answer:
20 minutes for 5 m/s 10 minutes for 10 m/s
A boy is sitting on the outside edge of a merry go round. If the angular velocity id doubled, then the linear speed of the boy will be
If the angular velocity id doubled, then the linear speed of the boy will be doubled.
As we know,
If a body is rotating in a circular path, it will have an angular as well as linear velocity.
The angular velocity is responsible for rotation and the linear velocity is responsible for changing the direction in a circular manner.
The angular velocity W is related to the linear velocity V as,
W = V/R
Let us say this is equation 1,
Where R is the radius of the merry go round,
Now if the angular velocity is doubled,
W' = V'/R
Let us say this is equation 2.
On dividing equation 1 and 2,
W/W' = V/V'
As we know, W' = 2W.
1/2 = V/V'
V' = 2V.
The linear velocity will increase by a factor of 2.
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A baseball player is dashing toward home plate with a speed of 5.6 m/s when she decides to hit the dirt. She slides for 1.5 s , just reaching the plate as she stops (safe, of course). Calculate the acceleration.
ANSWER:
-3.73 m/s^2
STEP-BY-STEP EXPLANATION:
Acceleration is the rate of change of velocity and we can calculate it as follows:
[tex]a=\frac{v_2-v_1}{\Delta t}[/tex]Replacing:
[tex]\begin{gathered} a=\frac{0-5.6}{1.5} \\ a=-3.73m/s^2 \end{gathered}[/tex]The baseball's player's deceleration is 3.73 m/s^2
From the third rule you know that if a rope passes over a pulley, the tension in the rope is unaffected. With this observation in mind, what is the magnitude of the tension in the second rope? Add the forces acting on the block to find the magnitude, T2 , of the tension in rope 2
Answer:F - mB g
Explanation: If two objects are connected by a rope, the tension is the same at both ends. Thus, the same magnitude T2 pulls both the block and the weight.
Find the magnitude of the force F exerted by the leg on the small pulley. (By Newton's third law, the small pulley exerts an equal and opposite force on the leg.) Let the mass m be 2.27 kg.
The magnitude of the force F exerted by the leg on the small pulley is, F = 38.53 N
We can calvulate the magnitude of the force by a mathematical calculation, which is as follows. As the small pulley and mass are not moving, they are stationary.
t2=t1 = mg = (2.27)(9.8)
= 22.246 N
Now resolve the force in the opposite direction.
F = 2Tcos30°
F = (2)(22.246)(cos30°)
F = 38.53 N
Therefore, the magnitude of the force is 38.53 N.
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Which phrase is an example of kinetic energy?
The radioactive isotope 14C has a half-life of approximately 5715 years. Now there are 50g of 14C.(1) How much of it remains after 1600 years? (Round your answer to three decimal places.) g Tries 0/99(2) How much of it remains after 16000 years? (Round your answer to three decimal places.) g Tries 0/99
Given
The half life is T=5715 years
The initial amount is N=50 g
To find
1) How much of it remains after 1600 years?
(2) How much of it remains after 16000 years?
Explanation
The amount of carbon remains after t yaers is
[tex]N^{\prime}=N(\frac{1}{2})^{\frac{t}{T}}[/tex]1. Thus putting t=1600 years
[tex]\begin{gathered} N^{\prime}=50(\frac{1}{2})^{\frac{1600}{5715}} \\ \Rightarrow N^{\prime}=41.208\text{ g} \end{gathered}[/tex]2.Putting t=16000
[tex]\begin{gathered} N^{\prime}=50(\frac{1}{2})^{\frac{16000}{5715}} \\ \Rightarrow N^{\prime}=7.184\text{ g} \end{gathered}[/tex]Conclusion
1.Amount remains after 1600 year is 41.208 g
2.Amount remains after 16000 year is 7.184 g
It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be latefor French class for the third time this week. She must get from one side of theschool to the other by hurrying down three different hallways. She runs downthe first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The secondhallway is filled with students, and she covers its 48.0-m length at an averagespeed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0-mlength at a speed of 5.00 m/s. a) Does Suzette make it to class on time or doesshe get detention for being late again? b) Draw a distance vs. time graph ofthe situation.
Answer:
a) She get detention for being late again
Explanation:
First, we need to identify how much time does she take on each hallway.
