The muzzle speed of the dart when the spring is released is 11.3 m/s
The given parameters are
k = spring constant = 400 N/m
m = mass = 20 g = 0.02 kg
Compression = x = 8 cm = 0.08 m
According to the question,
When the dart is loaded, the potential energy is converted into kinetic energy.
Potential Energy (P.E.)
[tex]P.E. =\frac{1}{2} kx^{2}[/tex]
Putting the values,
[tex]P.E. =\frac{1}{2} *400*(0.08)^{2} = 1.28[/tex]
Now, Kinetic Energy (K.E.)
K. E. = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2} *0.02*v^{2}= 0.01 v^{2}[/tex]
Now, P.E. = K.E.
[tex]0.01 v^{2} = 1.28 \\\\v^{2} =\frac{1.28}{0.01} \\\\v^{2} = 128\\ \\v = 11.3[/tex]
Hence, the muzzle speed of the dart when the spring is released is 11.3 m/s
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13 Dion is writing a survival guide for hikers who visit the nearby state park. Since the forest in the park is so dense, it is not unusual for hikers to get lost for days at a time before Dion and the other rangers can locate them. Dion creates a list of plant life that a lost hiker could eat to keep themselves going in an emergency. Which sentence will Dion MOST likely include about wild mushrooms? A. The forest floor in our park is very dense and wet so you will not find mushrooms to eat. B. If you look closely at the base of trees or fallen logs, you may find mushrooms to eat. C. It is best not to eat mushrooms that you find because some varieties are poisonous. D. Mushrooms must be heated to kill any harmful bacteria before they can be consumed.
Answer: C
Explanation: Some varieties of mushrooms are poisonous or may make you ill.
A rocket weighing 220,000 N is taking off from Earth with a total thrust of 500,000 N at an angle of 20 degrees, as shown in the image below. What is the approximate vertical component of the net force that is moving the rocket away from Earth?A. 400,000 NB. 350,000 NC. 250,000 ND. 300,000 N
ANSWER
[tex]C.250,000N[/tex]EXPLANATION
Parameters given:
Weight of rocket, W = 220,000 N
Thrust, F = 500,000N
Angle of thrust, θ = 20°
The vertical forces acting on the rocket are the thrust (acting at an angle) and its weight. This means that the sum of vertical forces is:
[tex]F_y=F\cos \theta-W[/tex]Note: W is negative since it acts opposite the motion of the rocket.
Therefore, we have that the net vertical force (vertical component of forces) is:
[tex]\begin{gathered} F_y=500000\cos 20-220000 \\ F_y=469,846.31-220000 \\ F_y=249,846.31N \\ F_y\approx250,000N \end{gathered}[/tex]The answer is option C.
What is the momentum of a 0.144kg baseball thrown at a speed of 46m/s? Give your answer in kgm/s.
ANSWER:
6.624 kgm/s
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 0.144 kg
Speed (v) = 46 m/s
We can calculate momentum using the following formula:
[tex]\begin{gathered} P=m\cdot v \\ \\ \text{ We replacing} \\ \\ P=0.144\cdot46 \\ \\ P=6.624\text{ kgm/s} \end{gathered}[/tex]The momentum is 6.624 kgm/s
. Car D is heading East at 25 m/s and Car T is heading West at 10 m/s. What is the relative velocity between car D and car T? How would the passenger in Car D describe Car T’s motion?
Cars D and T are travelling at a relative speed of 35 m/s.
Car D's velocity is u = + 25 m/s because it is travelling east at a speed of 25 m/s. (because it is travelling in the direction of positive x-)
Additionally, Car T is moving at v = -10 m/s and travelling west at a speed of 10 m/s. (because it is travelling opposite in the direction of positive x-)
Thus, relative velocity = velocity of D - velocity of T
relative velocity = 25 - (-10)
relative velocity = 35m/s
So, the relative velocity between Car D and Car T is 35 m/s.
Car D is travelling in the direction of the position axis' positive arrow, whereas Car T is moving in the direction of the axis' negative arrow, therefore the two automobiles are moving in the opposite direction.
