Answers
We can see that the order of magnitude of the problem is:
[tex]n=21[/tex]Related Questions
Determine the displacement x of the elevator above the starting point at the end of each 5-second interval?
Answers
Given data:
* The normal weight of the student is 500 N.
* The initial velocity of the student is 0 m/s at time 0 second.
Solution:
(a). For the time interval 0 to 5 seconds,
The scale reading is 500 N.
Thus, the force acting on the elevator is,
[tex]F=F^{\prime}-W[/tex]where F is the net force with which the elevator is moving in upward direction, F' is the scale reading (apparent weight), and W is the actual weight,
Substituting the known values,
[tex]\begin{gathered} F=500-500 \\ F=0 \end{gathered}[/tex]By the Newton's second law, the acceleration of the elevator is,
[tex]\begin{gathered} ma=F \\ ma=0 \\ a=0ms^{-2} \end{gathered}[/tex]By the kinematics equation, the displacement in the time interval 0 to 5 seconds is,
[tex]D=ut+\frac{1}{2}at^2[/tex]where D is the displacement, u is the initial velocity, a is the acceleration, and t is the time,
Substituting the known values,
[tex]\begin{gathered} D=0+0 \\ D=0\text{ m} \end{gathered}[/tex]Thus, the displacement in the 0 to 5 seconds time interval is zero.
(b). For 5 to 10 seconds interval,
As the scale reading is 700 N, thus, the acceleration of the elevator is,
[tex]\begin{gathered} F=F^{\prime}-W \\ ma=700-500 \\ ma=200\text{ N} \end{gathered}[/tex]where m is the mass,
From the weight, the value of mass is,
[tex]\begin{gathered} W=mg \\ m=\frac{W}{g} \end{gathered}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} m=\frac{500}{9.8} \\ m=51.02\text{ kg} \end{gathered}[/tex]Thus, the acceleration of the elevator is,
[tex]\begin{gathered} ma=200 \\ a=\frac{200}{m} \\ a=\frac{200}{51.02} \\ a=3.92ms^{-1} \end{gathered}[/tex]By the kinematics equation, the displacement of the elevator is,
[tex]D=ut+\frac{1}{2}at^2[/tex]Substituting the known values,
[tex]\begin{gathered} D=0+\frac{1}{2}\times3.92\times5^2 \\ D=49\text{ m} \end{gathered}[/tex]Thus, the displacement of elevator is 49 meters for the time interval 5 seconds to 10 seconds.
(c). For the time interval 10 seconds to 15 seconds,
The apparent weight (scale reading) is 500 N.
Thus, the acceleration of the elevator is,
[tex]\begin{gathered} F=F^{\prime}-W \\ ma=500-500 \\ ma=0 \\ a=0ms^{-2} \end{gathered}[/tex]As the scale count is same as the normal weight, thus, elevator must not be moving or it is stopped at some height after moving some displacement upto 10 seconds. Thus, there will be no displacement of the elevator.
(d). For the time interval of 15 seconds to 20 seconds,
The apparent weight of the student is 300 N.
Thus, the acceleration of the elevator is,
[tex]\begin{gathered} F=F^{\prime}-W \\ ma=300-500 \\ ma=-200 \\ a=\frac{-200}{51.02} \\ a=-3.92ms^{-2} \end{gathered}[/tex]Here, the negative sign indicates the elevator is moving in the downward direction.
By the kinematics equation, the displacement of the elevator in the time interval 15 to 20 seconds is,
[tex]D=ut+\frac{1}{2}at^2[/tex]Substituting the known values,
[tex]\begin{gathered} D=0+\frac{1}{2}\times(-3.92)\times5^2 \\ D=-49\text{ m} \end{gathered}[/tex]Here, the negatiev sign indicates the downward displacement,
Thus, the displacement of the elevator in the -49 meter.
