Given data:
[tex]\begin{gathered} V=12\text{ V} \\ Q=2\text{ mC} \end{gathered}[/tex]The work done is given as,
[tex]W=QV[/tex]Putting all known values,
[tex]\begin{gathered} W=2\times10^{-3}\text{ C}\times12\text{ V} \\ =24\text{ mJ} \end{gathered}[/tex]Therefore, 24 mJ of the work is done.
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.
Set the cannon to have an initial speed of 20 m/s. For which situation do you think the cannon ball will go father: if it is set at a 25-degree angle, or if it is set at a 35-degree angle?
Question 1 options:
35 degrees
25 degrees
The cannon would go further if it is projected at 35 degrees.
What is the range of the projectile?The term range has to do with the horizontal distance that is covered by the projectile. We know that the formula for the range of the projectile is;
R = u^2sin2θ//g
R - range of the projectile
θ = the angle of the projection
g = acceleration due to gravity
In the first case;
R = (20)^2sin2(35)/9.8
R = 38.4 m
In the second case;
R = (20)^2sin2(25)/9.8
R = 31.3 m
We would now have to compare the range of the two projectiles.
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He used an exquisitely sensitive balance to measure the force between two lead spheres whose centers were 0.23 m
m
apart. One of the spheres had a mass of 178 kg
k
g
, while the mass of the other sphere was 0.63 kg
k
g
.
The gravitational force exerted by the lighter sphere will be proportional to its lighter weight will be 2.24 * 10⁻⁷
The word "force" has a clear definition. At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains." One thing experiences a force from another. There is no distinction between living and non-living things in the concept of a force.
The force of attraction between any two bodies is directly inversely correlated to the square of the distance between them and is directly inversely correlated to the product of their masses, according to Newton's universal law of gravitation.
F = GM₁M₂ / R² --- (1)
W = M₁g ------ (2)
equation (1) is divided by equation (2)
⇒ F / W = GM₁M₂ / R²*M₁g
⇒ F / W = (6.67 * 10⁻¹¹ * 178) / (0.23)² = 2.24 * 10⁻⁷
Therefore, the ratio of the gravitational force between these spheres to the weight of the lighter sphere will be 2.24 * 10⁻⁷
COMPLETE QUESTION: The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance to measure the force between two lead spheres whose centers were 0.23 m apart. One of the spheres had a mass of 178 kg, while the mass of the other sphere was 0.63 kg. What was the ratio of the gravitational force between these spheres to the weight of the lighter sphere?
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Two ropes are pulling on a skater and they exert forces on her as shown. Calculate the magnitude and direction of the total force exerted by the ropes on the skater. with a force of 30N and an angle of 30
Magnitude of total force is found to be 54N.
30 N force is acting at 30° ,then the component in x direction is 30 cos(30°) = 25.98 N
Now the component in y direction = 30 sin(30°) = 15 N
Second, 40 N force is acting at -50°
So, the component in x direction is 40 cos(-50°) = 25.71 N
Also ,the component in y direction = 40 sin(-50°) = -30.64 N
Total force in x direction = 25.98 + 25.71 N = 51.69 N and the total force in y direction = 15 – 30.64 N = -15.64 N
Now the, Magnitude of total force is given by,
( 51.692 + 15.642) 1/2 = 54N.
The total number of forces exerted on an object is the force’s magnitude. The strength of a force increases when all forces are acting in the same direction. As a force is applied to an item from multiple angles, its strength diminishes. Force is a vector quantity since it has a magnitude and a direction.
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A worker drops a wrench from the top of a tower 101.3 m tall. What is the velocity when the wrench strikes the ground?
How to do calculate this the error? What are the rules?(Mark scheme Answer is B)
Here,
potential difference(V)= (2.00±0.02) V;
current (I)=( 5.3 ± 0.1) mA;
Resistance (X) will be given by
[tex]\begin{gathered} X=\frac{V}{I}; \\ X=\text{ }\frac{2.00±0.02}{5.3±0.1}\text{ }\times1000\begin{cases}V={(2.00±0.02)} \\ I=({5.3±0.1\text{ \rparen mA= \lparen5.3}\pm0.1)\times10^{-3\text{ }}}A\end{cases} \\ \\ \therefore X=(\text{ 377.358 }\pm0.001) \\ \text{ } \\ \end{gathered}[/tex]A 150 m long train entered the 450 m long bridge. From the entrance locomotives only after the last wagon passed 10 minutes. At what speed the train was going?
The train was going at the speed of 1m/s on the bridge.
The length L of the bridge is 450m.
The length B of the train is 150m.
After entering the bridge,
The last wagon comes out after it has passed ten minutes.
So, the speed of the train can be found out by using the formula,
Speed = Total Distance/total time
Here, total time is 10 minutes,
We know,
One minute = 60 seconds
10 minutes = 10×60 seconds.