As a result, car D is driving more quickly than car T, and both vehicles are travelling in the opposite direction.
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The radioactive isotope 239Pu has a half-life of approximately 24100 years. After 2000 years, there are 5g of 239Pu.(1) What was the initial quantity? (Round your answer to three decimal places.) g Tries 0/99(2) How much of it remains after 20000years? (Round your answer to three decimal places.) g Tries 0/99
Using the following formula:
[tex]\begin{gathered} N(t)=N_o(0.5)^{\frac{t}{t_{1/2}}} \\ where: \\ N(t)=Remaining_{\text{ }}quantity_{\text{ }}after_{\text{ }}time_{\text{ }}t \\ N_o=Initial_{\text{ }}quantity \\ t=time_{\text{ }}in_{\text{ }}years \\ t_{1/2}=half-life=24100 \end{gathered}[/tex](1)
[tex]\begin{gathered} t=2000 \\ t_{1/2}=24100 \\ N(2000)=5g \\ so: \\ 5=N_o(0.5)^{\frac{2000}{24100}} \\ N_o=\frac{5}{(0.5)^{\frac{2000}{24100}}} \\ N_o\approx5.296 \end{gathered}[/tex](2)
Using the initial quantity calculated previously:
[tex]\begin{gathered} t=20000 \\ N(20000)=5.296(0.5)^{\frac{20000}{24100}} \\ N(20000)=2.979 \end{gathered}[/tex]Answers:
For (1): 5.296
For (2): 2.979
A force gives energy to an object and that energy can:set the object in motion.stop its motion.change the speed and direction of its motion.All of the choices are correct.
To find:
Which of the given statements is the correct answer?
Explanation:
The work done is the measure of the transfer of energy through the application of force.
The force thus applied can either change the state of motion of the object or the speed and direction of the motion.
That is, when a force is applied to an object, the force can bring the object in motion to rest or set the object in rest, into motion. Or it can also change the speed or direction of motion of the object.
Final answer:
The correct answer is option D, All of the choices are correct.
please help me work through this, thank you for your time!
Given
The equation of the height is
[tex]h(t)=-16t^2+20t+950[/tex]To find
The velocity when the stone reach the ground
Explanation
When the stone reaches the ground
[tex]\begin{gathered} h(t)=0 \\ \Rightarrow-16t^2+20t+950=0 \\ \Rightarrow16t^2-20t-950=0 \\ \Rightarrow t=\frac{20\pm\sqrt{20^2-(4\times16\times(-950)}}{2\times16} \\ \Rightarrow t=\frac{20\pm247.38}{2\times16}=8.35\text{ s} \end{gathered}[/tex]Thus the time taken to reach the ground is 8.35s . (Here only the positive value is considered)
We know the velocity is the change in distance per unit time,
Thus,
[tex]\begin{gathered} v(t)=h^{\prime}(t) \\ \Rightarrow v(t)=-32t+20 \end{gathered}[/tex]At t=8.35 s
[tex]\begin{gathered} v(8.35)=-32\times8.35+20 \\ \Rightarrow v(8.35)=-247.2\text{ feet/s} \end{gathered}[/tex]Conclusion
The velocity is -247.20 feet/s
Which of the following X-Y tables agrees withthe information in this problem?A puck moves 2.35 m/s in a -22.0° direction. A hockeystick pushes it for 0.215 s, changing its velocity to 6.42m/s in a 50.0° direction. What was the acceleration?A)хYYYC)хV 0.8802.18-0.8802.35-2.18ViVE4.134.92B)XVi2.35VE6.42a?Ax 0.2156.42> > 04.924.13a???Ax0.215ΔΧ??t0.2150.215tt 0.2150.215H
let's find the components for the initial velocity
[tex]v_ix=2.35\cdot\cos (-22)=2.18[/tex][tex]v_iy=2.35\cdot\sin (-22)=-0.88_{}[/tex]then for the final velocity
[tex]v_fx=6.42\cdot\cos (50)=4.13[/tex][tex]v_fy=6.42\cdot\sin (50)=4.92[/tex]The table that agrees with the problem is A)
Then for the acceleration, we will use the next formula
[tex]a=\frac{v_f-v_i_{}}{t}[/tex]for the acceleration in x
[tex]a_x=\frac{v_fx-v_ix}{t}=\frac{4.13-2.18}{0.215}=9.06m/s^2[/tex]then for the acceleration in y
[tex]a_y=\frac{v_fy-v_iy}{t}=\frac{4.92-(-0.88)}{0.215}=26.97m/s^2[/tex]then we calculate the magnitude
[tex]a=\sqrt[]{9.06^2+26.97^2}=28.45\text{ m/s}^2[/tex]How much work is done to transfer 0.15 C of charge through a potential difference of 9.0 V?
