There is a distance of 3.60 * 10 ^ 7 m between the Earth and a satellite. The magnitude of the gravitational force of attraction between the and the Earth is 400 N. What would be the magnitude of the gravitational force of attraction if the distance between them changed to 1.80 * 10 ^ 7 meters?
Answers
Given:
The initial distance between the earth and the satellite, R₁=3.60×10⁷ m
The initial force between the earth and the satellite, F₁=400 N
The changed distance between the earth and the satellite R₂=1.80×10⁷ m
To find:
The changed force between the earth and the satellite.
Explanation:
From Newton's gravitational law, the gravitational force between two objects is directly proportional to the product of the mass of the objects and inversely proportional to the square of the distance between them.
Let us assume that the mass of the earth is M and the mass of the satellite is m.
Thus,
[tex]\begin{gathered} F_1=\frac{GMm}{R_1^2} \\ \Rightarrow F_1R_1^2=GMm\text{ }\rightarrow\text{ \lparen i\rparen} \end{gathered}[/tex]And, the changed force is given by,
[tex]\begin{gathered} F_2=\frac{GMm}{R_2^2} \\ \Rightarrow F_2R_2^2=GMm\rightarrow\text{ \lparen ii\rparen} \end{gathered}[/tex]From equation (i) and equation (ii),
[tex]\begin{gathered} F_1R_1^2=F_2R_2^2 \\ \Rightarrow F_2=\frac{F_1R_1^2}{R_2^2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} F_2=\frac{400\times\left(3.60\times10^7\right?^2}{\lparen1.80\times10^7)^2} \\ =1600\text{ N} \end{gathered}[/tex]Final answer:
The magnitude of the gravitational force of attraction between the earth and the satellite after the distance between them was changed is 1600 N.
Related Questions
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
Answers
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
How to calculate electric field on a charge
Answers
Let the charge be denoted by q.
Let the total electric force on the charge be denoted by F.
We have to the electric field on a charge.
Electric field on a charge is given by the formula
[tex]E=\text{ }\frac{F}{q}[/tex]a cannon sitting on top of a 40 m mound shoots a projectile with an initial velocity of 4.47 m/s parallel to ground. how far away did it land?
Answers
Given that,
The initial height of the projectile, y₀=40 m
The initial velocity of the projectile, v₀=4.47 m/s
The direction of the initial velocity is parallel to the ground. Therefore the x-component of the initial velocity v₀x=4.47 m/s.
And the y-component of the initial velocity is v₀y=0 m/s
From the equation of the motion we have,
[tex]y=y_0+v_{0y}t+\frac{1}{2}gt^2[/tex]Where y is the final height of the projectile which is zero as it finally hits the ground. And g =-9.8m/s² is the acceleration due to gravity. And t is the time interval of the flight of the projectile.
On substituting the known values in the above equation,
[tex]\begin{gathered} 0=40+0+\frac{1}{2}\times-9.8\times t^2 \\ =40-4.9t^2 \end{gathered}[/tex]On rearranging the above equation and simplifying it,
[tex]\begin{gathered} t=\sqrt{\frac{40}{4.9}} \\ =2.86\text{ s} \end{gathered}[/tex]The x-component of the velocity remains constant as there is no acceleration in that direction.
The horizontal distance travelled or the range of the projectile can be calculated using the formula,
[tex]R=v_{0x}t[/tex]On substituting the known values in the above equation,
[tex]R=4.47\times2.86=12.78\text{ m}[/tex]Therefore the projectile will land 12.78 meters away
A 49 lb piece of steel at 670 F is dropped into 95 lbs of water at 50 F. What is the final temperature of the mixture.
Answers
ANSWER
[tex]83.82F[/tex]EXPLANATION
Parameters given:
Mass of steel, M = 49 lb = 22.23 kg
Initial temperature of steel, ts = 670 F = 627.59 K
Mass of water, m = 95 lb = 43.09 kg
Initial temperature of water, tw = 50 F = 283.15 K
We want to find the final temperature of the mixture of the piece of steel and water.
The amount of heat released in the system must be equal to the total amount of heat absorbed by the system:
[tex]Q_R=Q_A[/tex]The piece of steel releases heat while the water absorbs the heat. This implies that the equation above becomes:
[tex]-MC(T_s-t_s)=mc(T_w-t_w)_{}[/tex]where C = specific heat capacity of steel = 468 J/kgK
c = specific heat capacity of water = 4184 J/kgK
Ts = final temperature of steel
Tw = final temperature of water
Since the final temperatures of both the steel and water are equal, it implies that:
[tex]T_s=T_w=T[/tex]Substituting the given values into the equation above, we have that:
[tex]-22.23\cdot468\cdot(T-627.59)=43.09\cdot4184\cdot(T-283.15)[/tex]Simplify the equation above:
[tex]\begin{gathered} -10403.64(T-627.59)=180288.56(T-283.15) \\ \Rightarrow-10403.64T+6529220.428=180288.56T-51048705.76 \\ \Rightarrow180288.56T+10403.64T=6529220.428+51048705.76 \\ \Rightarrow190692.2T=57577926.19 \\ T=\frac{57577926.19}{190692.2} \\ T=301.94K \end{gathered}[/tex]Let us now convert to Fahrenheit:
[tex]T=\frac{(301.94-273.15)\cdot9}{5}+32=83.82F[/tex]That is the final temperature of the mixture.
2. Connect an ammeter and voltmeter in the circuit below.
Answers
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.
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.
Answers
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
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.
Answers
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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A basketball player hit the basket eighteentimes during the game. He scored a totalof 30 points, two points for each field goaland one point for each free throw. If hethrew 6 more field goals than free throws,how many field goals did he make? Howmany free throws did he make?
Answers
Answer:
fxn g
Explanation:
ggnhbdfhbz
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.