With the distance and the speed, we can calculate the time as:
t = distance/speed
So, for each hallway, we get:
First hallway:
distance = 35 m
speed = 3.5 m/s
time = 35/3.5 = 10s
Second Hallway
distance = 48 m
speed = 1.2 m/s
time = 40s
Third Hallway
distance = 60 m
speed = 5 m/s
time = 60/5 = 12 s
Therefore, the total time that she takes was
10s + 40s + 12s = 62s
Since she takes more than 60 seconds, she will be late again.
Finally, we know that she takes 10s to run a distance of 35m, then another 40s to run a distance of 48 m, and another 12s to run a distance of 60 m. Therefore, the distance vs. time graph for this situation is
Match the graph with the correct description of that graph.
A. decreased speed; acceleration
B. at Rest, no motion
C. decreased speed; deceleration
D. increased speed; acceleration
D. increased speed; acceleration is the correct description of that graph.
A distance-time graph: what is it?Shown far an item has come in a certain amount of time is displayed on a distance-time graph. The graph that shows the results of the distance vs time analysis is a straightforward line graph. The Y-axis represents the distance.
A sloping line on a distance-time graph shows that an item is moving. In a distance-time graph, the object's speed is equal to the slope or gradient of the line. The line becomes steeper as the thing goes more quickly (and the greater the gradient).
The horizontal line on a graph of a distance vs time indicates that the item is at rest because the distance does not change over time.
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A 100 N force acts at a 90 degrees and a 60 N force acts at 150 degrees. Determine the magnitude and direction (include angle) of the resultant. Scale is 1cm = 10 N
HELPPP ASAPPPPP
The magnitude of the force is found to be 140N.
Force (F1)acting on θ1(90°) is 100N
Force (F2)acting on θ2(150°) is 60N
Therefore the angle between F1 and F2 is
θ2-θ1 = 150°-90° =60°
Now we calculate magnitude of force ,
Magnitude of force ,F= F1+F2
F. F = [tex]\sqrt{ ( F1+F2). (F1+F2) }[/tex]
|F|² = [tex]\sqrt{F1 ²+F2 ²+F1 F2 COS θ}[/tex]
F= [tex]\sqrt{ 100²+60²+2×100×60× cos60°}[/tex]
F= [tex]\sqrt{10000+3600+6000}[/tex]
F= [tex]\sqrt{19600}[/tex]
F= 140N.
Thus, the magnitude of force is found to be 140N.
The total amount of forces exerted on an object is referred to as the magnitude of the force. The strength of the force increases when all the forces are pulling in the same direction. When forces are exerted on an item from different angles, the force’s strength reduces.
There is magnitude and direction to force. When two forces of equal size are working in opposite directions one in the east and the other in the west the results of the two forces are not the same.
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Earth has a radius R of about 6.37 × 106 m. How many watts of power does Earth receive from the sun? Remember that even though the Earth is a sphere, its image with respect to the sun is a circle — in other words, Earth's shadow is circular, meaning that it captures a ‘circular amount’ of light. Use the observed value of S = 1361 W/m2.
If the earth has a radius R of about 6.37 × 10⁶ m, then the power received by the earth would be 2.722 × 10¹⁰ watts.
What is power?
The rate of doing work is known as power. The Si unit of power is the watt.
Power =work/time
As given in the problem Earth has a radius R of about 6.37 × 10⁶ m.
Area of the earth assuming it is a circle = 3.14 × 6.37 × 10⁶
= 2 × 10⁷ meter²
The power received by the earth = 2 × 10⁷ × 1361
= 2.722 × 10¹⁰ watts
Thus, the power received by the earth would be 2.722 × 10¹⁰ watts.
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What do these circuits all have in common?(A)(B)(C)(D)A. They all contain switchesB. They are all AC.They are all DCOC.D. They all contain resistors.
ANSWER:
STEP-BY-STEP EXPLANATION:
We have the following:
[tex]undefined[/tex]Fill in the blank question.
To send information through verbal speech, ____
cords send information encoded in _____
waves through the air.
To send information through verbal speech, transmitter cords send information encoded in radio waves through the air.
