Answer:
Palladium
Explanation:
Answer:
palladium-106
Explanation:
46 protons -46 electrons=no charge
46 electrons +60neutron = 106
Thus this is called palladium -106
A gas is collected from a radioactive material; upon inspection, the gas is identified as helium. The presence of the helium indicates the radioactive sample is most likely decaying by: A). alpha B). beta+ C). beta- D). gamma
Answer:
option (a) alpha I have doublt
A 35 grams bullet travels with a velocity of magnitude 126 km/h. What is the bullet's linear momentum?
The linear momentum of the bullet, given the data from the question is 1.225 Kg.m/s
What is momentum?Momentum is defined as the product of mass and velocity. It is expressed as
Momentum = mass × velocity
With the above formula, we can obtain the momentum of the bullet. Details below.
The following data were obtained from the question:
Mass of bullet = 35 g = 35 / 1000 = 0.035 KgVelocity = 126 Km/h = 126 / 3.6 = 35 m/sMomentum =?Momentum = mass × velocity
Momentum = 0.035 Kg × 35 m/s
Momentum = 1.225 Kg.m/s
From the calculation made above, we can conclude that the linear momentum of the bullet is 1.225 Kg.m/s
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A baton twirler in a marching band competition grabs one end of her 1.2 kg, 1.0 meter long baton. She throws her baton into the air such that it rises to a height of 5.0 meters while spinning end over end at a rate of 3.5 revolutions per second. How much work did she do on the baton?
Answer:
349 J
Explanation:
Length L of baton = 1.0 m
Mass m of baton = 1.2 kg
Weight W of baton = 1.2 kg x 9.81 m/[tex]s^{2}[/tex] = 11.772 N
Height h reached = 5.0 m
Angular speed ω = 3.5 rev/s = 2π x 3.5 (rad/s) = 21.99 rad/s
Total work done on baton will be the work done in taking it to a height of 5.0 m and the kinetic energy with which the baton rolls.
Work done to bringing it to the height of 5.0 m = weight x height above ground
W x h = 11.772 x 5 = 58.86 J
Velocity v of spinning baton = ω x L = 21.99 x 1 = 21.99 m/s
Kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] =
Total work done on baton = 58.86 + 290.14 = 349 J
Consider a system of an 85.0 kg man, his 14.5-kg dog, and the earth. The gravitational potential energy of the system increases by 1.85 103 J when the man climbs a spiral staircase from the first to the second floor of an apartment building. If his dog climbs a normal staircase from the same first floor to the second floor, by how much does the potential energy of the system increase (in J)
Answer:
Explanation:
Increase in gravitational potential energy = m x g x h
where m is mass , g is gravitational acceleration and h is height
In the first case when man climbs
increase in potential = 85 x g x h = 1.85 x 10³ J
gh = 21.7647
when dog climbs
increase in potential = 14.5 x g x h J
= 14.5 x 21.7647
= 315.6 J
An AC voltage is applied to a purely capacitive circuit. Just as the applied voltage is crossing the zero axis going negative, what is the value of capacitor current? Ic is at its positive peak Ic is zero Ic is at its negative peak
Answer:
The Ic will be zero.
Explanation:
Capacitors have a working principal as follows:
As the current flows through the circuit, they store the electrical energy according to certain attributes they have such as the area of the plates and the material's capacitence in between the plates.An AC voltage increases and decreases between certain maximum and minimum points periodically. So while the AC voltage is on the positive side, the capacitor charges up and when the AC voltage crosses to the negative side, the capacitor takes over and it's current starts increasing as the current coming from the AC source decreases.
So in this case, as the AC voltage crosses zero, the capacitor current was decreasing because the AC voltage was on the positive side and it was charging. The capacitor current will be zero as well and it will start to increase when AC voltage is on the negative.
I hope this answer helps.
The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.95 × 103 N with an effective perpendicular lever arm of 3.1 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia of the boxer's forearm?
Answer:
I = 0.483 kgm^2
Explanation:
To know what is the moment of inertia I of the boxer's forearm you use the following formula:
[tex]\tau=I\alpha[/tex] (1)
τ: torque exerted by the forearm
I: moment of inertia
α: angular acceleration = 125 rad/s^2
You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)
[tex]\tau=Fr=(1.95*10^3N)(0.031m)=60.45J[/tex]
Next, you replace this value of τ in the equation (1) and solve for I:
[tex]I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2[/tex]
hence, the moment of inertia of the forearm is 0.483 kgm^2
A long metallic wire is stretched along the x direction. The applied potential difference along
the wire is 12 volts. The resistance of the wire is estimated to be 8 Ohm.
a. Calculate the current in the wire and the passing charge in 2 seconds.
b. Calculate the electric power dissipated in the wire.
