Answer:
30
Explanation:
A cylindrical flask is fitted with an airtight piston that is free to slide up and down. A mass rests on the top of the piston. The initial temperature of the system is 313 K and the pressure of the gas is held constant at 137 kPa. The temperature is now increased until the height of the piston rises from 23.4 cm to 27.0 cm.What is the final temperature of the gas?
Answer:
T2 = 361.15 K
Explanation:
Since pressure is held constant, then we'll use Charles law
Charles law gives that V1/T1 = V2/T2
We are given;
Initial temperature; T1 = 313 K
Initial height; h1 = 23.4 cm
Final height; h2 = 27.0 cm
Now, formula for volume of a cylinder is V = πr²h
V1 = πr²h1 and V2 = πr²h2
Now, since V1/T1 = V2/T2 and we are looking for T2, let's make T2 subject;
T2 = (V2 × T1)/V1
Plugging in the relevant values gives;
T2 = (πr²h2 × 313)/πr²h1
πr² will cancel out to give;
T2 = (h2 × 313)/h1
Putting 23.4 for h1 and 27 for h2 gives;
T2 = (27 × 313)/23.4
T2 = 361.15 K
Calculate the angle θ between the radius-vector of the point and the positive x axis (measured counterclockwise from the positive x axis, within the limits of −180◦ to +180◦ ). Answer in units of ◦
The point obviously is in the 3rs quadrant
So
စ= tan^-1( y/x)-180
စ= -89.7°
The angle θ between the radius-vector of the point and the positive x axis is -89.7°
What is radius vector?Radius vector is defined as the length of the line that connects a stationary point to a moving point.
It can be defined as a line connecting a circle's center to a particle's location on its circumference.
It can also be defined as the straight line connecting two bodies moving in relative orbits, such as the path a planet takes to reach the sun at any given point along its orbit.
It is also known as position vector or location vector.
The unit vector pointing radially outward from the origin is the gradient of the position vector's length. It is typical for the spheres that are centered on the origin to have level surfaces.
Evidently, the location is in the third quadrant.
So, θ= tan⁻¹ (y / x) - 180
θ= - 89.7°
Thus, the angle θ between the radius-vector of the point and the positive x axis is -89.7°
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A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.700 m2 . At the window, the electric field of the wave has an rms value 0.0400 V/m .
How much energy does this wave carry through the window during a 30.0-s commercial?
Express your answer with the appropriate units.
Answer:
The value is [tex]E = 8.9 *10^{-5} \ J[/tex]
Explanation:
From the question we are told that
The area is [tex]A = 0.700 \ m^2[/tex]
The root mean square value is [tex]E_{rms} = 0.0400 \ V/m[/tex]
The time taken is [tex]t = 30.0 \ s[/tex]
Generally the energy is mathematically represented as
[tex]E = c * \sepsilon_o * A * t * E_{rms}^2[/tex]
=> [tex]E = 3.0*10^{8} * 8.85*10^{-12} * 0.700 * 30 * (0.04)^2[/tex]
=> [tex]E = 8.9 *10^{-5} \ J[/tex]
Consider a metal rod of uniform temperature. Is it possible that heat spontaneously flow from one end of the rod to the other end so that they end up in different temperatures? Why?
Answer:
No its not possible in that yes heat flow is due to temperature different but with time equilibrium in temperature is reached hence there is no temperature difference at both ends at this time
What mirror diameter gives 0.1 arc second resolution for infrared radiation of wavelength 2 micrometers?
Answer:
The diameter of mirror [tex] 5\times10^{-6}\ m[/tex]
Explanation:
Given that,
Wavelength = 2 μm
Resolution = 0.1
We need to calculate the diameter of mirror
Using formula of resolution
[tex]resolution=0.25\times\dfrac{wavelength}{diameter}[/tex]
[tex]diameter =0.25\times\dfrac{wavelength}{resolution}[/tex]
Put the value into the formula
[tex]diameter=0.25\times\dfrac{2\times10^{-6}}{0.1}[/tex]
[tex]diameter = 0.000005\ m[/tex]
[tex]diameter = 5\times10^{-6}\ m[/tex]
Hence, The diameter of mirror [tex] 5\times10^{-6}\ m[/tex]
the pressure of a gas is 100.0kpa and its volume is 500.0ml if the volume increase to 1000.0ml what is the new pressure of the gas
50Kpa
Explanation:
P1V1 = P2V2
Where;
P1= 100. P2= ?
V1 = 500. V2 = 1000
100 × 500 = P2 × 1000
50000 = 1000P2
50000/1000 = P2
50 = P2
P2 = 50Kpa
Manganese-52 has a half-life of 6 days. How many days would a scientist have to wait for the radioactivity to be 12.5% the starting amount?