Answer:
6.075J
Explanation:
The formula is 0.5X0.15X9X9. Hal of CxVxV. This results in 6.075J
why short circuit are dangrous
Short circuits occur when a high-volume electrical current is passed through a low-resistance path that isn't meant to carry electricity. Short circuits occur when a live wire makes contact with a conductive object that it should not have.
The most dangerous aspect of a short circuit is that it will cause a sudden and dramatic change in electrical resistance. Electrical systems can cause a large amount of resistance and will be constantly sending electricity to different areas when appliances are plugged in or outlets are being used. This has the possibility to cause bodily harm and appliance damage to those who are using electrical appliances at the time a short circuit happens, and can also cause structural damage to your home through fires and burns
Wires with faulty or unstable connections will create a situation in which the electrical current can take a detour down an unintended path if it is the one of least resistance. This can be unpredictable, and the route that the current travels could be through flammable materials or a human being.
Short circuits are liable to cause fires, electrical burns, and electrocution. Stray electrical currents can also cause considerable damage to appliances and a home’s electrical system.
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I need help filling out the acceleration for gravity due to the numbers in this chart
In order to calculate the acceleration of gravity, we can use the free fall formula:
[tex]d=\frac{gt^2}{2}\to g=\frac{2d}{t^2}[/tex]Where d is the distance travelled and t is the time.
So for each distance in the chart, we have:
[tex]\begin{gathered} d=0.2\colon \\ g=\frac{0.4}{0.184^2}=11.81 \\ \\ d=0.4\colon \\ g=\frac{0.8}{0.258^2}=12.02 \\ \\ d=0.6\colon \\ g=\frac{1.2}{0.4^2}=7.5 \\ \\ d=0.8\colon \\ g=\frac{1.6}{0.236^2}=28.73 \\ \\ d=1\colon \\ g=\frac{2}{0.33^2}=18.37 \end{gathered}[/tex](All gravity units in m/s²)
Now, calculating the average, we have:
[tex]g=\frac{11.81+12.02+7.5+28.73+18.37}{5}=15.69[/tex]Finally, calculating the percent error, we have:
[tex]\text{Percent error}=\frac{|9.81-15.69|}{9.81}=0.5994=59.94\text{\%}[/tex]URGENT!! ILL GIVE
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Answer:
Thermal energy flows from the flask on the left to the flask on the right
Explanation:
As the flask on the left is hot water the temperature of it will be higher meaning that the metal bar will conduct the heat at a faster rate than the flask on the left. Due to this, the metal bar will gradually begin to increase in temperature slowly throughout the whole bar. This means that the heat will travel through the bar reaching tho whole flask.
Hope this helps, have a great day!
How far did a runner travel if they ran 3 mph for 30 minutes?
Given:
Speed = 3 mph
Time = 30 minutes
Let's determine how far(distance) the runner travelled.
To find the distance, apply the formula:
[tex]Dis\tan ce=speed\ast time[/tex]Since the time is in minutes, let's convert to hours by dividing the time by 60 minutes.