Answers
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
What is the centripetal force exerted on a 1,600-kg car that rounds a 100-m curve at 12 m/s?Question 18 (2 points) What is the centripetal force exerted on a 1,600-kg car that rounds a 100-m radi curve at 12 m/s?
Answers
2304 Newton is the centripetal force exerted on a 1,600-kg car that rounds a 100-m curve at 12 m/s
centripetal force=mv^2/r
m=1,600-kg
v= 12 m/s
r= 100-m
centripetal force=mv^2/r
centripetal force=(1,600-kg×12 m/s×12 m/s)÷100-m
centripetal force=230400/100
centripetal force=2304 N
Any motion along a curved road is accelerated, necessitating the application of force to the path's center of curvature. Centripetal, which means "center seeking," is the name of this force.
The Centripetal Force Formula states that the centripetal force equals the product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters). This formula predicts that the centripetal force will double with a doubling of tangential velocity.
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URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answers
Answer:
LAST ONE!!!!
Explanation:
Answer:
the Temperature in each room will become equal over time
If a plane covers a distance of 700 miles in 1 hour and 43 minutes, what is its speed? A) Give your answer in mph. B) Give your answer in ft/second
Answers
Given:
Distance covered by the plane, d = 700 miles
Time, t = 1 hour and 43 minutes.
Let's find the speed of the plane.
To find the speed, apply the formula:
[tex]\text{Speed = }\frac{dis\tan ce}{time}[/tex]Where:
distance = 700 miles
time = 1 hour 43 minutes
(A) speed in mph.
mph is miles per hour.
Where:
60 minutes = 1 hour
Thus, to find the speed the time is to be in hours.
We have:
[tex]t=1\frac{43\text{ minutes}}{60\text{ minutes}}=1\frac{43}{60}=1.716666\text{ hrs}\approx1.72\text{ hrs}[/tex]Thus, to find the speed in mph, we have:
[tex]\text{Speed}=\frac{\text{distance}}{\text{time}}=\frac{700}{1.72}=407.8\text{ mph}[/tex]Therefore, the speed of the plane in mph is 407.8 mph
(B) To find the speed in ft/second
Let's first convert the distance from miles to feet
Where:
1 mile = 5280 feet
700 miles = 700 x 5280 = 3696000 feet
Also convert the time to seconds.
Where:
1 hour = 60 minutes x 60 seconds = 3600 seconds
1.7166 hours = 1.7166 x 3600 = 6180 seconds
Thus, we have:
Distance = 3696000 feet
Time = 6180 seconds
[tex]\text{Speed = }\frac{dis\tan ce}{\text{time}}=\frac{3696000}{6180}=598.06\text{ ft/second}[/tex]Therefore, the speed in ft/second is 598.06 ft/second
ANSWER:
(A) 407.8 mph
(B) 598.06 ft/second
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.
Answers
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).
I need help with all three questions for this problem
Answers
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]Identify each setup and describe the difference in the magnetic field created by each.
Answers
In A. we have a magnetic field created by a coil in physics, the term coil refers to a long, thin loop of wire, which produces a magnetic field when an electric current is passed through it. in this case, the core inside the coil is the air
In B. We have a magnetic field created by a magnet that is a piece of iron or other magnetic material that has its component with atoms so ordered that the material have properties of magnetism, such as attracting other iron-containing objects or aligning itself in an external magnetic field
In C. Also we have a coil but this time with a core of metal material, a core can increase the magnetic field to thousands of times the strength of the field of the coil alone.
What is the momentum of a 7.30 kg bowling ball going down the alley with a speed of 20.0 m/s?
Answers
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
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?
Answers
If you have a concave mirror whose focal length is 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object, where should you place the object?
Answers
given
[tex]\begin{gathered} focal\text{ length=100 cm} \\ let\text{ object distance is u form the pole of mirror } \\ so\text{ u=-u} \\ and\text{ v=?} \\ here\text{ v is the distance of image from the pole of mirror } \\ let\text{ height of the object h} \\ so\text{ h=+h} \\ according\text{ to the question image of the object is } \\ \text{ h}^{^{\prime}}=+10h \\ \end{gathered}[/tex]since the image is in the upright direction so the image is virtual.
and in the case of the mirror, an image can be virtual if and only if
the image is placed between the pole and the focus. so the object
must be placed between the pole and the focus.
[tex]\begin{gathered} the\text{ formula for magnification is given by;} \\ m=-\frac{v}{u}=\text{ }\frac{h^{^{\prime}}}{h} \\ and\text{ the mirror formula is given by the following} \\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \end{gathered}[/tex][tex]\begin{gathered} -\frac{v}{u}=\frac{h^{^{\prime}}}{h} \\ by\text{ putting all the values} \\ -\frac{v}{u}=\frac{10h}{h} \\ v=-10u \end{gathered}[/tex][tex]undefined[/tex]URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
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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.
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?
Answers
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
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
Answers
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.
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)
Answers
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
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?
Answers
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
An element with an atomic number of 90 and an atomic mass of 230 would have what atomic number after Alpha decay?88909186
Answers
Given data:
The atomic number of
Define vector and state two examples.
Answers
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]What is the wavelength of the electromagnetic wave emitted and absorbed by a phone?
Answers
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.
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?
Answers
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
In the calculation of the area of a rectangle with dimensions of 4.282 m by 0.050 m, which of the following answers has the correct number of significant figures?
Answers
We have
[tex]A=l\times w[/tex]in our case
l=4.282 m
w=0.050m
we substitute
[tex]A=(4.282)(0.050)=0.2141m^2=0.21m^2[/tex]ANSWER
0.21m^2
I need help on number 5. We need to use one of the four kinematics equations.
Answers
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.
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?
Answers
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.