Total time = 600 seconds.
Total distance travelled by the train is (L+B) when it comes out from the bridge,
So, putting all the values,
Speed = (450+150)/600
Speed = 600/600
Speed = 1m/s.
So, the speed of the train is 1m/s.
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a spring has a constant of 80 N/m. How much energy is stored in the spring when it is compressed 0.2 m past its natural length?
The amount of energy stored in the spring, given it has a spring constant of 80 N/m is 1.6 Joules
How to determine the energy stored in the springFirst, we shall list out the given parameters from the question. This is shown below:
Spring constant (K) = 80 N/mCompression (e) = 0.2 mEnergy stored (E) = ?The energy stored in the spring, given the above data can be obtained as follow:
E = ½Ke²
E = ½ × 80 × 0.2²
E = 40 × 0.04
E = 1.6 Joules
Thus, from the above calculation, we can conclude that the energy stored is 1.6 Joules
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A jogger jogs from one end to the other of a straight track in 1.17 min and then back to the starting point in 1.67 min. What is the jogger’s average speed in jogging to the far end of the track (assuming the track is 100 m long) in m/s?
ANSWER:
1.21 m/s
STEP-BY-STEP EXPLANATION:
Given:
One way time = 1.17 min
Return time = 1.67 min
1 minute is 60 seconds, therefore:
One way time = 1.17 min = 1.17 * 60 = 70.2 sec
Return time = 1.67 min * 60 = 100.2 sec
We calculate the speed for each journey, knowing that the distance traveled is 100 meters, like this:
[tex]\begin{gathered} v=\frac{d}{t} \\ \\ v_1=\frac{d}{t_1}=\frac{100}{70.2}=1.4245\text{ m/s} \\ \\ v_2=\frac{d}{t_2}=\frac{100}{100.2}=0.998\text{ m\/s} \end{gathered}[/tex]Therefore, the average speed would be:
[tex]\begin{gathered} v=\frac{v_1+v_2}{2}=\frac{1.4245+0.998}{2}=\frac{2.4225}{2} \\ \\ v=1.21\text{ m/s} \end{gathered}[/tex]The average speed is 1.21 m/s
Three blocks are sitting on a horizontal, frictionless table. They
are pushed from the left by an applied force F = 10 N, as
shown. How much force does block 3 exert on block 2, if
m₂ = 1 kg, m₂ = 2 kg, m₂ = 3 kg ?
a) 3 N
b) 5 N
c) 6 N
d) 8 N
Newton's laws of motion, which are composed of three basic principles, describe the interaction between an object's motion and the forces acting on it.
These laws can be summarized as follows: A body remains at rest or in motion in a straight line at a constant speed unless moved upon by a force.
The first law states that an object's motion cannot be changed until a force acts on it.
The second law states that an object's force is calculated by dividing its mass by its acceleration.
The third law states that when two objects come into contact, they apply pressures that are equal in size and direction to one another.
Let a represent the system's acceleration.
T1 = M1a .....(1)
T2 − T1 = M2a ....(2)
F2 − T2 = M3a ......(3)
When we combine (1), (2), and (3), we obtain
(M1 + M2 + M3)a = F
or (1 + 2 + 3) a= 6 a = 1 m/s²
Now , T2 = (M1 + M2)a = (1 + 2)(1) = 3N
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FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of inertia of the pulley is 2 kg m². (i) Sketch the free body diagram of the 1.5 kg block.(ii) When the mass is released from rest, calculate the angular velocity and number of revolutions of the pulley at t = 4.2 s.
Part (i)
Free body diagram of the 1.5 kg block;
Part (ii)
Only 1 force is acting on the pulley is the weight of the block attached with the sting. The torque acting on the pulley is given as,
[tex]\begin{gathered} \tau=F\times r \\ =Fr\sin \theta \\ =mgr\sin \theta \end{gathered}[/tex]Here, g is the acceleration due to gravity and the θ is the angle between force F and r (as force is acting tangentially hence θ=90°)
Substituting all known values,
[tex]\begin{gathered} \tau=(1.5\text{ kg})\times(9.8\text{ m/s}^2)\times(20\text{ cm})\times\sin (90\degree) \\ =(1.5\text{ kg})\times(9.8\text{ m/s}^2)\times(20\text{ cm})\times(\frac{1\text{ m}}{100\text{ cm}})\times1 \\ =2.94\text{ N}\cdot m \end{gathered}[/tex]In rotational dynamics torque is given as,
[tex]\tau=I\alpha[/tex]Here, I is the moment of inertia of the pulley (I=2 kg.m²) and α is the angular acceleration.
