Given:
The mass of the first sphere is: m1 = 3.699 g.
The mass of the second sphere is: m2 = 3.699 g
The distance between their centers is: d = 3.592 cm
The acceleration of each sphere is: a = 297.727 m/s^2
To find:
Since the spheres are identical in their masses, the force on each sphere is:
[tex]F=ma[/tex]Substitute the values in the above equation and simplify it, we get:
[tex]\begin{gathered} F=3.699\text{ g}\times297.727\text{ m/s}^2 \\ \\ F=3.699\text{ g }\times\frac{1\text{ kg}}{1000\text{ g}}\times297.727\text{ m/s}^2 \\ \\ F=3.699\times10^{-3}\text{ kg}\times297.727\text{ m/s}^2 \\ \\ F=1.1012\text{ N} \end{gathered}[/tex]This is the force experienced by each sphere and is has a magnitude equal to the magnitude of the electrostatic force.
The electrostatic force of attraction or repulsion between two charges is given by:
[tex]F=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}[/tex]Substitute the values in the above equation and simplify it, we get:
[tex]\begin{gathered} 1.1012\text{ N}=\frac{9\times10^9\text{ N}\cdot m^2\text{/C}^2\times q^2}{(3.592\text{ cm})^2} \\ \\ 1.1012\text{ N}=\frac{9\times10^9\text{ N}\cdot m^2\text{ / C}^2\times q^2}{(3.592\text{ cm}\times\frac{1\text{ m}}{100\text{ cm}})^2} \\ \\ 1.1012\text{ N}=\frac{9\times10^9\text{ N}\cdot m^2\text{ /C}^2\times q^2}{(3.592\times10^{-2})^2\text{ m}^2} \\ \\ 1.1012\text{ N}=\frac{9\times10^9\text{ N.m}^2\text{/C}^2\times q^2}{1.2902\times10^{-3}\text{ m}^2\text{ }} \\ \\ 1.1012\text{ N}=6.9757\times10^{12}\text{ N/C}^2\times q^2 \\ \\ \end{gathered}[/tex]Rearranging the above equation and simplify it, we get:
[tex]\begin{gathered} q^2=\frac{1.1012\text{ N}}{6.9757\times10^{12}\text{ N/C}^2} \\ \\ q=\sqrt{1.5786\times10^{-13}\text{ C}^2} \\ \\ q=0.3973\times10^{-6}\text{ C} \\ \\ q=0.3973\text{ }\mu\text{C} \end{gathered}[/tex]Final answer:
The magnitude of the charge on each sphere is 0.3973 microcolumns.
What is the momentum of a 7.30 kg bowling ball going down the alley with a speed of 20.0 m/s?
Given data
*The given mass of the bowling ball is m = 7.30 kg
*The given speed is v = 20.0 m/s
The formula for the momentum of a 7.30 kg bowling ball is given as
[tex]p=mv[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} p=(7.30)\times(20.0) \\ =146\text{ kg.m/s} \end{gathered}[/tex]Hence, the momentum of a 7.30 kg bowling ball is p = 146 kg.m/s
2. Connect an ammeter and voltmeter in the circuit below.
ANSWER and EXPLANATION
We want to connect a voltmeter and an ammeter in the given circuit.
In a circuit, a voltmeter is connected in parallel with a device in order to measure its voltage. Hence, the voltmeter will be connected in parallel with the resistor.
The ammeter is connected in a circuit in series in order to measure its current.
Let us connect the meters below:
That is the answer.
2. A parallel-plate capacitor has an area of 2.0 cm², and the plates are separated by 2.0 mm. a. What is the capacitance? b. How much charge does this capacitor store when connected to a 6.0 V battery?
Given data:
* The area of the parallel plate capacitor is,
[tex]\begin{gathered} A=2cm^2 \\ A=2\times10^{-4}m^2^{} \end{gathered}[/tex]* The distance between the plates is,
[tex]\begin{gathered} d=2\text{ mm} \\ d=2\times10^{-3}\text{ m} \end{gathered}[/tex]Solution:
(a). The capacitance of the capacitor in terms of area and distance between the plates is,
[tex]C=\frac{\epsilon_{\circ}A}{d}[/tex][tex]\text{where }\epsilon_{\circ}\text{ is the electrical permittivity of the fr}ee\text{ spaces}[/tex]Substituting the known values,
[tex]\begin{gathered} C=8.85\times10^{-12}\times\frac{2\times10^{-4}}{2\times10^{-3}} \\ C=8.85\times10^{-13}\text{ F} \end{gathered}[/tex]Thus, the value of the capacitanc is 8.85 times 10 power -13 Farad.