The basic principle is simple. At one end a transmitter encodes or modulates the message by varying the amplitude or frequency of the waves. This is a bit like Morse code. Meanwhile, a receiver tuned to the same wavelength receives the signal and decodes it into the desired format sound image data, etc.
Communications and satellite transmissions. Radio waves travel easily through the air. Radio waves are used for the wireless transmission of sound messages and information for communications, and for maritime and air navigation. Information is applied to an electromagnetic carrier wave as amplitude modulation AM frequency modulation FM or digital form pulse modulation.
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A force of magnitude Fx acting in the x-direction on a 2.35-kg particle varies in time as shown in the figure below. (Indicate the direction with the sign of your answer.)
The impulse is mathematically given as
v_f=3.61m/s
This is further explained below.
What is impulse?Generally, In classical mechanics, the integral of a force F over the time period t during which it operates is referred to as the impulse of the system.
It should come as no surprise that impulse is likewise a vector quantity given that force is one.
When an impulse is applied to an item, a corresponding vector change in the object's linear momentum occurs, also in the direction that the impulse causes.
In conclusion, Impulse = Area Under Curve
=2*4+1*4
[tex]m\left(v_f-v_i\right)=12 \\\\v_f=\frac{12}{2.35}=5.11 \mathrm{~m} / \mathrm{s}\\\\m\left(v_f-(-1.5)\right)=12 \\v_f=5.11-1.5 \\\\ =3.61 \mathrm{~m} / \mathrm{s} \\v_f=3.61 \mathrm{~m} / \mathrm{s}\end{gathered}$$[/tex]
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CQ
A force of magnitude Fx acting in the x-direction on a 2.35-kg particle varies in time as shown in the figure below.
(a) Find the impulse of the force.
____________kg
what is the center of mass?
a) the location that the object would concentrate to if we imagined it to shrink to a point.
b) the exact point of an object, measured from all its edges.
c) a point in space to which an object is attracted.
d) an arbitrarily chosen point in the object used to describe its motion.
e) a point in space that the object is moving toward.
The exact point of an object measured from all its edge center of mas.
What is center of mass?
the center of mass of a distribute of mass in space is the unique point where the weighted relatively position of the devided mass sums to 0. This is a point to which a force of may be applied to cause of a linear acceleration it without an angular acceleration.
Most of the time (not everytime)the centre of mass of an object lies within the object itself.
Eg-the center of mass of a ball in middle of a ball and the centre of mass of a copy is middle of a copy.
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Which of the following could be considered inertial frames of reference?1- A train speeding up to 54 m/s2- Both A & C3- A train traveling with a constant velocity of 54 m/s4- A train at rest
To determine if a frame is an inertial frame we just need to remember that they can only be at rest or moving with constant speed. Therefore, the inertial frames in this case are:
A train traveling with a constant velocity of 54 m/s
A train at rest
When exercising in the heat, you should __________.
A.
wear tight-fitting clothes to absorb your sweat
B.
dress in layers
C.
choose dark colored clothing
D.
vary the intensity and duration of the exercise
Please select the best answer from the choices provided.
When exercising in the heat, you should wear tight - fitting clothes to absorb your sweat, therefore the correct answer is option A.
What is exercise?Exercise is a physical activity that involves bodily movement that improves a person's health, fitness, and well-being while lowering their chance of contracting illnesses.
Dressing in layers is useful in extreme cold conditions and extremely cold weather.
Dark-colored clothing absorbs more heat as compared to light - colored clothing.
When exercising in the heat, you should wear tight-fitting clothes to absorb your sweat, therefore the correct answer is option A.
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The SPH3U class designs a cannon able to shoot a human being out of it. . If the human is launched at a velocity of 40 m/s, 37° from the ground, how far from the cannon should you place a mattress to catch the human if the muzzle of the cannon is 0.75 m from the ground?
Given:
Velocity = 40 m/s
Angle = 37°
Let's find the distance from the cnnon where you should place a matress to catch the human if the muzzle is 0.75 meters from the ground.
Let's first draw the projectile motion diagram:
To find the distance, first apply the formula to solve for the time taken for the human to reach the matress.