The siren of a burglar alarm system has a frequency of 960Hz . During a patrol a security officer, traveling in his car , hears the siren of the alarm of a house and approaches the house at a constant velocity. A detector in his car registers the frequency of the sound as 1000 Hz . Calculate the speed at which the patrol car approaches the house . Use the speed of sound in air as 340m.s
Answers
Answer:
14.2 m/s
Explanation:
When the source and receiver are getting closer, we can use the following equation:
[tex]f_o=(\frac{v+v_o}{v-v_s})f_s[/tex]Where fo is the observed frequency, fs is the emitted frequency, vo is the speed of the observed, vs is the speed of the source, and v is the speed of the sound. Solving for vo, we get:
[tex]\begin{gathered} \frac{f_o}{f_s}=\frac{v+v_o}{v-v_s} \\ \\ \frac{f_o}{f_s}(v-v_s)=v+v_o \\ \\ \frac{f_o}{f_s}(v-v_s)-v=v_o \\ \\ v_o=\frac{f_o}{f_s}(v-v_s)-v \end{gathered}[/tex]Then, replacing v = 340 m/s, vs = 0 m/s, fs = 960 Hz, and fo = 1000 Hz, we get:
[tex]\begin{gathered} v_r=\frac{1000}{960}(340-0)-340 \\ \\ v_r=14.2\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the patrol car is 14.2 m/s
Two 4-1 resistors are connected in parallel to a 12-V bat-tery. Use the fact that the voltage across each of the resis-tors is 12 V to find the total current through the battery.What single resistor, if connected to the battery alone(called the equivalent resistance), would draw this samecurrent?
Answers
Given data
*The given resistors are
[tex]R_1=R_{2_{}}_{}_{}=4\text{ ohm}[/tex]*The given battery voltage is V = 12 V
The equivalent resistance in parallel combination is calculated as
[tex]\begin{gathered} R_{eq}=\frac{R_1\times R_2}{R_1+R_2} \\ =\frac{4\times4}{4+4} \\ =\frac{16}{8} \\ =2\text{ ohm} \end{gathered}[/tex]The total current flows through the battery is calculated as
[tex]\begin{gathered} I=\frac{V}{R_{eq}} \\ =\frac{12}{2} \\ =6\text{ A} \end{gathered}[/tex]When will the force be balance and when will the force won’t be balance [unbalance]
Answers
A force is a pull or push of an object that causes it to accelerate. When there are more than one force acting on a body then we call this a system of forces.
Suppose that we have the following system of two forces:
Forces "a" and "b" are acting in opposite directions. If the forces have the same magnitude then the total acceleration in the object will be zero. When a system of forces produces no acceleration on an object we call this balanced force.
If the magnitude of the forces is different then by Newton's second law we have;
[tex]F_a-F_b=ma[/tex]Where "m" is mass and "a" is acceleration. A system where the sum of forces is the product of mass by acceleration is called an unbalanced force.
Bella is approaching some congested traffic and wisely slows her Audi from 19.6 m/s to 3.7 m/s at a constant rate of -3.33 m/s/s. How much time elapses before the car is traveling at 3.7 m/s?
Answers
The Time that elapses before the car is traveling at 3.7 m/s is 4.8 s.
What is time?Time is a measure of non-stop, consistent change in our surroundings, usually from a specific viewpoint.
To calculate the time require for the to be traveling at 3.7 m/s, we use the formula below.
Formula:
t = (v-u)/a........... Equation 1Where:
t = Timev = Final velocityu = Initial velocitya = AccelerationFrom the question,
Given:
v = 3.7 m/su = 19.6 m/sa = - 3.33 m/s²Substitute these values into equation 1
t = (19.6-3.7)/-3.33t = -15.9/-3.33t = 4.8 sHence, the time need for the car to start traveling at 3.7 m/s is 4.8 s.
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Find the angular velocity (in rad/s) of a motor that is developing 37.0 W of power with a torque of 5.5 Nm.
Answers
ANSWER
[tex]6.72\text{ rad/s}[/tex]EXPLANATION
Parameters given:
Power, P = 37.0 W
Torque, T = 5.5 Nm
To find the angular velocity, we apply the relationship between torque and power:
[tex]P=\tau\cdot\omega[/tex]where ω = angular velocity
Therefore, the angular velocity of the motor is:
[tex]\begin{gathered} \omega=\frac{P}{\tau} \\ \omega=\frac{37}{5.5} \\ \omega=6.72\text{ rad/s} \end{gathered}[/tex]That is the answer.