Thus, we have:
[tex]\begin{gathered} D=3\ast\frac{30}{60} \\ \\ D=3\ast\frac{1}{2} \\ \\ D=1.5\text{ m} \end{gathered}[/tex]Therefore, the runner travelled 1.5 miles
A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the force of friction is not significant.
The given problem can be exemplified in the following diagram:
Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:
[tex]\Sigma F=ma[/tex]Replacing the values:
[tex]mg\sin 40=ma[/tex]We may cancel out the mass:
[tex]g\sin 40=a[/tex]Using the gravity constant as 9.8 meters per square second:
[tex]9.8\frac{m}{s^2}\sin 40=a[/tex]Solving the operations:
[tex]6.3\frac{m}{s^2}=a[/tex]Therefore, the acceleration is 6.3 meters per square second.
The Taipei Tower is a 508 meter, 101 story skyscraper. If you were to toss a orange of 0.13 kg off the top, how much kinetic energy would it have when it hits the sidewalk? Ignore air resistance .
Answer:
647.19 J
Explanation:
By the conservation of energy, the potential energy when you toss the orange is converted to kinetic energy when it hits the sidewalk, so
Ef = Ei
KE = PE
KE = mgh
Where m is the mass, g is 9.8 m/s², and h is the height of the Tower. Replacing m by 0.13 kg and h by 508 meters, we get
KE = (0.13kg)(9.8 m/s²)(508 m)
KE = 647.19 J
So, the orange would have 647.19 J of kinetic energy when it hits the sidewalk.
Question 2
An object has a momentum of 500 kg mls. If its mass is 20 kg, its speed must be
O 25 m/sO 20 m/s50 m/sO 500
Given:
The momentum of the object is p = 500 kg m/s
The mass of the body is m = 20 kg
To find the speed of the object.
Explanation:
Speed can be calculated by the formula
[tex]\begin{gathered} p=mv \\ v=\frac{p}{m} \end{gathered}[/tex]On substituting the values, the speed of the object will be
[tex]\begin{gathered} v=\frac{500}{20} \\ =25\text{ m/s} \end{gathered}[/tex]Thus, the speed of the object is 25 m/s
In the table below place the different forms of electromagnetic radiation and visible light in order of lowest energy to highest energy
Answer:
Step-by-step explanation:
The electromagnetic spectrum is the range of all types of EM radiation. I will write the electromagnetic spectrum from lowest energy/longest wavelength to highest energy/shortest wavelength:
*Visible light refers to the visible region of the electromagnetic spectrum, that is the range of wavelengths that trigger brightness and color perception in humans. It lies between Ultraviolet and Infrared radiation
Our galaxy, the Milky Way, contains approximately 4.0 x 1011 stars with anaverage mass of 2.0 X 1030 kg each. How far away is the Milky Way from ournearest neighbor, the Andromeda Galaxy, if Andromeda contains roughly thesame number of stars and attracts the Milky Way with a gravitational force of2.4 x 1030 N?
We will have the following:
First, we remember:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Then, from the problem we will have that:
[tex]2.4\ast10^{30}N=\frac{G(2.0\ast10^{30}kg)(2.0\ast10^{30}kg)}{r^2}[/tex][tex]\Rightarrow r^2=\frac{G(2.0\ast10^{30}kg)^2}{2.4\ast10^{30}N}\Rightarrow r\approx1.1116666667\ast10^{20}m[/tex]So, the Andromeda galaxy is approximately 1.1*10^20 meters from the milky way.
The acceleration of the Andromeda galaxy towards the milky way is:
[tex]2.4\ast10^{30}N=(2.0\ast10^{30}kg)a\Rightarrow a=1.2m/s^2[/tex]So, the acceleration towards the milky ways is 1.2m/s^2.
Calculate the Mach number for sound given the temperature and the speeda. 332 m/s at 30°Cb. 340 m/s at -10°Cc. 6000 km/h at 13°Cd. 6000 km/h at -13°0
The Mach number indicates how many times a speed is greater than the speed of sound.