The angular acceleration is given as,
[tex]\alpha=\frac{\tau}{I}[/tex]Substituting all known values,
[tex]\begin{gathered} \alpha=\frac{2.94\text{ N.m}}{2\text{ kg.m}^2} \\ =1.47\text{ rad/s}^2 \end{gathered}[/tex]The angular velocity is given as,
[tex]\omega=\alpha t[/tex]Here, t is the time.
Substituting all known values,
[tex]\begin{gathered} \omega=(1.47\text{ rad/s}^2)\times(4.2\text{ s}) \\ =6.174\text{ rad/s} \end{gathered}[/tex]Therefore, the angular velocity of the pulley is 6.174 rad/s.
The angular displacement of the pulley in 4.2 s is given as,
[tex]\Theta=\omega t[/tex]Substituting all known values,
[tex]\begin{gathered} \Theta=(6.174\text{ rad/s})\times(4.2\text{ s}) \\ =25.9308\text{ rad} \end{gathered}[/tex]The number of revolutions of the
What is the “time constant” for a capacitor, and why is it important?
We know that in a RC circuit the voltage in the capacitor when is charging is given by:
[tex]V_C(t)=V_0(1-e^{-\frac{t}{RC}})[/tex]when this is happening the voltage in the resistor is given by:
[tex]V_R(t)=V_0e^{-\frac{t}{RC}}[/tex]In both equations V0 denotes the voltage given by the source, R is the resistance of the resistor and C is the capacitance of the capacitor.
We notice that in both expressions the product RC appear, this product is what we call the time constant of the capacitor; and it is important since it determines the time intervals in which the voltage, charges and currents chage in a RC circuit. This means that while the capacitor is charging or discharging the variables mentioned will always have the time constant in their expressions.
An ant can crawl 72 inches every 6 minutes. The table below shows the distance the ant can travel for different amounts of time at this rate.According to the table, if the ant moves at a constant speed, what is its speed? (Remember that speed is a unit rate).
Given,
The ant crawls 72 inches every 6 minutes.
And the table of the time and distance covered by the ant.
From the table, the values of distance are,
d₁=6 inches
d₂=12 inches
d₃=18 inches
And the values of corresponding time are,
t₁=0.5 min
t₂=1 min
t₃=1.5 min
The speed of an object is given by the ratio of the distance covered by the object to the time it takes the object to cover the distance.
Thus the speed of the ant is given by,
[tex]v=\frac{d_1}{t_1}=\frac{d_2}{t_2}=\frac{d_3}{t_3}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\frac{6}{0.5}=\frac{12}{1}=\frac{18}{1.5} \\ =12\text{ imches/min} \end{gathered}[/tex]Thus the speed of the ant is 12 inches/min
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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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
this is a 2 part question28) A 1100-kg car coasts on a horizontal road with a speed of 19 m/s. After crossing anunpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net work done on the car positive, negative, or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section.
ANSWER:
(a) negative
(b) -120.3 N
STEP-BY-STEP EXPLANATION:
(a)
The net work done on the car is negative because its speed decreases from 19 m/s to 12 m/s
(b)
The acceleration is found from the equationof motion that:
[tex]v^2-u^2=2\cdot a\cdot s[/tex]Replacing and solving for a:
[tex]\begin{gathered} 12-19=2\cdot a\cdot32 \\ -7=64a \\ a=-\frac{7}{64}\frac{m}{s^2} \end{gathered}[/tex]Therefore, the force would be:
[tex]\begin{gathered} F=m\cdot a \\ F=1100\cdot-\frac{7}{64} \\ F=-120.3\text{ N} \end{gathered}[/tex]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]consider the graph at the right. the object whose motion is represented by this graph is…
Here, the slope of the graph is negative and constant.
Hence, acceleration is constant and is in opposite direction to its motion.
So, The object whose motion is represented by this graph is moving with constant acceleration.
Motion is defined in physics as the phenomenon through which an object changes its location with respect to time. Motion is mathematically characterized in terms of displacement, distance, velocity, acceleration, speed, and frame of reference to an observer, with the change in position of the body relative to that frame measured as time passes. Kinematics is the branch of physics that studies forces and their effects on motion, whereas dynamics is the branch that studies forces and their effects on motion.
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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
Steve takes his car out for a joy ride and travels 400 meters north. He then travels 100 meters east and picks up his buddy Frank. They then stop at a 7-11 which is 200 meters south from Frank’s house. If the total trip takes 10 minutes, determine the average velocity of Steve’s car. (Draw a picture!) NEED HELPP ASAPPPP !!!!
The average velocity of Steve's car is 1.17m/s.
How to calculate average velocity?Average velocity is the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.
It can be calculated by dividing the total distance of a moving body by the time taken.
According to this question, Steve takes his car out for a joy ride and travels 400 meters north. He then travels 100 meters east and picks up his buddy Frank. They then stop at a 7-11 which is 200 meters south from Frank’s house.