1. An object 2.25 mm high is 8.5 cm to the left of a convex lens of 5.5-cm focal length.
Answers
According to problem ( considering sign convention for convex lens)
height of object = 2.25 mm;
object distance(u)= - 8.5 cm
focal length (f)= 5.5 cm
image distance(v)= ?
Using lens formula
[tex]\begin{gathered} \frac{1}{f}=\text{ }\frac{1}{v}-\frac{1}{u}; \\ \therefore\frac{1}{5.5}=\text{ }\frac{1}{v}\text{ -}\frac{1}{-8.5} \\ \frac{1}{5.5}=\frac{1}{v}\text{ +}\frac{1}{8.5}; \\ \frac{1}{v}=\frac{1}{5.5}-\frac{1}{8.5}=\frac{3}{5.5\times8.5}=\frac{3}{46.75} \\ v=\frac{46.75}{3}=15.58\text{ cm} \end{gathered}[/tex]a) Answer is :- Location of image = 15.58cm
Using magnification formula
[tex]\begin{gathered} Magnification\text{ =}\frac{height\text{ }of\text{ }image}{height\text{ of object}}=\frac{v}{u} \\ \therefore Magnification=\text{ }\frac{15.58}{8.5}\text{ =1.83} \\ \therefore\frac{height\text{ }ofimage}{2.25}\text{ =1.83;} \\ \therefore height\text{ of image = 1.83}\times2.25=4.12\text{ mm} \end{gathered}[/tex]b) Height of image = 4.12 mm
C) Magnification= 1.83
physics need help picture below. Consider the position vs time graph for objects A and B below.
Answers
The motion of the object B is uniform and having constant velocity and, the motion of the object A is non-uniform throughout. Object A doesn't having constant velocity.
A 0.7kg ball Is thrown and accelerated by 62 m/s. How much force is applied to it? Round your answer to the nearest whole number
Answers
Given:
the mass of the ball is m=0.7 kg.
The acceleration in the ball is
[tex]a=62\text{ m/s}^2[/tex]Required: the force applied on the ball
Explanation:
The force is given by
[tex]F=ma[/tex]here, F is force, m is the mass, and a is the acceleration.
Plugging all the values in the above relation, we get:
[tex]\begin{gathered} F=0.7\text{ kg}\times62\text{ m/s}^2 \\ F=43.4\text{ N} \\ F\approx43\text{ N} \end{gathered}[/tex]Thus, the force on the ball is 43 N.
How much potential energy does a person of mass 50 kg have if they are standing at the top of a building that is 54.8 m high? Submit your answer in exponential form.
Answers
ANSWER:
2.69 x 10^4 J
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 50 kg
Height (h) = 54.8 m
The potential energy is calculated by the following formula:
[tex]E=m\cdot g\cdot h[/tex]We replace and determine the value of the energy:
[tex]\begin{gathered} E=50\cdot9.8\cdot54.8 \\ \\ E=26852=2.69\cdot10^4\text{ J} \end{gathered}[/tex]The potential energy is equal to 2.69 x 10^4 joules.
What is one way that the gravitational force between the moon and the earth would increase?a. If the size of the moon decreased while the distance remained constant.b. If the size of the moon increased while the distance remained constantc. If the size of the earth increased while the distance remained constant.d. both b and c
Answers
Answer:
d. both b and c
Explanation:
If the mass of either object increases, the gravitational attraction between them increases.
F = G m1*m2 / r^2
Where:
F: Force
G = constant
m = masses
r = distance between objects
Mass is directly proportional to the gravitational force.
You read about the Air Canada flight, where the grounds crew used the wrong units
and conversion factors to fill the plane's fuel tank leading to it running out of fuel
and having to make a crash landing.