[tex]y=y_o+v_{oy}t-\frac{1}{2}gt^2[/tex]Where:
Height from the ground to the muzzle = yo = 0.75
y = 0
Velocity = vo = 40 m/s
initial vertical velocity = voy = vo sinθ = 40 sin37
We have:
[tex]v_{oy}=v_o\sin \theta=40\sin 37=24.07\text{ m/s}[/tex]g is the acceleration due to gravity = 9.8 m/s²
Now, let's solve for the time t:
[tex]\begin{gathered} y=y_o+v_{oy}t-\frac{1}{2}gt^2 \\ \\ 0=0.75+24.07t-\frac{1}{2}\times9.8\times t^2 \\ \\ t=4.94\text{ seconds} \end{gathered}[/tex]To sove for the horizontal distance, x, apply the formula:
[tex]\begin{gathered} x=v_{ox}t \\ \\ \text{Where:} \\ v_{ox}=v_o\cos \theta=40\cos 37=31.94\text{ m/s} \end{gathered}[/tex]Hence, we have:
[tex]\begin{gathered} x=31.94\times4.94 \\ \\ x=157.8\text{ m} \end{gathered}[/tex]Therefore, the matress should be placed at 157.8 meters from the cannon in order to catch the human.
ANSWER:
157.8 meters
A 4.5-m-wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 20° above the horizon.Part AHow deep is the pool?
ANSWER
[tex]4.5\text{ }m[/tex]EXPLANATION
First, let us make a sketch of the problem:
where r = angle of refraction
d = depth of the pool
First, we have to find the angle of refraction, r, by applying Snell's law:
[tex]\frac{n_1}{n_2}=\frac{\sin r}{\sin i}[/tex]where n1 = incident refractive index = 1
n2 = refracted index = 1.33
i = angle of incidence = 70°
Therefore, solving for r, we have that:
[tex]\begin{gathered} \frac{1}{1.33}=\frac{\sin r}{\sin70} \\ \\ \sin r=\frac{\sin70}{1.33}=0.7065 \\ \\ r=\sin^{-1}(0.7065) \\ r=45.0\degree \end{gathered}[/tex]Now, we can solve for the depth of the pool by applying trigonometric ratios for right triangles for tangent:
[tex]\begin{gathered} \tan45=\frac{4.5}{d} \\ \\ d=\frac{4.5}{\tan45} \\ \\ d=4.5m \end{gathered}[/tex]That is the depth of the pool.
Find the direction of the sum
of these two vectors:
16.3 m
7.70 m
20.0°
magnitude (m)
A
27.8°
The resultant of the two given vectors with magnitudes 16.3 m and 7.7 m at an angle of 137.8 degree is equal to 23.87 meters.
What is the parallelograms law of vector addition?
If two vectors are acting simultaneously at a point, then it can be represented both in magnitude and direction by the adjacent sides drawn from a point.
What is the trigonometric ratio?
Trigonometric Ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle. The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle.
Given:
Vector 1 of magnitude 7.7 m at an angle of 27.8 degree to x-axis.
vector 2 of magnitude 16.3 m at an angle of 20 degree to minus y axis.
Using the law of parallelogram vector addition
i.e.
resultant vector = √((v₁) ² + (v₂) ² + 2 × v₁ × v₂ × cos (angle between two vectors))
substituting given value in parallelogram vector addition we get,
resultant vector = √ (570)
resultant vector = 23.87 meters
therefore, the resultant of the two given vectors with magnitudes 16.3 m and 7.7 m at an angle of 137.8 degree is equal to 23.87 meters.
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Answer:
11.79
Explanation:
Pasagot po need ko na po ng answer yung maayos po sana at yung tamang sagot salmat po. plszz lang po naiiyak na po ako kasi po ang dami ko pa po gagawin ang hirap na po talaga
Answer:
the analysis from are there 331
A projectile is launched at an angle of 30° above the horizontal with an initial velocity (vo) of 42.2 m/s. After some time passes, the projectile reaches the peak of its trajectory.Its horizontal velocity (vx) at the peak of its trajectory is ____ m/s.
First, for us to start, let us draw the problem:
When we analyse this problem, we can see that the only velocity that changes overtime is the vertical velocity (due to gravity), so the horizontal speed is the same throughout the whole movement. This gives us the following:
[tex]v_x=42.2*cos(30°)=36.546\frac{m}{s}[/tex]Then, the horizontal speed at the peak of the trajectory (and also during the whole movement) is v=36.546m/s
Two 3.09 cm by 3.09 cm plates that form a parallel-plate capacitor are charged to +/- 0.617 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.784 mm?