(b). The voltage across the battery is,
[tex]V=6\text{ Volts}[/tex]The charge stored in the capacitor in terms of the voltage and the capacitance is,
[tex]\begin{gathered} C=\frac{Q}{V} \\ Q=CV \end{gathered}[/tex]where Q is the charge stored in the capacitor
Substituting the known values,
[tex]\begin{gathered} Q=8.85\times10^{-13}\times6 \\ Q=53.1\times10^{-13}\text{ Coulomb} \end{gathered}[/tex]Thus, the charge stored in the parallel plate capacitor is 53.1 times 10 power -13 coulomb.
Calculate the current needed to carry the 2 000 MW generated at alarge power station if the distribution voltage is kept at 22 kV, thegenerator output voltage.
Given data
*The given power is P = 2000 MW = 2000 × 10^6 W
*The given distribution voltage is V = 22 kV = 22 × 10^3 V
The formula for the current needed to carry the 2000 MW generated at a large power station is given as
[tex]I=\frac{P}{V}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} I=\frac{2000\times10^6}{22\times10^3} \\ =90.90\times10^3\text{ A} \end{gathered}[/tex]Hence, the current needed to carry the 2000 MW generated at a large power station is I = 90.90 × 10^3 A
A grocery shopper tosses a 6 kg bag of rice into a stationary 27 kg grocery cart. The bag hits the cart with a horizontal speed of 8.0 m/s toward the front of the cart. What is the final speed of the cart and the bag? Round to the hundredths place
Given,
The mass of the bag, m=6 kg
The mass of the cart, M=27 kg
The initial velocity of the cart, u=0 m/s
The initial velocity of the bag, v=8.0 m/s
From the law of conservation of momentum, the sum momentum of the bag and the momentum of the cart before the bag hits the cart must be equal to the momentum of the bag and the cart after the bag hits the cart.
Thus,
[tex]mv+Mu=(M+m)V[/tex]Where V is the velocity of the bag and the cart after the bag hits the cart.
On rearranging,
[tex]V=\frac{mv+Mu}{(M+m)}[/tex]On substituting the known values,
[tex]\begin{gathered} V=\frac{6\times8.0+0}{(27+6)} \\ =\frac{48}{33} \\ =1.45\text{ m/s} \end{gathered}[/tex]Thus the final speed of the cart and the bag is 1.45 m/s
Suppose a 1.90 N force can rupture an eardrum having an area of 1.14 cm².(a) Calculate the maximum tolerable gauge pressure inside the eardrum (in the middle ear) in N/m². (Pressures in themiddle ear may rise when an infection causes a fluid buildup. Use 13.6 x 10³ kg/m³ as the density of mercury.) submit answer in N/m²(a) part 2: Convert this value to mm Hg.mm Hg(b) At what depth in fresh water would this person's eardrum rupture, assuming the gauge pressure in the middle ear iszero?Submit Answer in m
Given:
The force is
[tex]F=1.90\text{ N}[/tex]The area of the eardrum is
[tex]\begin{gathered} A=1.14\text{ cm}^2 \\ =1.14\times10^{-4}\text{ m}^2 \end{gathered}[/tex]To find:
The maximum tolerable gauge pressure inside the eardrum
a) the pressure in mm of Hg
b) At what depth in freshwater would this person's eardrum rupture
Explanation:
The pressure at the eardrum is
[tex]\begin{gathered} P=\frac{F}{A} \\ =\frac{1.90}{1.14\times10^{-4}} \\ =16.67\times10^3\text{ N/m}^2 \end{gathered}[/tex]Hence, the pressure is
[tex]16.67\times10^3\text{ N/m}^2[/tex]a)
We know,
[tex]1\text{ N/m}^2=0.0075\text{ mm of Hg}[/tex]So,
[tex]\begin{gathered} 16.67\times10^3\text{ N/m}^2=0.0075\times16.67\times10^3\text{ mm of Hg} \\ =125.02\text{ mm of Hg} \end{gathered}[/tex]Hence, the pressure is 125.02 mm of Hg.
b)
The depth of fresh water is,
[tex]\begin{gathered} h=\frac{P}{dg} \\ Here,\text{ d=1000 kg/m}^3 \\ g=9.8\text{ m/s}^2 \end{gathered}[/tex]So,
[tex]\begin{gathered} h=\frac{16.67\times10^3}{1000\times9.8} \\ =1.70\text{ m} \end{gathered}[/tex]Hence, the depth of water is 1.70 m.