There are many other scenarios besides filling up a gas tank that require accuracy to
prevent serious injury. Several of these were discussed in the lessons. Even you may
have to make critical measurements that could result in a big problem if not done
accurately.
Think of a different scenario (NOT FUEL TANK RELATED) where it would be
important to measure or calculate something correctly.
Describe what would be measured and why it would be important to measure with
accuracy.
Answers
Answer:
no I don't understand it oo
How much heat is necessary to change 480 g of ice at -19°C to water at 20°C? answer in:____kcal
Answers
ANSWER
[tex]18.05\text{ }kcal[/tex]EXPLANATION
The amount of heat necessary to change the ice to water is given by:
[tex]Q=m(c_{ice}\Delta T_{ice}+L+c_{water}\Delta T_{water})[/tex]where m = mass of ice = 480 g = 0.48 kg
c(ice) = specific heat capacity of ice = 2108 J/kg/K
ΔTice = change in temperature of ice = 0 - (-19) = 19 K or 19 °C
L = latent heat of fusion of ice = 33600 J/k
c(water) = specific heat capacity of water = 4186 J/kg/K
ΔTwater = change in temperature of water = 20 - 0 = 20 K or 20 °C
Therefore, the heat necessary is:
[tex]\begin{gathered} Q=0.48([2108*19]+33600+[4186*20]) \\ \\ Q=0.48(40052+33600+83720)=0.48*157372 \\ \\ Q=75538.56\text{ }J \end{gathered}[/tex]Convert this to kcal:
[tex]\begin{gathered} 1\text{ }J=\frac{1}{4184}\text{ }kcal \\ \\ 75538.56\text{ }J=\frac{75538.56}{4184}\text{ }kCal=18.05\text{ }kcal \end{gathered}[/tex]That is the answer.
How does the velocity of a falling object change with time? How would you describe the mathematical relationship between velocity and time of a falling object? Explain the answer using your data, graphs, and the kinematic equations.The image below shows the ball stopping at 2 4 6 8 and 10 and how long it took it too stop there
Answers
We will have the following:
First, we can see that the data presented is the following:
Now, from this information we can construct the following graphical relationship between time and distance drawn by the object:
From this we can infer that the relationship between time and distance is linear in nature, and the path of best descripton by linear regression is given by:
Where the approximation is:
[tex]f(x)=2.47204029x-0.4796893103[/tex]We can see the linear relationship that fits the model, from this and the fact that in kinematics the following is true:
[tex]d=v\ast t[/tex]Where "d" is the distace, "v" is the velocity and "t" the time.
From the data, the values of "x" represent the time and thus, the ratio of the function found is the velocity of the object.
In summary, the relationship is linear as per in kinematics, time and distace change proportionally in the data provided; and the approximated velocity of the object was 2.47m/s.
A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat's speed in still water is 20 miles per hour. Find the stream of the current.
Answers
The equation of the boat upstream is 20-s.
The equation downstream is 20+s.
Where s is the speed of the current.
Then, we know that the time spent is equal in both cases, and we also know that time equals distance divided by the speed. So, we express the following.
[tex]\frac{40}{20-s}=\frac{60}{20+s}[/tex]Solve for s.
[tex]\begin{gathered} 40(20+s)=60(20-s) \\ 800+40s=1200-60s \\ 60s+40s=1200-800 \\ 100s=400 \\ s=4 \end{gathered}[/tex]Therefore, the speed of the current is 4 miles per hour.
Calculate the resistance in milliOhms of a copper wire 2.84 m long and 0.04 m diameter? The resistivity of copper is 1.7 x 10^-8.