Given:
The charge on the capacitor is Q = 0.617 nC
The distance between plates is d =1.784 mm
The area is
[tex]\begin{gathered} A=\text{ 3.09}\times3.09 \\ =9.5481\text{ cm}^2 \end{gathered}[/tex]To find the electric field strength.
Explanation:
The electric field strength can be calculated by the formula
[tex]E=\frac{Q}{\epsilon_oA}[/tex]The constant is
[tex]\epsilon_o=8.85\times10^{-12}\text{ C}^2\text{ / N m}^2[/tex]On substituting the values, the electric field strength will be
[tex]undefined[/tex]
A pot holds water. The water and pot and air are all at the same temperature, therefore no heat is being transferred. Which law of thermodynamics supports this statement?
According to the zeroth law of thermodynamics, no heat is transferred between the two objects if both the objects are at the same temperature.
Thus, no heat transfer between the water, pot, and air occurs according to the zeroth law of thermodynamics.
Hence, the zeroth law of thermodynamics is the correct answer.
On July 19, 1969, the lunar orbit of Apollo 11 was adjusted to an average height of 172 kilometers above the Moon's surface. The radius of the Moon is 1840 kilometers, and the mass of the Moon is 7.3 x 1022 kilograms. At what speed did the spacecraft orbit the Moon? Include units in your answer. Answer must be in 3 significant digits.
In order to calculate the speed of the spacecraft, use the following formula;
[tex]v=\sqrt[]{\frac{GM}{r}}[/tex]where,
M: mass of the moon = 7.3*10^22 kg
G: Cavendish constant = 6.67*10^-11 Nm^2/kg^2
r: distance in between the spacecraft and the center of the Moon =
172 km + 1840 km = 2012 km
Replace the previous values of the parameters into the formula for v and simplify:
[tex]v=\sqrt[]{\frac{(6.67\cdot10^{-11}N\frac{m^2}{\operatorname{kg}^2})(7.3\cdot10^{22}kg)}{2012\operatorname{km}}}\approx4.91\cdot10^4\frac{m}{s}[/tex]Hence, the speed of the spacecraft is approximately 4.91*10^4 m/s
A 55 kg motorcyclist is flying through the air at 72 km/h at the apex of the jump 8.3 m above the ground. Find his landing speed assuming energy is conserved.
Answer: 85.4 km/h
Explanation:
We would apply the law of conservation of kinetic energy. The formula is expressed as
KE1 + PE1 = KE2 + PE2
where
KE1 and KE2 are the initial and final kinetic energies
PE1 and PE2 are the initial and final potential energies
The formula for calculating kinetic energy is
KE = 1/2mv^2
where
m is the mass of the object
v is the velocity
The formula for calculating potential energy is
PE = mgh
where
g is the acceleration due to gravity = 9.8m/s^2
h is the height of the object
From the information given,
m = 55
h = 8.3
Potential energy at the top = 55 x 9.8 x 8.3 = 4473.7
At the top, v = 72km/h
Recall, 1 km = 1000m
1 hour = 3600s
72km/h = 72 x 1000/3600 = 20m/s
kinetic energy at the top = 1/2 x 55 x 20^2 = 11000
KE1 + PE1 = 4473.7 + 11000 = 15473.7
At the point of landing,
PE2 =0
KE = 1/2 x 55 x v^2 = 27.5v^2
Thus,
27.5v^2 = 15473.7
v^2 = 15473.7/27.5
v = √(15473.7/27.5)
v = 23.72 m/s
his landing speed is 23.72 m/s
Converting to km/h, we would multiply 23.72 by 18/5
Landing speed = 85.4 km/h
How old is an artifact if four half-lives have occurred and the half-live of carbon-14 is 5730 years?
22920 years old is an artifact if four half-lives have occurred and the half-live of carbon-14 is 5730 years.
Why does carbon-14 have a short half-life?Due to radioactive decay to nitrogen-14, carbon-14 has a very short half-life of 5,730 years, which means that over this time, the proportion of carbon-14 in a sample will have decreased by 50%.