1. The pupil brought the needle to one pole of the magnetic pointer. The needle attracted the pole of the pointer. Can this be evidence that the needle is magnetized? Why?
Yes, it is an evidence that the needle is magnetized.
Explanation:Magnetism describes the ability of an object to be attracted or repelled by a magnet.
Since the needle was attracted when brought to one pole of the magnet, we can conclude that the pin has properties that make it to be magnetized.
Therefore, the above is an evidence that the needle is magnetized.
I need help on number 5. We need to use one of the four kinematics equations.
ANSWER
[tex]35.02m[/tex]EXPLANATION
Parameters given:
Initial velocity, u = 26.2 m/s
When the vase reaches its maximum height, its velocity becomes 0 m/s. That is the final velocity.
We can now apply one of Newton's equations of motion to find the height:
[tex]v^2=u^2-2as[/tex]where a = g = acceleration due to gravity = 9.8 m/s²
Therefore, we have that:
[tex]\begin{gathered} 0=26.2^2-2(9.8)s \\ \Rightarrow19.6s=686.44 \\ s=\frac{686.44}{19.6} \\ s=35.02m \end{gathered}[/tex]That is the height that the vase will reach.
Force can be given as ?
The force is, momentum/time.
option 1
Romeo and Juliet are sitting on a balcony 1.5 meters apart. If Romeo has a mass of 61.6 Kg and Juliet has a mass of 48.8 kg. What is the attractive force between them?
hello I need help with question 7 a and b please
Given data:
* The mass of the child is 40kg.
* The reading of the scale is 30 N.
Solution:
(a). The free body diagram of the given system is,
(b). The weight of the child is,
[tex]\begin{gathered} W=40\times9.8 \\ W=392\text{ N} \end{gathered}[/tex]The Scale reading is,
[tex]N=30\text{ Newton}[/tex]The net force acting on the box in terms of the weight of the box is,
[tex]\begin{gathered} F_{\text{net}}=N-mg \\ F_{\text{net}}=30-392 \\ F_{\text{net}}=-362\text{ N} \end{gathered}[/tex]According to the Newton's second law,
[tex]undefined[/tex]Explain what inertia is. What causes inertia in an object that is at rest? In an object that is moving?
Inertia is the ability of the body to remain in the state or in uniform motion unless some external force acts on the body.
The inertia of the body is direly proportional to the mass of the body.
Thus, the greater the mass of the body greater is its inertia.
Three types of inertia:
In the state of rest, the inertia of the body resists the cause of motion. This inertia is known as resting inertia.
In the state of uniform motion, the inertia resists the cause of changing of motion. This inertia is known as motion inertia.
Some external force is applied to the body to change the direction of the motion of the body, the inertia that opposes this change in direction is known as directional inertia.
The position of a particle moving along the x axis is given by x = (21 + 22t - 6.0t 2)m, where t is in s. What is the average velocity during the time interval t = 1.0 s to t = 3.0 s?
Using the concept of linear motion, we got the average velocity as -2m/s.
Since the particle is moving in the x direction and we are given an equation
x = (21 + 22t - 6.0t 2)m
using the concept of differentiation we got
velocity (v) = dx/dt
Now distance x for t=1 s
x=21+22(1)-6
x=37m .....equation (1)
similarly distance at t=3 s
x=21+22(3)-54
x=33m ....equation (2)
so average velocity is
average velocity = total distance/total time
v=(33-37)/(3-1)
v=-2m/s
So average velocity for a given time interval is -2m/s
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Find the magnitude of the vector sum A→+B→+C→ . Each grid square is 2.00 N on a side. If the vector sum is to the west, enter a negative value. If the vector sum is to the east, enter a positive value.