Answers
therefore, the resistivity is 0.019210 miliOhms
ExplanationStep 1
The resistance of a cylindrical segment of a conductor is equal to the resistivity of the material times the length divided by the area
[tex]\begin{gathered} R=\rho\frac{L}{A} \\ where\text{ }\rho\text{ is the resistivity of the material} \\ L\text{ is the lenght} \\ A\text{ is the area} \end{gathered}[/tex]so
Step 1
a)let
[tex]\begin{gathered} \rho=1.7*10^{-8} \\ L=2.84 \\ diameter=\text{ 0.04 m} \\ Area\text{ = unknown} \end{gathered}[/tex]bI find the area
the area of a circle is given by:
[tex]area=\pi(\frac{diameter^2}{4})[/tex]so, replace
[tex]\begin{gathered} area=\pi(\frac{0.04m^2}{4}) \\ Area=0.0025132741\text{ m}^2 \end{gathered}[/tex]c) now, we can replace the values in the formula
[tex]\begin{gathered} R=\rho\frac{L}{A} \\ R=1.7*10^{-8}\frac{2.84\text{ m}}{0.0025132741\text{ m}^2} \\ R=1.9*10^{-5}\text{ ohm} \\ to\text{ convert into miliOhms , multiply by 1000} \\ R=R=\frac{1.9\text{ohm}}{10^{5}}*\frac{1000\text{ miliOhms}}{1\text{ Ohms}}=0.01921\text{ ohms} \end{gathered}[/tex]therefore, the resistivity is 0.019210 miliOhms
I hope this helps yo u
Newton's third law of motion states that:an object in motion will tend to stay in motion unless outside forces act upon it.objects will accelerate in the direction of the force exerted upon it.an object's inertia will always remain unchanged.all forces come in pairs.
Answers
Let's state the Newton's Third Law of Motion.
The Newton's Third Law of Motion states that when two bodies interact, the forces they apply to each other are equal in magnitude and opposite in direction.
Therefore, since the forces the bodies exert equal forces in opposite direction on each other, it means that forces come in pairs.
The forces can be said to be action-reaction force pairs
Therefore, Newton's third Law of Motion states that all forces come in pairs.
ANSWER:
All forces come in pairs.
help please asap !looking for the answer to number 1 Its assigned homework for online that I don’t understand
Answers
Let's explain what a simple machine is.
Definition:
A simple machine can be defined as a mechanical device used for applying force and doing work.
It changes the direction and magnitude of force.
In a simple machine there is no combination of parts and they have very few movable parts.
Types of Simple machines:
• Screw
,• Pulley
,• Wedge
,• Wheel and axel
,• Lever
,• Inclined plane
Example of Screw is a screw.
Example of a pulley is an elevator
Example of a wedge is
How do I solve number 5? Answer hing in radians
Answers
Angular velocity is the speed at which an object travels in terms of radians or degrees, not any unit of length.
One full revolution around the sun: 2*pi radians, 365 days
2*pi/365 = 0.0172 rad/day
1 day = 86400 seconds, so to convert from rad/day to rad/s, simply divide by 86400.
0.0172 rad/day = 1.99238 * 10^-7 rad/s
A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 15.0 min at 90.0 km/h, 7.0 min at 65.0 km/h, and 40.0 min at 40.0 km/h and spends 40.0 min eating lunch and buying gas.(a) Determine the average speed for the trip. km/h(b) Determine the distance between the initial and final cities along the route. km
Answers
a)
To determine the average speed we need to determine the time the person travels and the distance. Since we need the average speed in km/h we need to convert all the times given into hours, let's do this:
[tex]\begin{gathered} 15\text{ min}\cdot\frac{1\text{ h}}{60\text{ min}}=0.25\text{ h} \\ 7\text{ min}\cdot\frac{1\text{ h}}{60\text{ min}}=\frac{7}{60}\text{ h} \\ 40\text{ min}\cdot\frac{1\text{ h}}{60\text{ min}}=\frac{2}{3}\text{ h} \end{gathered}[/tex]Now, that we have the time in each interval of the motion we can determine the distance is travel in each of them:
[tex]\begin{gathered} d_1=90(0.25)=22.5 \\ d_2=65(\frac{7}{60})=7.583 \\ d_3=40(\frac{2}{3})=26.667 \end{gathered}[/tex]This means that the total distance the person traveled is 56.8 km (we use three significant figures as shown in the problem).