Answer:
Let N = N0 where N0 is the number of atoms originally present.
In 6 days N = N0 / 2
In 12 days N = N0 / 4
In 18 days N = N0 / 8 = .125 N0
So it would take 18 days.
Manganese-52 has a half-life of 6 days. a scientist has to wait 18 days for the radioactivity to be 12.5% of the starting amount
What is radioactivity?The ability of some unstable atoms to emit nuclear radiation spontaneously, typically in the form of alpha or beta particles frequently accompanied by gamma rays, is known as radioactivity. This radiation is released when a nucleus undergoes radioactive decay and transforms into a different isotope that may be radioactive (unstable) or non-radioactive depending on the number of neutrons and protons in the nucleus (stable).
As given in the problem statement Manganese-52 has a half-life of 6 days. this means after 6 days Manganese-52 will reduce to its half value
similarly, after 12 days it further reduces to 50% of the previous value which would be 25 %
Similarly, after 18 days, it will reduce 50% from the previous value which comes out to be 12.5 %
Thus, The half-life of manganese-52 is 6 days. It takes 18 days for the radioactivity to drop to 12.5% of the initial level.
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all digits shown on the measuring device, plus one estimated digit. are consided what
Answer:
The significant figures in a measurement consist of all the certain digits in that measurement plus one uncertain or estimated digit. In the ruler illustration below, the bottom ruler gave a length with 2 significant figures, while the top ruler gave a length with 3 significant figures.
Explanation:
Daniel has a bill of 2750 on his credit card. If interest is charged at a rate of 15% p.a., calculate the amount of interest that Daniel must pay for the month?
Answer:
412.5
Explanation:
all you have to do is multiply .15
An athlete completes a round of circular tract with diameter 150 m in 5 min. What is the average speed V of the athlete in m/s? Answer with two decimal places
Answer:
Explanation:
From the question we are told that
The diameter is [tex]d = 150 \ m[/tex]
The time taken is [tex]t = 5 \ min = 300\ s[/tex]
Generally the circumference is mathematically represented as
[tex]C = d * \pi[/tex]
[tex]C = 150 * 3.142[/tex]
[tex]C = 471.3 \ m[/tex]
The average speed is mathematically represented as
[tex]v = \frac{C}{ t}[/tex]
[tex]v = \frac{471.3}{300}[/tex]
[tex]v = 1.571 \ m/s[/tex]
A softball player is testing her physics knowledge and tosses a softball upward from the top of a building. The building is 100 m tall and the ball starts with a velocity of 10 m/s.a. What is the maximum height the ball reaches and at what time? b. How much time does it take for the ball to go from h=50 m to h=0 m on the way down? c. What is the velocity when h=50 m?
Answer:
Explanation:
We shall apply newton's laws formula
a )
initial velocity in upward direction u = 10 m/s
acceleration due to gravity g = 9.8/ m .s²
Let h be the maximum height where v = o
v² = u² - 2gh
0 = 10² - 2 gh
h = 10² / 2g
= 10² / 2 x 9.8
= 5.10 m
Since the ball was thrown from height of 100 m , total maximum height of ball
= 100 + 5.10
= 105.10 m
Let t be the time taken
v = u - gt
0 = 10 - gt
t = 10 / 9.8
= 1.02 s
b )
when h = 50 on its way downwards , velocity
v² = u² + 2 g s
v² = 0 + 2 x 9.8 x ( 105.10 - 50 )
[ distance travelled by ball at this point from top = 105.1 - 50 = 55.10 ]
v = 32.86 m / s
Let us find out final velocity of touching the ground . For it distance travelled = 105.10
v² = u² + 2gh
v² = 0 + 2 x 9.8 x 105.1
v = 45.39 m /s
Now velocity at h = 50 is 32.86
velocity at h = 0 is 45.39
time taken to travel fro h = 50 to h = 0
v = u + gt
45 .39 = 32.86 + 9.8 x t
t = 1.28 s .
Bacteria vary somewhat in size, but a diameter of 1.9 m is not unusual.A. What would be the volume (in cubic centimeters) of such a bacterium, assuming that it is spherical?
B. What would be the surface area (in square millimeters) of such a bacterium, assuming that it is spherical?
Answer:
Volume = 3.6 x 10^-12 cm^3
Area = 1.1 x 10^-5 mm^2
Explanation:
The diameter of the bacteria that is in the shape of spherical = 1.9 m
Now we have to find the volume of the bacteria. So, use the below formula to find the volume of bacteria.