Given data:
* The magnitude of the vector C is,
[tex]\begin{gathered} C=2\times2 \\ C=4\text{ N} \end{gathered}[/tex]* The magnitude of the vector B is,
[tex]\begin{gathered} B=4\times2 \\ B=8\text{ N} \end{gathered}[/tex]Solution:
From the given diagram, the magnitude of the vector A is,
[tex]\begin{gathered} A=\sqrt[]{Base^2+Perpendicular^2} \\ A=\sqrt[]{(3\times2)^2+(4\times2)^2} \\ A=\sqrt[]{6^2+8^2} \\ A=\sqrt[]{36+64} \\ A=\sqrt[]{100} \\ A=10\text{ N} \end{gathered}[/tex]The count of square grid in the hypotenuse is,
[tex]\begin{gathered} n=\frac{A}{2} \\ n=\frac{10}{2} \\ n=5 \end{gathered}[/tex]The angle of the vector A with the x-axis is,
[tex]\begin{gathered} cos(\theta)=\frac{Base}{\text{Hypotenuse}} \\ \cos (\theta)=\frac{3}{5} \\ \theta=53.13^{\circ} \end{gathered}[/tex]Thus, the value of vector A is,
[tex]\begin{gathered} \vec{A}=A\cos (53.13^{\circ})+A\sin (53.13^{\circ}) \\ \vec{A}=10\times\cos (53.13^{\circ})i+10\times\sin (53.13^{\circ})j \\ \vec{A}=6\text{ i + 8 j} \end{gathered}[/tex]The value of vector B is,
[tex]\vec{B}=-8\text{ j}[/tex]The value of vector C is,
[tex]\vec{C}=-4\text{ i}[/tex]Thus, the sum of the vectors is,
[tex]\begin{gathered} \vec{A}+\vec{B}+\vec{C}=6\text{ i+8 j-8 j-4 i} \\ \vec{A}+\vec{B}+\vec{C}=2\text{ i} \\ |\vec{A}+\vec{B}+\vec{C}|=\sqrt[]{2^2} \\ |\vec{A}+\vec{B}+\vec{C}|=\text{ 2 N} \end{gathered}[/tex]Thus, the magnitude of the sum of three given vectors is 2 N towards the east (positive of the x-axis).
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It can't be A. because the food isn't touching the burner, it can't be C. because food can't be completely cooked that way and that's not how ovens work, and it can't be D. because cool air doesn't rise.
B is the answer because it accurately describes how a oven is supposed to work.
E. O 58.456 S48. A bullet is fired from the ground making an angle of 20 degwith the horizontal with a speed of 1500 m/s.Calculate the maximum height reached? (1 point)A. O20831.097 mB. 3642.875 mC. O8232.474 mD. O15803.894 mE. 13428.572 m9. A bullet is fired from the ground making an angle of 20 degwith the horizontal with a speed of 1500 m/s.How far (horizontally) will it fall? (1 point)
The maximum height of a projectile is given by:
[tex]h=\frac{v_0^2\sin^2\theta}{2g}[/tex]In this case, the initial velocity is 1500 m/s, the angle is 20°; then we have:
[tex]\begin{gathered} h=\frac{(1500^)^2(\sin20)^2}{(2)(9.8)} \\ h=13428.572 \end{gathered}[/tex]Therefore, the maximum height is 13428.572 m
An airplane flying horizontally at a velocity of 138 m/s and at an altitude of 1500 meters when one of its wheels falls off.What horizontal distance (in meters) will the wheel travel before it strikes the ground?
ANSWER:
2415 meters
STEP-BY-STEP EXPLANATION:
Given:
Initial horizontal velocity (ux) = 138 m/s
Initial vertical velocity (uy) = 0 m/s
Height (h) = 1500 meters
The first thing is to calculate the time it takes for the airplane to reach the ground, just like this:
[tex]\begin{gathered} h=u_yt+\frac{1}{2}at^2 \\ \\ \text{ we replacing} \\ \\ 1500=0\cdot t+\frac{1}{2}(9.8)t^2 \\ \\ 4.9t^2=1500 \\ \\ t^2=\frac{1500}{4.9} \\ \\ t=\sqrt{\frac{1500}{4.9}} \\ \\ t=17.5\text{ sec} \end{gathered}[/tex]Therefore, the horizontal distance would be:
[tex]\begin{gathered} x=u_x\cdot t \\ \\ \text{ we replacing} \\ \\ x=138\cdot17.5 \\ \\ x=2415\text{ m} \end{gathered}[/tex]Therefore, the horizontal distance is 2415 meters
‼️‼️IF ANSWERED W WORK SHOWN I WILL GIVE BRAINLIEST‼️‼️The electric field intensity between two large, parallel metal plates is 6000 N/C. The plates are 0.05 mapart. What is the electric potential difference between them?