So, to find the Mach number, we just need to divide the speed by the speed of sound.
a) at 30°C, the speed of sound is 349 m/s, so we have:
[tex]\frac{332}{349}=0.95[/tex]b) at -10°C, the speed of sound is 325 m/s, so we have:
[tex]\frac{340}{325}=1.05[/tex]c) at 13°C, the speed of sound is 339 m/s. First, let's convert the speed from km/h to m/s:
[tex]\begin{gathered} 6000\text{ km/h}=\frac{6000}{3.6}\text{ m/s}=1666.67\text{ m/s} \\ \frac{1666.67}{339}=4.92 \end{gathered}[/tex]d) at -13°C, the speed of sound is 323 m/s, so we have:
[tex]\frac{1666.67}{323}=5.16[/tex]URGENT!! ILL GIVE
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Answer: Your answer will be D
Explanation:
Scenario 3: a clock attached to the wall.
1. State the object stationary.
2. State the outside, external, unbalanced force acting on the object.
answer: this object is stationary because it is in rest position until or unless the external force act on it we can say this by netwons 1st law
and the stationaty object has relatively zero velocity.
netwons 1st law: Newton's first law states that unless compelled to change its state by the action of an external force, every object will remain at rest or in uniform motion in a straight line.
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2. here the object has contact with the wall hence there is a normal force between clock and wall
the clock is at rest this means that the forcee is balanced and normal force is balanced by gravitational force .the normal force is always perpendicular to the surface of the object.
there is no external force is acting on the clock.
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What is a limitation to using nuclear fusion for energy?
Answer:
Explanation:
The radiation of components in a fusion reactor is not much enough for the materials to be reused or recycled within centuries
Write 8.0090 x 10^2 in standard form . (Example of standard form: 4550)Blank 1:
We are given the following number in scientific notation:
[tex]8.0090\times10^2[/tex]To convert this number into standard form we will move the decimal point a number of times equal to the exponent of the ten. Since the exponent is positive we will move the point to the right, like this:
[tex]8.0090\times10^2=800.90[/tex]Therefore, the standard form of the number is 800.90
Two equally charged, 3.699 g spheres are placed with 3.592 cm between their centers. When released, each begins to accelerate at 297.727 m/s2. What is the magnitude of the charge on each sphere? Express your answer in microCoulombs.
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 speed (m/s) of a caterpillar traveling 45m along a sidewalk in 50hr?
We will have the following:
[tex]v=\frac{45m}{50h}\cdot\frac{1h}{60s}\Rightarrow v=\frac{3}{200}m/s[/tex][tex]\Rightarrow v=0.015m/s[/tex]So, the speed is 0.015 m/s.
A dog barks with an intensity level of 80 decibels. Two barking dogs produce what intensity level?
We are given that a dog barks with an inetensity level of 80 dB and asked to find out the intensity level produced by two barking dogs.
The combined intensity level of both dogs is the sum of each dog's intensity level.
[tex]\begin{gathered} I_{net}=I_1+I_2 \\ I_{net}=2I_{} \\ I_{net}=2\cdot I_0\cdot10^{\frac{\beta}{10}}_{} \end{gathered}[/tex]Where β is 80 dB and I0 is the reference intensity (1x10^-12 W/m^2)
[tex]\begin{gathered} I_{net}=2\cdot10^{-12}\cdot10^{\frac{80}{10}}_{} \\ I_{net}=2\cdot10^{-12}\cdot10^8_{} \\ I_{net}=2\cdot10^{-12+8} \\ I_{net}=2\cdot10^{-4} \end{gathered}[/tex]The net β is given by
[tex]\begin{gathered} \beta_{\text{net}}=10\log (\frac{I_{net}}{I_0}) \\ \beta_{\text{net}}=10\log (\frac{2\cdot10^{-4}}{10^{-12}}) \\ \beta_{\text{net}}=10\log (2\cdot10^{-4+12}) \\ \beta_{\text{net}}=10\log (20^8) \\ \beta_{\text{net}}=10(8.301) \\ \beta_{\text{net}}=83\; dB \end{gathered}[/tex]Therefore, two barking dogs produce 83 dB intensity level.