This wire has a length of 100 m and the material of the wire has a resistivity of 1.6 x 10 -8 Ωm.
c. Calculate the cross-section area of the wire.
d. Calculate the conductivity of the wire.
e. What is the capacitance of the capacitor, that is connected to a series resistor of 8 MΩ in an RC
circuit that has a time constant of one second.
Answer:
a) I = 1.5 A , b) P = 18 W , c) A = 8 10⁻⁷ m², d) σ = 0.625 10⁸ (Ω m)⁻¹,
e) C = 0.125 10⁻⁶ F
Explanation:
a) for this exercise let's use Ohm's Law
V = I R
I = V / R
I = 12/8
I = 1.5 A
b) the power is given by the expression
P = V I
P = V V / R = V² / R
P = 12²/8
P = 18 W
c) the resistance of the wire is given by
R = ρ l / A
A = ρ l / R
A = 1.6 10⁻⁸ 100/8
A = 8 10⁻⁷ m²
d) conductivity is the inverse of resistivity
σ = 1 / ρ
σ = 1 / 1,610⁻⁸
σ = 0.625 10⁸ (Ω m)⁻¹
e) In an RC circuit the response time is
τ = RC
C = τ / R
C = 1/8 10⁶
C = 0.125 10⁻⁶ F
list and discuss how the nature of a rural settlements affect the type and expanse of agricultural activities
Answer:
Availability of land for Agricultural activities- The rural areas are known for a lesser degree of development which means lesser factories and other work buildings. The area has undeveloped lands which are usually used for a commercial type of agricultural activities.
Bad road networks: Bad road networks are mainly associated with rural settlements. This hinders to an extent the agricultural activities of planting and harvesting of crops due to difficulties in moving of the crops.
Consider five charged particles: A,B,C,D,and E.
(A attracts B),(C attracts D ), (B repels C), and (D repels E). If C is Positive,what is the charge of the other particles?
Answer:
A_negative
B_positive
C_positive
D_positive
E_negative
Explanation:
according to the law of electrostatic which is
like charge repels and unlike charge attract.
Answer:A=negative, B=positive, C=positive, D=negative, E=negative
Explanation:
Like poles or charges repels and unlike poles or charges attract each other
Which action is due to field forces?
A. an apple falling from a tree
B. a moving car stopping when the brakes are applied
C. the rowing of a boat
D. pushing a chair against the wall
Answer:
a
an apple falling from a tree
Answer an apple falling from a tree
Explanation:
Silver and Copper rods of equal areas are placed end to end with the free end of the silver rod in ice at 0.00 degrees Celsius and the free end of the copper rod in steam at 100. degrees Celsius. The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.
a) What is the temperature of the junction between copper and sliver when they have come to equilibrium?
b) How much ice (in grams) melts per second?
Answer:
A.) The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius
B.) ice (in grams) melts per second = 0.078 kg/s
Explanation:
A.) Given that the two material are of the same area.
The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.
Silver temperature = 0 degree Celsius
Copper temperature = 100 degree Celsius
Thermal conductivity k of silver = 429 W/m•K
Thermal conductivity k of copper = 385W/m.k
Rate of energy transferred P in the two materials can be expressed as
P = k.A.dT/L
dT = change in temperature
Since the rate and the area are the same
429 ( T -0 )/0.15 = 385( 100 - T )/0.25
2860T = 1540(100 - T)
Open the bracket
2860T = 154000 - 1540T
Collect the like terms
2860T + 1540T = 154000
4400T = 154000
T = 154000/4400
T = 35 degree Celsius
The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius
B.) Rate of energy transferred P will be
P = 2860 × 35 = 100100
P = Q/t ..... (1)
Where Q = energy transferred
But Q = mcØ .....(2)
And specific heat capacity c of water = 4182J/k.kg
Substitutes Q into formula 1.
P = mcØ/t
Make m/t the subject of formula
m/t = P/cØ
m/t = 100100/ 4182( 35 + 273 )
m/t = 100100/1288056
m/t = 0.078 kg/s
A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45 degrees. With what speed was it thrown?