Once we know this we need to find the time it takes to make this, from our previos discussion and adding the 40 minutes the person took to eat we have that the total time was 1.7 h, then the average speed is:
[tex]v=\frac{56.8}{1.7}=33.4[/tex]Therefore, the average speed on the trip was 33.4 km/h
b)
From point a we know that the distance was 56.8 km
A gymnast of mass 52.0 kg is jumping on a trampoline. She jumps so that her feet reach a maximum height of 2.98 m above the trampoline and, when she lands, her feet stretch the trampoline 70.0 cm down. How far does the trampoline stretch when she stands on it at rest? Assume that the trampoline is described by Hooke’s law when it is stretched.
Answers
When the gymnast is at 2.98 meters above the trampoline, she has potential gravitational energy in relation to the trampoline:
[tex]\begin{gathered} PE_g=m\cdot g\cdot h \\ PE_g=52\cdot9.81\cdot2.98 \\ PE_g=1520.16\text{ J} \end{gathered}[/tex]When she lands on the trampoline, all this energy will be converted into potential elastic energy:
[tex]\begin{gathered} PE_e=\frac{kx^2}{2} \\ 1520.16=\frac{k\cdot0.7^2}{2} \\ 0.49k=3040.32 \\ k=6204.73\text{ N/m} \end{gathered}[/tex]Now, to find the stretch in the trampoline when the gymnast is at rest, let's use the gymnast weight force in the formula for the force in a string:
[tex]\begin{gathered} F=k\cdot x \\ m\cdot g=k\cdot x \\ 52\cdot9.81=6204.73\cdot x \\ 510.12=6204.73 \\ x=\frac{510.12}{6204.73} \\ x=0.0822\text{ m} \end{gathered}[/tex]Therefore the trampoline stretches 8.22 cm.
Part 2/2 please.The answer to Part 1/2 is 4.5 m
Answers
In order to calculate the wave speed, we can use the formula below:
[tex]v=\lambda f[/tex]Where lambda is the wavelength and f is the frequency.
If the wavelength is 4.5 meters and the frequency is 3 Hz, the speed is:
[tex]v=4.5\cdot3=13.5\text{ m/s}[/tex]5. Law of Acceleration: A 1400 kg car is initially at rest. It is then accelerated by a 3000 N force for 10 seconds.What is the car's acceleration?How fast is it going after 10 seconds?
Answers
The force acting on the car can be expressed as,
[tex]F=ma[/tex]Plug in the known values,
[tex]\begin{gathered} 3000\text{ N=(1400 kg)a} \\ a=\frac{3000\text{ N}}{1400\text{ kg}}(\frac{1kgm/s^2}{1\text{ N}}) \\ =2.14m/s^2 \end{gathered}[/tex]Thus, the acceleration of the car is 2.14 m/s2.
The final velocity of the car is given as,
[tex]v=u+at[/tex]Substitute the known values,
[tex]\begin{gathered} v=0m/s+(2.14m/s^2)(10\text{ s)} \\ =21.4\text{ m/s} \end{gathered}[/tex]Thus, the final velocity of the car is 21.4 m/s.
An object is placed 10.0 cm in front of a thin concave lens with a focal length of 18 cm. Calculate the imageposition
Answers
The focal length of a concave lens is expressed as,
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]Here, f is the focal length, u is the distance of object and v is the distance of image.
Plug in the known values,
[tex]\begin{gathered} \frac{1}{18\text{ cm}}=\frac{1}{10.0\text{ cm}}+\frac{1}{v} \\ \frac{1}{v}=\frac{1}{18\text{ cm}}-\frac{1}{10.0\text{ cm}} \\ v=\frac{(18\text{ cm)(10.0 cm)}}{10.0\text{ cm-18 cm}} \\ =-22.5\text{ cm} \end{gathered}[/tex]Thus, the distance of image is -22.5 cm where negative sign indicates that the image will be at the same side of object.