How much force must I lift with to lift a 30kg object off the ground?
The force required to lift the object is 294 N.
Given data:
The mass of object is m=30 kg.
The force applied to lift an object will be equal to the weight of the object.
The weight of the object can be calculated as,
[tex]\begin{gathered} W=mg \\ W=(30)(9.8) \\ W=294\text{ N} \end{gathered}[/tex]The weight of the object is 294 N; therefore, the force required to lift the object is 294 N.
If a ball is thrown with an initial horizontal velocity of 2.3m/s, from a tall building, how far away from thebuilding does the ball land if it takes 4s to land?Referring to the ball above, how tall is the building? (2 sig figs)
Given,
The initial velocity of the ball, u=2.3 m/s
The time interval in which the ball hits the ground, t=4 s
As the ball was thrown horizontally, the angle of projection is θ=0°
The distance traveled by the ball in the horizontal direction is given by,
[tex]\begin{gathered} x=u_xt \\ \Rightarrow x=u\cos \theta\times t \end{gathered}[/tex]Where uₓ is the x-component of the initial velocity.
On substituting the known values,
[tex]\begin{gathered} x=2.3\times\cos 0^{\circ}\times4 \\ =9.2\text{ m} \end{gathered}[/tex]Therefore the distance traveled by the ball in the x-direction is 9.2 m. That is the ball landed 9.2 meters away from the building.
Applying the equation of the motion in the y-direction,
[tex]\begin{gathered} y=y_0+u_yt+\frac{1}{2}gt^2 \\ =y_0+u\sin \theta\times_{}t+\frac{1}{2}gt^2 \end{gathered}[/tex]Where y is the final height of the ball which is zero meters, y₀ is the initial height of the ball, i.e., the height of the building, uy is the y-component of the initial velocity.
Let us consider that the upward direction is positive while the downward direction is negative. This makes the acceleration due to gravity, g, a negative value, and the height of the building a positive value.
On substituting the known values,
[tex]\begin{gathered} 0=y_0+2.3\times\sin (0^{\circ})\times4+\frac{1}{2}\times-9.8\times4^2 \\ \Rightarrow-y_0=0-4.9\times4^2 \\ y_0=78.4\text{ m} \\ \approx78\text{ m} \end{gathered}[/tex]Therefore the height of the building is 78 m
What is the resistance of a lamp which draws 250 milliamperes when connected toa 12.6 volt battery
V = I*R
R = V / I
250 ma = 0.25 A
R = 12.6 / 0.25
R =50.4 ohm
When the length of a pendulum is doubled, its frequency will be cut in half. Is this true or false?
The period T of a pendulum with length L is given by the formula:
[tex]T=2\pi\cdot\sqrt[]{\frac{L}{g}}[/tex]Where g is the acceleration of gravity:
[tex]g=9.81\frac{m}{s^2}[/tex]If the length of the pendulum is doubled, the period will be:
[tex]\begin{gathered} T^{\prime}=2\pi\cdot\sqrt[]{\frac{2L}{g}} \\ =\sqrt[]{2}\times2\pi\cdot\sqrt[]{\frac{L}{g}} \\ =\sqrt[]{2}\times T \end{gathered}[/tex]On the other hand, the frequency is the reciprocal of the period. Then:
[tex]\begin{gathered} f=\frac{1}{T} \\ f^{\prime}=\frac{1}{T^{\prime}}=\frac{1}{\sqrt[]{2}\times T}=\frac{1}{\sqrt[]{2}}\times\frac{1}{T}=\frac{1}{\sqrt[]{2}}\times f \end{gathered}[/tex]Then, if the length of the pendulum is doubled, the frequency is cut by a factor of 1/√2:
[tex]f^{\prime}=\frac{1}{\sqrt[]{2}}\times f[/tex]This is not the same as if the frequency was cut by a factor of 1/2.
Therefore, the statement is:
[tex]\text{False}[/tex]