Answer:
Explanation:
This is the case of horizontal projection from a height:
Time, t = sqrt ( 2h / g )
= sqrt ( 2 * 20 / 9.8 )
= 2.02 s
Vfx = V
Vfy = g* t = 2.02 g
theta (θ)= 45 deg
tan theta (tan θ) = Vfy / Vfx
tan 45 = 2.02 g / V
V = 2.02 * 9.8
= 19.8 m/s
≅ 20m/s
A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.11 m and a mass of 0.010 kg. A uniform magnetic field of magnitude 0.055 T is directed from the ceiling to the floor. When a current of I = 29 A exists in the wire, the wire swings upward and, at equilibrium, makes an angle φ with respect to the vertical, as the drawing shows. Find (a) the angle and (b) the tension in each of the two strings.
Answer:
Explanation:
The magnetic force acting horizontally will deflect the wire by angle φ from the vertical
Let T be the tension
T cosφ = mg
Tsinφ = Magnetic force
Tsinφ = BiL , where B is magnetic field , i is current and L is length of wire
Dividing
Tanφ = BiL / mg
= .055 x 29 x .11 / .010 x 9.8
= 1.79
φ = 61° .
Tension T = mg / cosφ
= .01 x 9.8 / cos61
= .2 N .
Two identical metal balls of radii 2.50
cm are at a center to center distance of
1.00 m from each other. Each ball is
charged so that a point at the surface of
the first ball has an electric potential of
+1.20 x 103 V and a point at the surface
of the other ball has an electric
potential of -1.20 x 103 V. What is the total charge on each ball?
Answer:
+1.33 × [tex]10^{-7}[/tex] C and -1.33 × [tex]10^{-7}[/tex] C respectively.
Explanation:
Electric potential (V) is the work done in moving a unit positive charge from infinity to a reference point within an electric field. It is measured in volts.
V = [tex]\frac{kq}{r}[/tex] ............. 1
where: k is a constant = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex], q is the charge and r is the distance between the charges.
From equation 1,
q = [tex]\frac{Vr}{k}[/tex] ............... 2
The charge on each ball can be determined as;
given that; V = 1.2 × [tex]10^{3}[/tex], k = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex] and r = 1.00 m.
From equation 2,
q = [tex]\frac{1.2*10^{3} * 1.0}{9*10^{9} }[/tex]
= 1.33 × [tex]10^{-7}[/tex] C
Thus, the charge on the first ball is +1.33 × [tex]10^{-7}[/tex] C, while the charge on the second ball is -1.33 × [tex]10^{-7}[/tex] C.
A certain radio wave has a wavelength of 6.0 × 10-2m. What is its frequency in hertz?
Answer:
The frequency of the wave is 5 x 10⁹ Hz
Explanation:
Given;
wavelength of the radio wave, λ = 6.0 × 10⁻²m
radio wave is an example of electromagnetic wave, and electromagnetic waves travel with speed of light, which is equal to 3 x 10⁸ m/s².
Applying wave equation;
V = F λ
where;
V is the speed of the wave
F is the frequency of the wave
λ is the wavelength
Make F the subject of the formula
F = V / λ
F = (3 x 10⁸) / (6.0 × 10⁻²)
F = 5 x 10⁹ Hz
Therefore, the frequency of the wave is 5 x 10⁹ Hz
Two resistors, A and B, are connected in parallel across a 8.0 V battery. The current through B is found to be 3.0 A. When the two resistors are connected in series to the 8.0 V battery, a voltmeter connected across resistor A measures a voltage of 2.4 V. Find the resistances A and B.
Answer:
R_A = 2.67 ohms
R_B = 1.14 ohms
Explanation:
When the resistors are connected in parallel, the voltage will be the same across both resistors A and B.
Thus, we now have the current and the voltage across B and so we can use Ohm's Law to find the resistance.
V/I = R
Thus, resistance of B; R_B = 8/3
R_B = 2.67 ohms
Now, when the resistors are connected in series, the voltage drop across B is;
V = 8V - 2.4V = 5.6V
Since we now have the resistance of B , we can find the current using Ohm's Law. Thus;
I = V/R
I = 5.6/2.67
I = 2.1 A
Now, current is the same for all resistances in a series circuit because this is the same current through resistors A and B. So, we can use Ohm's law again to find the resistance across A.
So, R = V/I
R_A = 2.4/2.1
R_A = 1.14 ohms
a body of 2.0kg mass makes an elastic collision with another at rest and continues to move in the original direction but with 1/4 of it's original speed.what is the mass of the struck body?
2.0 kg into the bifactor of 1/4 so the mass struck would be 32
The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click initialize to set up the animation and start to run it.