We have the next information
E= 6000 N/C
d=0.05m
In order to find the electric potential difference we have the next formula
[tex]\Delta V=Ed[/tex]then we substitute the data given in the formula
[tex]\Delta V=(6000\frac{N}{C})(0.05m)=300\frac{J}{C}=300V[/tex]The electric potential difference between them is 300 V
In a bike race, a sport person starts his bike from rest and maintains a constant acceleration of 3 m/s2 for 8 seconds, and then a constant acceleration of 2 m/s2 for another 8 seconds. Determine the acquired velocity of the bike.
Answer:
40 m/s
Explanation:
First, we need to calculate the velocity after 8 seconds, so we will use the following equation:
v₂ = v₁ + at
Where v₁ is the initial velocity, so v₁ = 0 m/s, a is the acceleration a = 3 m/s² and t is the time, so t = 8s. Replacing the values, we get:
v₂ = 0 m/s + (3 m/s²)(8s)
v₂ = 0 m/s + 24 m/s
v₂ = 24 m/s
Then, it accelerates for another 8 seconds with a = 2 m/s², so the acquired velocity v₃ is equal to:
v₃ = v₂ + at
v₃ = 24 m/s + (2 m/s²)(8 s)
v₃ = 24 m/s + 16 m/s
v₃ = 40 m/s
Therefore, the acquired velocity of the bike is 40 m/s
A gardener pushes a lawn roller through a distance of 20m if he applies a force of 20kg weight in a direction inclined at 70 degree to ground find the work done by himG=9.8m/s^2
The total force applied to the roller can be obtained as,
[tex]\begin{gathered} F=(20\text{ kg-wt)(}\frac{9.8\text{ N}}{1\text{ kg-wt}}) \\ =196\text{ N} \end{gathered}[/tex]The work done by the gardner can be given as,
[tex]W=Fd\cos \theta[/tex]Substitute the known values,
[tex]\begin{gathered} W=(196N)(20m)(\frac{1\text{ J}}{1\text{ Nm}})cos70^{\circ} \\ =(3920\text{ J)(}0.342) \\ =1340.64\text{ J} \end{gathered}[/tex]Thus, the work done by the gardener is 1340.64 J.
Define vector and state two examples.
Vector is a mathematical entity that has a magnitude and a direction.
It is graphically represented by an arrow. The size of the arrow indicates the magnitude, and the tip of the arrow indicates the direction.
For example, we have the vector below:
Some forms of writing this vector are:
[tex]\begin{gathered} \vec{v}=ai+bj=4i+3j\\ \\ \vec{v}=<4,3> \end{gathered}[/tex]Now, let's write a second example of a vector with 6 units in horizontal direction and -4 units in vertical direction, starting at the origin:
The magnitude and angle of a vector are given by:
[tex]\begin{gathered} \vec{v}=ai+bj\\ \\ magnitude:\\ \\ |\vec{v}|=\sqrt{a^2+b^2}\\ \\ angle:\\ \\ \theta=\tan^{-1}(\frac{b}{a}) \end{gathered}[/tex]A wheel of radius 30.0 cm is rotating at a rate of 2.20 revolutions every 0.0910 s. What is the linear speed of a point on the wheels rim?
Explanation:
First, we need to calculate the frequency of the wheel, so if it is rotating at a rate of 2.20 revolutions every 0.0910 seconds, the frequency is:
[tex]f=\frac{2.20\text{ revolutions}}{0.0910\text{ seconds}}=24.16\text{ Hz}[/tex]Then, the magnitude of the angular velocity is equal to:
[tex]w=2\pi f=2(3.14)(24.16)=151.9\text{ rad/s}[/tex]Finally, we can calculate the linear speed as the angular velocity times the radius, so:
[tex]v=w\cdot r=(151.9\text{ rad/s)(30 cm) = 4557.04 m/s}[/tex]Therefore, the linear speed is 45.47 m/s
Consider the diagram of a combination circuit below on the left. In the middle, the resistors in the two parallel branches have been replaced by a single resistor (R4) with an equivalent resistance to the overall branch resistors. On the right, all three resistors have been replaced by a single resistor (R5) with an equivalent resistance as all three original resistors. Suppose that you know that:R1 = 24.8ΩR2 = 24.8ΩR3 = 12.7Ω What must R4 and R5 be in order for the two circuits to have the same equivalent resistance? R4 = ------- Ω R5 = ----- Ω
Given:
• R1 = 24.8Ω
,• R2 = 24.8Ω
,• R3 = 12.7Ω
From the diagram, let's find R4 and R5.
