Electricity is the phenomenon of the flow of electric charges due to the potential difference.
The flow of electric charges across the conductor is known as electric current.
Unit of electric current is Ampere.
Potential difference is the difference of potential between two charges.
There are two charges namely positive and negative.
Uneven distribution of charges causes the potential difference.
Suppose A and B are two points on a conductor where A has a 5 C charge and B has a 3 C charge.
Thus, the current will flow from higher potential to lower potential which is from A to B.
Resistance is the opposing property of the conductor which opposes the flow of current.
The unit of resistance is ohm.
Hi I just answered a question for electricity grade 12 could you please if it is correct
F = k*(q1)*(q2)/r^2
We know that:
F = 71 N
k = coulomb's constant = 9*10^9
q1 = ?
q2 = ?
We know that q1 = q2, since in the question it states that the charges are identical.
r = 0.99 m
F = k*(q1)*(q1)/r^2
F = k*(q1)^2/r^2
71 = 9*10^9 * (q1)^2 / 0.99^2
Isolating (q1)^2,
71 * 0.99^2 / (9*10^9) = (q1)^2
Taking square root of both sides to get q1 to the power of 1,
sqrt( 71 * 0.99^2 / (9*10^9) ) = q1
Now, simplifying the left side in a calculator, we get that
q1 = 8.793 * 10^-5 C = q2
Two 4.137 cm by 4.137 cm plates that form a parallel-plate capacitor are charged to +/- 0.531 nC.What is the electric field strength inside the capacitor if the spacing between the plates is 2.198 mm?
The electric field of two parallel plates is given by:
[tex]\begin{gathered} E\cdot d=\frac{Q}{d\cdot A\epsilon o} \\ so: \\ E=\frac{Q}{A\epsilon o} \end{gathered}[/tex]Where:
[tex]\begin{gathered} \epsilon o=8.8542\times10^{-12} \\ A=0.04137^2=1.71\times10^{-3}m^2 \\ Q=0.531\times10^{-9}C \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} E=\frac{0.531\times10^{-9}}{(8.8542\times10^{-12})(1.71\times10^{-3})} \\ E\approx35071V/m \end{gathered}[/tex]An element with an atomic number of 90 and an atomic mass of 230 would have what atomic number after Alpha decay?88909186
Given data:
The atomic number of
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
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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66) The fundamental wavelength for standing sound waves in an empty pop bottle is 0.88 m. (a) What is the length of the bottle? (b) What is the frequency of the third harmonic of this bottle?
Explanation
Step 1
Let
[tex]\begin{gathered} \text{fundamental wavelength=0.88 ,} \\ \lambda=2L \\ \text{hence, replace} \\ 0.88m=2\text{ L} \\ \text{divide both sides by 2} \\ \frac{0.88m}{2}=\frac{2\text{ L}}{2} \\ 0.44=L \end{gathered}[/tex]hence, the length of the bottle is 0.44 meters
Step 2
What is the frequency of the third harmonic of this bottle?
For the third harmonic, the length of the string is equivalent to three-halves of a wavelength,so
[tex]\begin{gathered} third\text{ armonci= }\frac{\text{3}}{2}\cdot0.88\text{ m} \\ third\text{ armonci= }1.32\text{ m} \\ \end{gathered}[/tex]so
[tex]\begin{gathered} 1.32=2L \\ \end{gathered}[/tex]replace in the quation and solve for
Let speed of sound = 343 m/s
[tex]\begin{gathered} f=\frac{v}{\lambda} \\ \text{replacing} \\ f=\frac{343}{0.88} \\ f=389.77\text{ Hz} \end{gathered}[/tex]I hope this helps
you
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?
Answer:
fxn g
Explanation:
ggnhbdfhbz
Identify each setup and describe the difference in the magnetic field created by each.
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.
Which formula is equivalent to
v = m / D
v = mD / v
v = D x m
v = D / m
Answer:
i don't know.
Explanation: I = me Don't = Don't know = know
Object A, which has been charged to +13.13 nC, is at the origin. Object B, which has been charged to -26.7 nC, is at x=0 and y=2.72 cm. What is the magnitude of the electric force on object A?
ANSWER:
0.004259 N
STEP-BY-STEP EXPLANATION:
Given:
qA = +13.13 nC
qB = -26.7 nC
d = y - x = 2.72 - 0 = 2.72 cm = 0.0272 m
We use Coulomb's law to calculate the force:
[tex]F=k\cdot\frac{q_A\cdot q_B}{d^2}[/tex]We substitute each value and calculate the force:
[tex]\begin{gathered} F=9\times10^9\cdot\frac{(13.13\times10^{-9})(26.7\times10^{-9})}{(0.0272)^2} \\ \\ F=0.004259\text{ N} \end{gathered}[/tex]The magnitude of the electric force on object A is 0.004259 N
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?