Ghosts are left by the ball once per second. The animation can also be paused and moved forward in single frame mode using the step button. The cursor can be used to read the (x,y) coordinates of a position in the grid by holding down the left mouse button. Assume the grid coordinates read out in meters. When entering components, presume that x is positive to the right and y is positive upwards. Note that this ball is NOT being kicked on Earth. Do not expect an acceleration of 9.80 m/s2 downward, though you can presume that gravity is acting straight down. Use this animation to answer the following questions. Note that there are a number of different ways to go about each of the following questions. Your answer needs to be within 5% of the correct answer for credit. Please enter your answer to 3 significant digits.
What is the maximum height which the ball reaches? 42.24 m
What is the horizontal component of the initial velocity of the ball? 5.57 m/s
What is the vertical component of the initial velocity of the ball? 16.18 m/s
What is the vertical component of the acceleration of the ball? _____????
Answer:
The acceleration of the ball is [tex]a_y = - 0.3672 \ m/s^2[/tex]
Explanation:
From the question we are told that
The maximum height the ball reachs is [tex]H_{max} = 42.24 \ m[/tex]
The horizontal component of the initial velocity of the ball is [tex]v_{ix} = 5.57 \ m/s[/tex]
The vertical component of the initial velocity of the ball is [tex]v_{iy} = = 16.18 m/s[/tex]
The vertically motion of the ball can be mathematically represented as
[tex]v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}[/tex]
Here the final velocity at the maximum height is zero so [tex]v_{fy} = 0 \ m/s[/tex]
Making the acceleration [tex]a_y[/tex] the subject we have
[tex]a_y = \frac{v_{iy} ^2}{2H_{max}}[/tex]
substituting values
[tex]a_y = - \frac{5.57^2}{2* 42.24}[/tex]
[tex]a_y = - 0.3672 \ m/s^2[/tex]
The negative sign shows that the direction of the acceleration is in the negative y-axis
A 12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s2. What is the magnitude of the contact force exerted on the block by the surface?
Answer:
157.36 N
Explanation:
Contact force is the force which is created due to contact and it is applied on the contact point . The force applied by body on the surface is its weight .
If R be the reaction force of the ground
R = mg + F son30
= 12 x 9.8 + 75 sin 30
= 117.6 + 37.5
= 155.10 N .
friction force = f
Net force in forward direction = F cos 30 - f = ma
75cos 30 - f = 12 x 3.2
f = 65 - 38.4
= 26.6 N
Total force on the surface =√( f² + R² )
√ (26.6² + 155.1²)
= √707.56 + 24056²
=√ 24763.57
= 157.36 N.
contact force = 157.36 N .
A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it is released and explain how you know. b. If work is done on the particle after its release, is the work positive or negative
Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.
A neutron star has about one and a half times the mass of our Sun but has collapsed to a radius of 10 kmkm . Part A What is the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface?
Answer:
gₓ = 1.36 x 10¹³ g
Explanation:
The value of acceleration due to gravity at a certain place is given by the following formula:
gₓ = GM/R²
where,
gₓ = acceleration due to gravity on the surface of neutron star
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the star = 10 * Mass of sun = (10)(2 x 10³⁰ kg) = 2 x 10³¹ kg
R = 10 km = 10⁴ m
Therefore,
gₓ = (6.67 x 10⁻¹¹ N.m²/kg²)(2 x 10³¹)/(10⁴)²
gₓ = 1.334 x 10¹⁴ m/s²
Hence, comparing it with the free-fall acceleration at Earth's Surface:
gₓ/g = (1.334 x 10¹⁴)/9.8
gₓ = 1.36 x 10¹³ g
The acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].
The given parameters:
Mass of the neutron star, m = 1.5 MRadius of the neutron star, R = 10 kmkmThe acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is calculated as follows;
[tex]F = mg = \frac{GM_sm}{R^2} \\\\(1.5 M_s)g = \frac{GM_s(1.5 M_s)}{R^2} \\\\g = \frac{GM_s}{R^2} \\\\[/tex]
where;
[tex]M_s[/tex] is the mass of the Sun = 1.989 x 10³⁰ kg.
[tex]g = \frac{6.67 \times 10^{-11} \times 1.989 \times 10^{30} }{(10,000,000)^2} \\\\g = 1.326 \times 10^{6} \ m/s^2[/tex]
In terms of gravity of Earth [tex](g_E)[/tex];
[tex]= \frac{1.326 \times 10^6}{9.81} = 1.35 \times 10^5 \\\\= 1.35 \times 10^5 \ g_E[/tex]
Thus, the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].
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Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 52.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.936 m/s2. Calculate her mass.
Answer:
55.56kg
Explanation:
Given:
F= 52N
a=0.936m/s²
Applyinc Newton's second law, that states: force is equal to mass times acceleration.
F = ma
m=F/a =>52 / 0.936
m=55.56kg