Rearrange the formula for V1 get the formula for V1
Answers
To isolate v₁ from the given equation, subtract aΔt from both members of the equation and simplify:
[tex]\begin{gathered} v_2=v_1+a\cdot\Delta t \\ \Rightarrow v_2-a\cdot\Delta t=v_1+a\cdot\Delta t-a\cdot\Delta t \\ \Rightarrow v_2-a\cdot\Delta t=v_1 \\ \therefore v_1=v_2-a\cdot\Delta t \end{gathered}[/tex]Therefore, the formula for v₁ is:
[tex]v_1=v_2-a\cdot\Delta t[/tex]A model of a helicopter rotor has four blades, each 2.60 m long from the central shaft to the bladetip. The model is rotated in a wind tunnel at 477 rev/min. What is the radial acceleration of theblade tip expressed as a multiple of g?Answer:Choose...Next page
Answers
The radial acceleration in terms of the frequency and the radius is given by:
[tex]a_c=4\pi^2rf^2[/tex]To use this formula we need to convert the frequency to revolution per second:
[tex]477\frac{rev}{\min}\cdot\frac{1\text{ min}}{60\text{ s}}=7.95\text{ Hz}[/tex]Plugging the values we have:
[tex]\begin{gathered} a_c=4\pi^2(2.6)(7.95)^2 \\ a_c=6487.35 \end{gathered}[/tex]hence the acceleration is 6487.35 m/s^2. To get the acceleration in terms of g we divide it by its value:
[tex]\begin{gathered} a_c=\frac{6487.35}{9.8} \\ a_c=661.97 \end{gathered}[/tex]This means that the acceleration is approximately 662 times the acceleration of gravity, that is:
[tex]a_c=662g[/tex]11. [0/10 Points]
A rectangular block has dimensions 2.9 cm x 2.6 cm x 10.0 cm. The mass of the block is 605.0 g
What is the volume of the block?
4.0
DETAILS
x cm³
What is the density of the block?
4.0
X g/cm³
Submit Answer
PREVIOUS ANSWERS
Answers
Volume of rectangular block is 75.4 cm^3
Density of the rectangular block is 8.02 g/cm^3
Volume is simply defined as the space occupied within the boundaries of an object in three-dimensional space.
It is also known as the capacity of the object.
Volume of rectangular block = length× breadth× height
=2.9 cm × 2.6 cm × 10.0 cm
=75.4 cm^3
Density is defined as the substance's mass per unit of volume.
Mathematically ,density is defined as mass divided by volume.
Density of the block = Mass of block / volume of block
=605.0 g / 75.4 cm^3
=8.02 g/cm^3
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Keplers third law of planetary motion describes the relationship between the semi-major axis of a planet’s orbit and its (Fill in the blank)
Answers
Kepler's third law states that:
The squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of their orbits.
Therefore, it states the relationship between the semi major axis of the planet's orbit and its period. And the correct choice is C
After traveling for 6.0 seconds, a runner reaches a speed of 12 m/s after starting from rest. What is the runner’s acceleration?A. .5 m/s2B. 6 m/s2C. 72 m/s2D. 2 m/s2
Answers
We can find the acceleration as follows:
[tex]\begin{gathered} a=\frac{\Delta v}{\Delta t} \\ so: \\ a=\frac{12-0}{6-0} \\ a=\frac{12}{6} \\ a=2m/s^2 \end{gathered}[/tex]Answer:
D. 2 m/s^2
Win Blonehare and Kent Swimtashore are sail boating in Lake Gustastorm. Starting from rest near the shore, they accelerate with a uniform acceleration of 0.136 m/s/s. How far are they from the shore after 13.6 seconds?
Answers
The distance they from the shore after 13.6 seconds is 12.58 m.
What is distance?Distance is the total movement of an object without any regard to direction
To calculate the distance traveled from the shore, we use the formula below.
Formula:
s = ut+at²/2......... Equation 1Where:
s = Distancet = timeu = Initial velocitya = AccelerationFrom the question,
Given:
u = 0 m/s (from rest)a = 0.136 m/s²t = 13.6 sSubstitute these values into equation 1
s = (0×13.6)+(0.136×13.6²/2)s = 12.58 mHence, the distance traveled from the shore is 12.58 m.
Learn more about distance here: https://brainly.com/question/26550516
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