Volume = (4πr^3) / 3
Volume = (4/3)* π*(1.9*10^-6 / 2)^3
Volume = 3.6 x 10^-18 m^3
Volume = 3.6 x 10^-12 cm^3
Now find the surface area.
The surface area = 4πr^2
Area = 4 π (1.9/2)62
Area = 1.1 x 106-11 m^2
Area = 1.1 x 10^-5 mm^2
change-22^0f to kelvin?
Answer:
The formula is kelvin = (tempature in farenheit-32) ÷ 1.8 +273.15
So it will be (-22-32) ÷ 1.8 +273.15= 243.15 kelvin
Answer:
T(K) = (T(°F) + 459.67)× 5/9
answer is 243.15
Kassie is 7. She completes exactly two of the 15 arithmetic problems she has been assigned for homework
before starting a video game. Kassie's behavior BEST exemplifies:
Answer: Procrastination, Defiance, Laziness
Explanation:
Answer: impulsiveness
Explanation:
In a ballistic pendulum experiment, suppose the digital timer shows 0.02 s for the time of flight of the projectile. The manufacturer information about the precision of the timer is nowhere to be found. What error would you quote on your measurement
Answer:
The value is [tex]\Delta t = 0.01 \ s[/tex]
Explanation:
From the question we are told that
The time of flight is [tex]T = 0.02 \ s[/tex]
Given that the value of the time of flight is in three decimal place then the error quote is
[tex]\Delta t = 0.01 \ s[/tex]
Which of these statements partially defines law?
Answer: I dont see the option choices
Explanation:
A metre rule balances when the 50 cm mark is directly above a pivot. State where in the rule its centre of mass is located?
Answer:
The ruler's center of mass is located on the vertical line defined by the 50 cm mark.
Explanation:
Recall that in order to have equilibrium as that described by this example, the net Torque in the ruler should be zero. which means in our case that the weight due to the right half of the ruler, should equal in absolute value of the weight of the left side of the ruler, thus defining where the ruler's center of mass is.
Since this happens at the mark 50 cm, then the center of mass of the ruler is located somewhere inside the ruler somewhere above where the pivot is located (50 cm mark)
3)What are the "bright stars" that seem to move against the sky? Are they really stars? What are they?
Suppose you are asked to compute the tangent of 5.00 meters. Is this possible? Why or why not?
Answer:
Generally it is only possible to compute the tangent of angle in their various unit but because 5 meter is not an angle then it is impossible to compute the tangent of 5 meters.
Explanation:
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F3=17 N. F4=15N and F5=9N
2)In the situation illustrated below, two blocks of Ma and Mb masses are suspended between two walls using ropes. If Ma=2kg, calculate the tension developed in the segments AB, BC and CD and the mass of block B for the equilibrium condition
3)In the figure below, the crane and its AB jib have together 390kg and center of mass in G1. If the 90kg BCD cage and the 80kg man have center of mass located in G2 and G3, respectively, (a) calculate the angle θ of inclination of the boom for which the crane is on the eminence of tumbling. (b) for which size the boom should be reduced (retracting its point B) so that it can be positioned horizontally (θ=0) and the crane does not tip over. Consider that the cage always stays in the horizontal position
4) A 1.7kg plate with center of mass in G is supported by a bar and three cables, as shown in the figure below. Determine the voltage developed in the AB, AC and DE cables and the reaction (there is only one) at point O for the equilibrium condition
Explanation:
1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis. The angle between F₁ and the +z axis is 30°. Therefore, the vector is:
<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)
<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)
<F₁> = 5 i + 5√3 j + 10√3 k
F₂ is in the xy plane. Its slope is -24/7. The vector is:
<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)
<F₂> = -3.36 i + 11.52 j
F₃ is parallel to the +x axis. The vector is:
<F₃> = 17 (i + 0 j + 0 k)
<F₃> = 17 i
F₄ is parallel to the -z axis. The vector is:
<F₄> = 15 (0 i + 0 j − k)
<F₄> = -15 k
F₅ is in the xy plane. It forms a 15° angle with the -y axis. The vector is:
<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)
<F₅> = -9 sin 15° i − 9 cos 15° j
The resultant vector is therefore:
<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k
<F> = 16.31 i + 11.49 j + 2.32 k
2) Sum of forces at point B in the x direction:
∑F = ma
Tbc cos 40° − ¹⁵/₁₇ Tab = 0
Tbc cos 40° = ¹⁵/₁₇ Tab
Tbc = 1.15 Tab
Sum of forces at point B in the y direction:
∑F = ma
Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0
Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)
(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N
1.21 Tab = 20 N
Tab = 16.52 N
Tbc = 19.02 N
Sum of forces at point C in the x direction:
∑F = ma
Tcd sin 25° − Tbc cos 40° = 0
Tcd sin 25° = Tbc cos 40°
Tcd = 1.81 Tbc
Tcd = 34.48 N
3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F. Relative to point A:
3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)
2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm
1336 kgm = 1530 kgm cos θ
θ = 29.17°
3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)
2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm
1336 kgm = (170 kg) x
x = 7.86 m
4) Find the lengths of the cables.