The total distance traveled by Steve is 400m + 100m + 200m = 700m.
Average velocity = 700m ÷ 600s
Average velocity = 1.17m/s
Therefore, 1.17m/s is the average velocity of the car.
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A boat heads East across a South flowing river. The river is 246 m wide and flows at a rate of 6.4 m/s. The boat sails across at 5.3 m/s..
How long will it take the boak to cross the river?
If a boat heads East across a South flowing river. The river is 246 m wide and flows at a rate of 6.4 m/s. The boat sails across at 5.3 m/s, then the time taken by the boat to cross the river would be 46.41 seconds.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.
As given in the problem a boat heads East across a South flowing river. The river is 246 m wide and flows at a rate of 6.4 m/s. The boat sails across at 5.3 m/s.
Time taken by the boat to cross the river = 246/5.3
=46.41 seconds
Thus, the time taken by the boat to cross the river would be 46.41 seconds.
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Select all that apply
Which conditions are necessary for the diffusion of a substance to occur across a membrane?
a concentration gradient
a supply of energy
membrane permeability
Answer:
a concentration gradient.
membrane permeability.
Explanation:
diffusion occurs down a concentration gradient.
and is the free net movement of particles therefore, does not require energy.
membrane permeability is necessary to ensure that the particles can pass through the cell membrane.
Which atoms were cations?
Alkali and alkaline earth metals invariably produce cations.
What kind of compounds are cations?Calcium (Ca2+), potassium (K+), and hydrogen (H+) are a few examples of cations.
What does common cation mean?Positively charged ions are referred to as cations. Less electrons than protons make up cations. An ion can be made up of a single atom of an element (a monatomic ion, monatomic cation, or monatomic anion) or many atoms that are chemically connected to one another (a polyatomic ion or polyatomic cation or anion)
Alkali and alkaline earth metals always create cations, whereas halogens always produce anions. Most nonmetals normally create anions, while the majority of other metals typically produce cations (such as iron, silver, and nickel) (e.g. oxygen, carbon, sulfur)
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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.
what is constant angular speed
Answer:
in optical storage, constant angular velocity (CAV) is a qualifier for the rated speed of any disc containing information, and may also be applied to the writing speed of recordable discs. A drive or disc operating in CAV mode maintains a constant angular velocity, contrasted with a constant linear velocity (CLV)
Terrance pulls up at an angle on a rope attached to a crate to move the crate across a horizontal floor. The mass of the crate is 20.6 kilograms, Terrance applies 115 newtons of force at an angle of 26.7 degrees, and the coefficient of kinetic friction between the crate and the floor is 0.349.
a. How hard does the floor push up on the crate?
b. What is the acceleration of the crate?
The floor push up on the crate by a force of 51.75 N and the acceleration of the crate is 4.14m/s².
The angles at which the forces is applied is 26.7°. The coefficient of kinetic friction between the crate and the floor is 0.349. The force applied by terrance is 115 Newton. Total mass of the crate is 20.6 kilograms.
When the force F is applied at an angle A. The force will have two components, one along horizontal, FcosA and one along vertical, FsinA.
As the crate is pulled, the the floor will apply a normal reaction against the crate.
Normal force N is equal to FsinA.
Putting the values,
N = 115×sin(26.7°)
N = 115×0.45
N = 51.75 Newton.
The net acceleration can be calculated as,
Net force on the crate = Ma
Where a is the net acceleration,
Force applied(horizontal) - Friction force = Ma.
FcosA - uN = Ma
Where u is coefficient of friction,
Putting all the values,
115(0.9) - 0.349(51.75)= (20.6)a
103.5 - 18.06 = (20.6)a
a = 4.14m/s².
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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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40. 41. Two dentical springs are used to support a load of 80 N as shown below. 80 N If each spring extends by 12 cm, determine the spring's constant in N/m.
The spring constant for a spring with load of 80 N and which extends by 12 cm is -333.34 N/m for each springs.
What is spring constant?
According to Hooke's Law, when a spring is stretched, the force applied is proportional to the lengthening from the equilibrium length. The formula k = -F/x, where k is the spring constant, can be used to get the spring constant. F stands for force, and x for the variation in spring length.
The spring constant is calculated by dividing the force required to stretch or compress a spring by the lengthening or shortening of the spring. It is used to identify whether a spring is stable or unstable, and consequently, what system it should be employed in.
According to Hooke's law
k = -F/x
Where, k = spring constant
F = force,
x = variation in spring.
As given in the question,
There are two springs which support 80 N force so one will support 40 N
F = 40 N, x = 12 cm, k needs to be found out
putting the values in Hooke's law,
we get, k = -333.34 N/m
Therefore, the spring constant here is -333.34 N/m for each springs
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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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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