We can see that the 3 resistors R1, R2, and R3 are connected in parallel.
Where:
R1 + R2 = R4
To solve for R4, we have:
[tex]\frac{1}{R_4}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]Thus, we have:
[tex]\begin{gathered} \frac{1}{R_4}=\frac{1}{24.8}+\frac{1}{24.8} \\ \\ \frac{1}{R_4}=\frac{1+1}{24.8} \\ \\ \frac{1}{R_4}=\frac{2}{24.8}=\frac{1}{12.4} \\ \\ R_4=12.4\text{ \Omega} \end{gathered}[/tex]Now, to solve for R5 since R3 and R4 are now in series, we have:
[tex]\begin{gathered} R_5=R_3+R_4 \\ \\ R_5=12.7+12.4 \\ \\ R_5=25.1Ω \end{gathered}[/tex]Therefore, we have:
R4 = 12.4 Ω
R5 = 25.1 Ω
ANSWER:
• R4 = 12.4 ,Ω
,• R5 = 25.1 ,Ω
I need help with all three questions for this problem
The given force is:
[tex]F=(2.2N)x+(1.0N)y[/tex]Part A
If the displacement of the dog is:
d = (0.29m)x
The work done is the dot product of the force and displacement
[tex]\begin{gathered} W=F.d \\ \\ W=\text{ \lparen2.2\rparen\lparen0.29\rparen+\lparen1.0\rparen\lparen0\rparen} \\ \\ W=0.638Joules \end{gathered}[/tex]Part B
If the displacement of the dog is:
d = (0.29m)y
The workdone will be calculated as:
[tex]\begin{gathered} W=F.d \\ \\ W=(2.2)(0)+(1.0)(0.29) \\ \\ W=0.29\text{ Joules} \end{gathered}[/tex]2. Two charges are repelled by a force of 9.0 N. If the distance between them triples, what is the force between the charges?
F = k q1q2 / d^2
F= 9 N
The force is inversely related to the square of the distance. If d is 3 times larger, F is 9 times smaller.
F= 1 N
What is the wavelength of the electromagnetic wave emitted and absorbed by a phone?
The wavelengths of the electromagnetic waves emitted and absorbed by a phone is approximately 10-1000 m. Radio waves are those waves that are used for the cell phones to communicate with the towers.
A car starts from rest and accelerates uniformly for a distance of 137 m over an 9.6-second time interval. The car's acceleration ism/s².
In order to solve this question, we will need to use kinematics
Let's see what is given to us:
Distance traveled is 137 meters
Time elapsed is 9.6 seconds
and inital velocity is 0 m/s
Since we are trying to find acceleration, we can use this formula
[tex]\Delta x=v_0t+\frac{1}{2}at^2[/tex]Where Δx is the distance traveled, v0 is the inital velocity, t is time, and a is acceleration
Plugging in what we have, we get
137 = 0(9.6) + 1/2(a)(9.6)^2
Solving for a, we get 2.97 m/s^2
What would happen to the pressure and temperature of a box as gas is added to it
When gas is added to the box, the number of molecules in the box will increase.
This will result in an increase in pressure as volume is fixed or constant.
As the box has a fixed volume, this will be an isochoric process.
In an isochoric process, the pressure is directly proportional to temperature, so the temperature will also increase.
Thus, pressure and temperature will increase when gas is added to a box.
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Answer:
LAST ONE!!!!
Explanation:
Answer:
the Temperature in each room will become equal over time
When an ice cube melts in your hand, it changes from a solid to a liquid.
H2O (s) → H2O (l)
This is an example of ________________ reaction, because heat is _____________ the system.
The reaction is endothermic as heat is taken into the system.
What is an endothermic reaction?We know that in an endothermic reaction. Heat is taking in by the reaction. The heat that is taken into the system is used to break apart the chemical bonds that are in the system and this would facilitate the conversion of the reactants to the products.
In this case, there is the intake of heat by the solid water and this would lead to the breakage of all the rigid hydrogen bonds that hold the molecules of the water in the solid state and when that occurs we can now see that we would have liquid water.
Thee main point here is that heat is taken into the system thus heat is a reactant and ought to appear at the left hand side of the reaction equation and leads to the formation of the liquid water product.
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