Lab = √((2 m)² + (3 m)² + (5 m)²)
Lab = √38 m
Lac = √((2 m)² + (3 m)² + (5 m)²)
Lac = √38 m
Lde = √((2 m)² + (3 m)²)
Lde = √13 m
Sum of forces in the x direction:
∑F = ma
-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0
Sum of forces in the y direction:
∑F = ma
2/√38 Fab − 2/√38 Fac = 0
Fab = Fac
Sum of forces in the z direction:
∑F = ma
3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0
Sum of moments about the y-axis:
∑τ = Iα
(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0
Substitute Fab = Fac and simplify:
6/√38 Fab + 3/√13 Fde − mg = 0
30/√38 Fab + 6/√13 Fde − 2mg = 0
Double first equation:
12/√38 Fab + 6/√13 Fde − 2mg = 0
Subtract from the second equation:
28/√38 Fab = 0
Fab = 0
Fac = 0
Solve for Fde:
3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0
3/√13 Fde = mg
3/√13 Fde = (1.7 kg) (10 m/s²)
Fde = 20.43 N
Solve for Rx:
-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0
Rx = 2/√13 Fde
Rx = 11.33 N
Suppose the U.S. national debt is about $15 trillion. If payments were made at the rate of $1,500 per second, how many years would it take to pay off the debt, assuming no interest were charged? Note: Before doing these calculations, try to guess at the answers. You may be very surprised. yr (b) A dollar bill is about 15.5 cm long. How many dollar bills attached end to end would it take to reach the Moon? The Earth-Moon distance is 3.84 108 m. dollar bills
Answer:
This question has already been answered.
Explanation:
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The unknown vector v satisfies b - v = λ and b x v-c, where λ, b, and c are fixed and known. Find v in terms of λ, b, and c.
A physics student runs along a line tangent to the edge of a motionless merry-go-round and jumps on at the very outside. The merry-go-round has the shape of a uniform disk.
A) Make an angular momentum chart to help you keep track of what is and is not changing. Consider the two objects, the student and merry-go-round, with the total angular momentum of these two objects conserved. Must the student have angular momentum before jumping on the merry-go-round?
B) Is linear momentum conserved in this interaction if you just consider the merry-go-round and the student?
Answer:
We know that angular momentum is mvr
So that of the students initially will be
A1= mvr
and the final of the student and the merrygo round will be
A2= (1/2Mr²+mr)omega
So
Angular velocity (omega)= mvr/(1/2Mr²+mr)
This is the final angular velocity of the system so yea the student must have angular momentum before jumping on the merry-go-round
B. Because of directional changes at every edge of the Merry go round that result in changes in velocity, this results in changes in momentum too thus linear momentum is not conserved
A cross-country skier can complete a 7.5 km race in 45 min. What is the skier’s average speed?
Answer:
Speed = 10 km/hr
Explanation:
Speed = Distance / Time
Since we have to convert it into minutes to hours ( because it is km/hr )
45 minutes = 0.75 hours
Therefore,
Speed = 7.5 / 0.75
= 10 km/hr
Hope it helps!
Answer:
10
Explanation:
You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building. Before the state legislature votes on funding for the project, they would like you to prepare a report on the benefits of upgrading the elevators. One of the numbers that they have requested is the time it will take the elevator to go from the ground floor to the 102nd floor observatory. They are unlikely to approve the project unless the new elevators make the trip much faster than the old elevators. If state law mandates that elevators cannot accelerate more than 2.40 m/s2 or travel faster than 18.8 m/s , what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor
Answer:
t_total = 23.757 s
Explanation:
This is a kinematics exercise.
Let's start by calculating the distance and has to reach the limit speed of
v = 18.8 m / s
v = v₀ + a t₁
the elevator starts with zero speed
v = a t₁
t₁ = v / a
t₁ = 18.8 / 2.40
t₁ = 7.833 s
in this time he runs
y₁ = v₀ t₁ + ½ a t₁²
y₁ = ½ a t₁²
y₁ = ½ 2.40 7.833²
y₁ = 73.627 m
This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.
x_total = x₁ + x₂
x₂ = x_total - x₁
x₂ = 373 - 73,627
x₂ = 299.373 m
this distance travels at constant speed,
v = x₂ / t₂
t₂ = x₂ / v
t₂ = 299.373 / 18.8
t₂ = 15.92 s
therefore the total travel time is
t_total = t₁ + t₂
t_total = 7.833 + 15.92
t_total = 23.757 s
In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a proton, assume the radius of the orbit is 5.29 10-11 m.