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
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
How to calculate electric field on a charge
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 ball is thrown vertically upward with a speed of 36.0 m/s.(a) How high does it rise? m(b) How long does it take to reach its highest point? s(c) How long does the ball take to hit the ground after it reaches its highest point? s(d) What is its velocity when it returns to the level from which it started? m/s
ANSWER:
(a) 66.1 m
(b) 3.67 s
(c) 3.67 s
(d) -36 m/s
STEP-BY-STEP EXPLANATION:
Given:
Initial velocity (u) = 36 m/s
Final velocity (v) = 0 m/s
(a)
We can calculate the height using the following formula:
[tex]\begin{gathered} h=\frac{v^2-u^2}{2g} \\ \\ \text{ We replacing:} \\ \\ h=\frac{0^2-(36)^2}{(2)(-9.8)} \\ \\ h=66.1\text{ m} \end{gathered}[/tex](b)
Now, we calculate the time as follows:
[tex]\begin{gathered} v=u+gt \\ \\ t=\frac{v-u}{g} \\ \\ \text{ we replacing} \\ \\ t=\frac{0-36}{-9.8} \\ \\ t=3.67\text{ s} \end{gathered}[/tex](c)
The time it takes for the ball to hit the ground after reaching its highest point is the same as the time it takes to reach it, which means that it is equal to 3.67 seconds
(d)
The velocity it takes to return is also the same but being in the opposite direction, it has the same magnitude but with the opposite sign, that is, with a minus sign.
So the velocity when it returns to the level it started from is -36 m/s
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]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
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
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.
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.
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.
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
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
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.
5. Draw the ray diagram to show the image formed by a concave lens with focal length 5cm whenthe object is placed 15 cm from the lens. What is the magnification?
Given:
The focal length of the concave lens is f = -5 cm
The object distance from the lens is u = -15 cm
Required:
Ray diagram
Magnification
Explanation:
The rules related to incident ray and reflected ray are:
1) Any incident ray moving parallel to the principal axis will pass through focus after reflection.
2) Any incident ray passing through the focus will move parallel to the principal axis after reflection.
3) Incident ray passing through the optical center will reflect undeviated.
The ray diagram is shown below
Here, yellow lines represent incident rays passing parallel to the principal axis and the reflected ray passing through the focus.
The maroon line shows the incident ray passing through the optical center and the undeviated reflected ray.
The green color is the object and image.
In order to find magnification, first, we need to calculate the image distance given as
[tex]\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ =-\frac{1}{5}-\frac{1}{15} \\ v=-3.75\text{ cm} \end{gathered}[/tex]The magnification will be
[tex]\begin{gathered} m\text{ = }\frac{v}{u} \\ =\frac{-3.75}{-15} \\ =0.25\text{ } \end{gathered}[/tex]Final Answer: Thus, the magnification is 0.25 which means the size of the image is 0.25 times the object.
The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 903 Hz?
ANSWER:
0.38 meters.
STEP-BY-STEP EXPLANATION:
Given:
The speed of sound in air (v) = 340 m/s
Frequency (f) = 903 Hz
We calculate the wavelength using the following formula:
[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \\ \text{ We replacing:} \\ \\ \lambda=\frac{340}{903} \\ \\ \lambda=0.37652\cong0.38\text{ m} \end{gathered}[/tex]The wavelength is equal to 0.38 meters.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
LAST ONE!!!!
Explanation:
Answer:
the Temperature in each room will become equal over time
The Round Up carnival ride below has a radius of 3.90 meters and rotates 0.527 times per second. As shown, riders can be held up by only friction. What coefficient of friction is needed to keep the riders from sliding down? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The radius of the carnival ride is r = 3.9 m.
* The linear frequency of oscillation is f = 0.527 Hz.
Solution:
The angular frequency (angular speed) of the oscillation is,
[tex]\begin{gathered} \omega=2\pi f \\ \omega=2\pi\times0.527 \\ \omega=3.31\text{ rad/s} \end{gathered}[/tex]The frictional force acting on the ride is,
[tex]F=\mu mg[/tex]The force acting on the ride in terms of the angular speed is,
[tex]F=m\omega^2r[/tex]The frictional force acting on the ride is equal to the force acting on the ride for circular motion,
[tex]\begin{gathered} \mu mg=m\omega^2r \\ \mu g=\omega^2r \\ \mu=\frac{\omega^2r}{g} \end{gathered}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} \mu=\frac{(3.31)^2\times3.9}{9.81} \\ \mu=4.36 \end{gathered}[/tex]Thus, the coefficient of friction is 4.36.
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?
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
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 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?
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]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?
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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A sports car with a mass of 2000 kg travels at a velocity of 20 m/s. What would be true of its momentum if the same sports car was traveling at a velocity of 40 m/s?
The true statement about the sports car is that the momentum of the sports car would be increased.
What is momentum?The term momentum is defined as the product of the mass and the velocity. The momentum of an object is a vector quantity. What it means is that both the magnitude and the direction of the momentum is important.
Now we are told that the sports car has a mass of 2000 kg travels at a velocity of 20 m/s. The momentum at this point is 40000 Kgm/s. If the velocity of the sports car was increased to 40 m/s, the momentum would increase to 80,000 Kgm/s.
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