Why is the term "half-life" used?A half-life is the amount of time it takes for something to go from 100% to 50%. The phrase is most frequently used in reference to radioactive decay, which happens when energetic atomic particles that are unstable lose momentum.
Briefing:The half-life of carbon-14 is 5730 years. Accordingly, a sample of 100 carbon-14 atoms has a half-life of 5730 years, or 50 carbon-14 and 50 carbon-12 (i.e., 100 carbon-14, one half-life = 50 carbon-14, 50 carbon-12).
4 half lives
=5730×4
=22,920 years old.
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You are driving your 1800 kg car at 21 m/s over a circular hill that has a radius of 150 m. A deer running across the road causes you to hit the brakes hard while right at the summit of the hill, and you start to skid. The coefficient of kinetic friction between your tires and the road is 0.75.
What is the magnitude of your horizontal acceleration as you begin to slow?
The magnitude of your horizontal acceleration as you begin to slow is 7.35m/s².
Applying work-energy theorem,
Total work done on the body by all the forces = change in kinetic energy.
So, here,
Work done by friction = ∆KE
-uMgd = 1/2M(v²-u²)
u is the coefficient of friction,
M is the mass of the car,
g is the acceleration due to gravity,
v is the final velocity of car,
u is the final velocity of car,
d is the horizontal distance moved by car.
(-0.75)(1800)(9.8)d = -1/2(1800)(21)²
d= 30 meters.
Applying Equation of motion,
V²-U²=2ad
Where a is the horizontal acceleration,
-(21×21) = 2(30)a
a = 7.35m/s².
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Given three electrically charged spheres A,B and C. Sphere A is brought near Sphere B and both are attracted to each other electrically. Sphere B is brought near Sphere C and both repel each other electrically. What happens when Sphere A is brought near Sphere C? A)electrical repulsion B)both attraction and repulsion C)electrical attraction D)neither attraction nor repulsion E)nothing
Given
Three electrically charged spheres A,B and C. Sphere A is brought near Sphere B and both are attracted to each other electrically. Sphere B is brought near Sphere C and both repel each other electrically.
To find
What happens when Sphere A is brought near Sphere C?
Explanation
Sphere A and sphere B are attracted to each other.
Thus, both A and B are of opposite charge.
Let A be a positive charge and B be a negative charge.
When sphere B is brought near sphere C both repel.
Thus, sphere C is also of negative charge.
So A and C are then of opposite charges and would both attract each other.
Conclusion
When sphere A is brought near sphere then C)electrical attraction occurs.
Do two bodies have to be in physical contact to exert a force upon one another?
a) No, the gravitational force is a field force and does not require physical contact to exert
a force.
b) No, the gravitational force is a contact force and does not require physical contact to
exert a force.
c) Yes, the gravitational force is a field force and requires physical contact to exert a force.
d) Yes, the gravitational force is a contact force and requires physical contact force to exert
a force
The correct answer to the statement " Do two bodies have to be in physical contact to exert a force upon one another " is:
No, the gravitational force is a field force and does not require physical contact to exert
a force.
The correct option is a.
Why two bodies do not have to be in physical contact to exert a force upon one another as a result of gravitational force.It has been practically proven that two bodies can exert a force upon each other even if there is no physical contact between them. This can as a result of gravity.
That being said, a magnetic attraction can also exert a force between two different bodies upon one another.
So therefore, it can be deduced from above that two different bodies do not have to be in physical contact before they exert a force upon one another.
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220 g of water at 90 °C is added to a certain unknown mass of water at 10 °C The final temperature of the mixture is 33 °C. Calculate the unknown mass of the water.
The unknown mass of water is 545.21
What is mass?
mass is a quantitative measure of inertia , a fundamental property of all matter .
mass depends on the number and kind of atoms.
It is a dimensionaless quantity.
Sol-
As per the given question
220 g of water added to 90°c
Mass of water=10°C
Tempreture of the mixture =33°C
Unknown mass
m1 (n-t)= m2(t-b)
220(90-33)=m2(33-10)
:- (we can also do multiple of cp but as both side of the cp fluid unit are same so it's not needed)
220×57=m2 (23)
m2= 220×57/23
m2 =545.21
Thus the unknown mass of water is 545.21 .
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