A) Find the magnitude of the electric force exerted on each particle.B) If this force causes the centripetal acceleration of the electron, what is the speed of the electron?
Answer:
Explanation:
Force between two charged particle having charge q₁ and q₂ at distance r
F = k q₁ x q₂ / r²
Putting the values
F = 9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹ / 5.29² x 10⁻²²
= .823 x 10⁻⁷
= 823 x 10⁻¹⁰ N
This force will apply on both the charges and will be attractive in nature .
B )
Let speed of electron be v
centripetal force on electron = F
m v² / r = 823 x 10⁻¹⁰
putting the values for electron
9.1 x 10⁻³¹ x v² / 5.29 x 10⁻¹¹ = 823 x 10⁻¹⁰
v² = 478.42 x 10¹⁰
v = 21.87 x 10⁵ m /s
PLEASE HELP! BRAINLIEST! The three main areas of science each contribute to forensic science in unique ways. Using your own words, write a paragraph (five or more sentences) to describe how biological sciences, chemical sciences, and physical sciences can contribute to the analysis of a crime scene. Please provide specific examples for each. Proper grammar, spelling, and sentence structure are expected.
Explanation:
Explanation:Those interested in working with trace evidence, such as glass, hairs, and gunshot residue, should focus on instrumentation skills and take courses in geology, soil chemistry, and materials science. If forensic biology, such as DNA analysis, is preferred, take microbiology, genetics, and biochemistry courses. Those interested in the toxicological aspects of this work, such as obtaining and interpreting toxicology reports, should study physiology, biochemistry, and chemistry.
A plastic wire is wind on a wheel on a radius 3 cm. The whole system is initially at rest. A force is applied on the plastic wire and the wire unwinds from the wheel from its axis of rotation. The wheel is given an angular acceleration of 92.00 rad/s2 for:______.
Answer:
The torque of the wheel is 0.92 N-m.
Explanation:
Given that,
Radius of wire = 3 cm
Angular acceleration = 92.00 rad/s²
Suppose, Mass of wheel is 4 kg and radius is 5 cm. find the torque of the wheel.
We need to calculate the torque of the wheel
Using formula of torque
[tex]\tau=I\alpha[/tex]
[tex]\tau=MR^2\times\alpha[/tex]
Where, M = mass of wheel
R = radius of wheel
[tex]\alpha[/tex] = angular acceleration
Put the value into the formula
[tex]\tau=4\times(5\times10^{-2})^2\times92[/tex]
[tex]\tau=0.92\ N-m[/tex]
Hence, The torque of the wheel is 0.92 N-m.
A particle moves along the x-axis according to the equation
S = 4+6t-2t^2
where S is in meters and t is in seconds. At t = 3.0 s, calculate
(a) the position of the particle
(b) its instantaneous velocity
(c) its instantaneous acceleration.
◈ A particle is moving along the x-axis according to the equation
S = 4 + 6t - 2t²____________________________
✪ Position at t = 3s :
➝ S = 4 + 6t - 2t²
➝ S = 4 + 6(3) - 2(3)²
➝ S = 4 + 18 - 18
➝ S = 4m
✪ Instantaneous velocity at t = 3s :
➝ v = dx/dt = d(4 + 6t - 2t²)/dt
➝ v = 6 - 4t
➝ v = 6 - 4(3)
➝ v = 6 - [tex]\sf{12}[/tex]
➝ v = -12m/s
✪ Instantaneous acc. at t = 3s :
➝ a = dv/dt = d(6 - 4t)/dt
➝ a = -4m/s²
[Acceleration does not depend on time]
1. A signal source has an open-circuit voltage of 1V, and a short-circuit current of10mA. What is the source resistance
Answer:
100 Ω
Explanation:
Given that
Open circuit voltage, V = 1 V
Short circuit current, I = 10 mA
Source resistance R, = ?
This is rather a straight forward question. Remember Ohms Law? Current being directly proportional to the voltage and inversely proportional to the resistance?
Yeah, that's the formula we'd be using.
Ohms Law states that V = IR, and thus, if we make R subject of the formula, we have
R = V / I, on substituting the values, we have
R = 1 / 10*10^-3
R = 1 / 0.01
R = 100 Ω