Answer:
[tex]\tau = 6.35 s[/tex]
A = 3.2 cm
Explanation:
given,
Amplitude of the vibration of the top of lamppost = 6.4 cm
after 8.8 s the amplitude is 1.6 cm
time constant for damping of oscillation = ?
Amplitude at 4.4 second= ?
using formula
[tex]A = A_0e^{-\dfrac{T}{\tau}}[/tex]
[tex]1.6 = 6.4\times e^{-\dfrac{8.8}{\tau}}[/tex]
taking ln both side
[tex]ln (1.6) = ln(6.4)-\dfrac{8.8}{\tau}[/tex]
[tex]\tau = 6.35 s[/tex]
[tex]A = A_0e^{-\dfrac{T}{\tau}}[/tex]
[tex]A =6.4\times e^{-\dfrac{4.4}{6.35}}[/tex]
A = 3.2 cm
The amplitude of the oscillation 4.4 s after the quake stopped is:
A= 3.2cm
This refers to the maximum length which an object can attain when it oscillates or vibrates.
Given that the amplitude from the top of the lamppost is given as:
6.4cm
We need to find the time constant for the damping of the oscillation.
We use the formula
[tex]A= Aoe[/tex]^-T/t
We would input the values
1.6 = 6.4 x [tex]e[/tex]^-8.8/t
Taking Ln on both sides
Ln(1.6) = ln(6.4)^-8.8/t
t=6.35₈
A= [tex]Aoe[/tex]^T/t
A= 3.2cm
Read more about amplitude here:
https://brainly.com/question/25699025
Find the volume of a slab whose dimensions are L=45.24 cm, W=23.981cm, H=2.33002 cm in proper significant digits
Answer:
The volume of the slab is 2527.84 cm
Explanation:
Only the first two decimals are significant digits because so are the data that the problem gives us.
The top of a cliff is located a distance (H) above the ground. At a distance of H/2 a bird nest sits on a branch that juts out from the wall of the cliff. One (bad) child throws a rock up from the ground as a second (bad) child throws a rock down from the top of the cliff. Which rock hits the nest first if they are thrown at the same velocity, at the same time? (ignore air resistance)
Answer:
The rock thrown from the top of the cliff.
Explanation:
This is more of a conceptual question. The rock thrown from the top of the cliff will be accelerated downwards, that means, accelerated towards the bird nest, will the rock thrown from the bottom will be accelerated downwards too, but, in this case, this means that it will be accelerated against the direction of the bird nest.
The rock thrown downwards from the top of the cliff will hit the bird nest located at H/2 first, as it is continuously accelerated by gravity, unlike the upward-thrown rock which initially decelerates.
Which Rock Hits the Nest First?
When considering the rocks thrown by the two children - one upwards from the ground and one downwards from the top of the cliff - the rock that hits the bird nest located at H/2 first would be the one thrown downwards. This is based on the principles of kinematics in physics, which describe the motion of objects without considering the forces that cause the motion. The initial velocities of both rocks are the same in magnitude but opposite in direction; however, gravity only decelerates the upward-thrown rock and accelerates the downward-thrown rock, resulting in the latter reaching the nest quicker.
The rock thrown upwards will slow down as it reaches a height of H/2 and has to combat gravity's pull, which is not the case for the downward-throwing rock. Although both rocks are subjected to the same acceleration due to gravity, the downward-thrown rock will cover the distance to the nest faster due to its uninterrupted acceleration. Ignoring air resistance, we don't have to consider any opposing forces which might otherwise affect the time it takes for each rock to reach the nest.
A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3–36)? Why are there two different angles? Sketch the two trajectories.
Answer:
17.72° or 72.28°
Explanation:
u = 6.5 m/s
R = 2.5 m
Let the angle of projection is θ.
Use the formula for the horizontal range
[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]
[tex]2.5=\frac{6.5^{2}Sin2\theta }{9.8}[/tex]
Sin 2θ = 0.58
2θ = 35.5°
θ = 17.72°
As we know that the range is same for the two angles which are complementary to each other.
So, the other angle is 90° - 17.72° = 72.28°
Thus, the two angles of projection are 17.72° or 72.28°.
A flea walking along a ruler moves from the 45 cm mark to the 27 cm mark. It does this in 3 seconds. What is the speed? What is the velocity?
(Define increasing numbers to be the positive direction and decreasing numbers to be the negative direction.)
Answer:
Speed= 6cm/s and velocity= 6cm/s in the negative direction
Explanation:
the change in position is from 45cm to 27 cm (moving towards the negative x direction)
[tex] \Delta x = 45 cm - 27 cm = 18 cm[/tex]
And the change in time:
[tex] \Delta t= 3 s[/tex]
Now we must define the difference between speed and velocity:
Speed is a scalar quantity, which means that it is a number. Velocity is also a number but you must also indicate the direction of the movement.
Thus, the speed is:
[tex] speed= \Delta x/ \Delta t = 18cm/3s=6cm/s[/tex]
An the velocity is:
6cm/s in the negative direction
If the electron has a speed equal to 9.10 x 10^6 m/s, what is its wavelength?
Answer:
[tex]\lambda=8.006\times 10^{-11}\ m[/tex]
Explanation:
Given that,
The speed of an electron, [tex]v=9.1\times 10^6\ m/s[/tex]
We need to find the wavelength of this electron. It can be calculated using De -broglie wavelength concept as :
[tex]\lambda=\dfrac{h}{mv}[/tex]
h is the Planck's constant
[tex]\lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 9.1\times 10^6}[/tex]
[tex]\lambda=8.006\times 10^{-11}\ m[/tex]
So, the wavelength of the electron is [tex]8.006\times 10^{-11}\ m[/tex]. Hence, this is the required solution.
Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart. What is the electric field (in N/C) at the center of the region between the plates? (Enter the magnitude.)
Answer:
E = 5291.00 N/C
Explanation:
Expression for capacitance is
[tex]C = \frac{\epsilon A}{d}[/tex]
where
A is area of square plate
D = DISTANCE BETWEEN THE PLATE
[tex]C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}[/tex]
[tex]C = 24.06 \epsilon[/tex]
[tex]C = 24.06\times 8.854\times 10^{-12} F[/tex]
[tex]C =2.1\times 10^{-10} F[/tex]
We know that capacitrnce and charge is related as
[tex]V = \frac{Q}{C}[/tex]
[tex]= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}[/tex]
v = 9.523 V
Electric field is given as
[tex]E = \frac{V}{d}[/tex]
= [tex]\frac{9.52}{1.8*10^{-3}}[/tex]
E = 5291.00 V/m
E = 5291.00 N/C
Two point charges of -7uC and 4uC are a distance of 20
cmapart. How much work does it take to move these charges outto a
separation of 90 cm apart @ a constant speed?
Answer:
Approximately 0.979 J.
Explanation:
Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy [tex]\mathrm{EPE}[/tex].
[tex]\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}[/tex],
where
The coulomb's constant [tex]k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2}[/tex],[tex]q_1[/tex] and [tex]q_2[/tex] are the sizes of the two charges, and[tex]r[/tex] is the separation of (the center of) the two charges.Note that there's no negative sign before the fraction.
Make sure that all values are in SI units:
[tex]q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C[/tex];[tex]q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C[/tex];Initial separation: [tex]\rm 20\; cm = 0.20\; cm[/tex];Final separation: [tex]\rm 90\; cm = 0.90\; cm[/tex].Apply Coulomb's law:
Initial potential energy:
[tex]\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}[/tex].
Final potential energy:
[tex]\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}[/tex].
The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.
[tex]\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}[/tex].
Answer:
Answer is c
Explanation:
trust me.
How many nanoseconds does it take light to travel a distance of 3.80 km in vacuum? Express your answer numerically in nanoseconds.
Answer:
t=12600ns
Explanation:
We use the relation between distance and velocity to solve this problem:
d=v*t
d=3.8km=3.8*10^3m
v=3*10^8 m/s^2 light's speed
we solve to find t:
t=d/v=(3.8*10^3)/(3*10^8)=1.26*10^(-5)s=12600*10^(-9)s=12600ns
Answer:
1.27 × 10⁴ ns
Explanation:
Given data
Distance (d): 3.80 km = 3.80 × 10³ mSpeed of light (v): 3.00 × 10⁸ m/sWe can find the time (t) that it takes to the light to travel 3.80 km using the following expression.
v = d/t
t = d/v
t = (3.80 × 10³ m)/(3.00 × 10⁸ m/s)
t = 1.27 × 10⁻⁵ s
We know that 1 s = 10⁹ ns. Then,
1.27 × 10⁻⁵ s × (10⁹ ns/1s) = 1.27 × 10⁴ ns
The Sun appears to move relative to the stars. For example, right now the Sun is in front of the constellation Leo, but in a month it will have moved to Virgo. Using the length of the year (365 days), calculate how fast (degrees/day) this motion is.
Answer:
0.98°
Explanation:
The Sun appears to move across various constellation because of the Earth's revolution around it. This apparent motion is essentially same as the actual motion of the Earth. The orbit of Earth around the Sun is almost circular.
Time taken to complete one revolution = 365 days
Degrees traveled in 365 Days = 360°
The motion per day is
[tex]=\frac{360}{365}[/tex]
= 0.98° per day
Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at the angle 75° above horizontal. What ball will have bigger speed at the highest point of their trajectory? O Ball A O Ball B O Their will have equal speeds O Impossible to answer without knowing their actual initial speeds.
Answer:
Ball A
Explanation:
Let the initial speed of the balls be u .
Angle of projection for ball A = 20°
Angle of projection for ball B = 75°
As we know that at highest point, the ball has only horizontal speed which always remains constant throughout the motion because the acceleration in horizontal direction is zero.
Speed of ball A at highest point = u Cos 20° = 0.94 u
Speed of ball B at highest point = u Cos 75° = 0.26 u
So, the ball A has bigger speed than B.
A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across from the original and 2.5 meters lower. How far (in m) will they fall in that time? (use g = 10 m/s^2)
Answer:
They will meet at a distance of 7.57 m
Given:
Initial velocity of policeman in the x- direction, [tex]u_{x} = 10 m/s[/tex]
The distance between the buildings, [tex]d_{x} = 2.0 m[/tex]
The building is lower by a height, h = 2.5 m
Solution:
Now,
When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.
Thus
From the second eqn of motion, we can write:
[tex]h = ut + \frac{1}{2}gt^{2}[/tex]
[tex]h = \frac{1}{2}gt^{2}[/tex]
[tex]2.5 = \frac{1}{2}\times 10\times t^{2}[/tex]
t = 0.707 s
Now,
When the policeman was chasing across:
[tex]d_{x} = u_{x}t + \frac{1}{2}gt^{2}[/tex]
[tex]d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m[/tex]
The distance they will meet at:
9.57 - 2.0 = 7.57 m
The policeman will fall a distance of 0.2 meters while jumping to the next building, calculated using the kinematic equation for vertical displacement under gravity.
Explanation:The question relates to the topic of physics, specifically to the calculation of the distance fallen by an object under the influence of gravity, which is a fundamental kinematic concept. We want to find out how far the policeman will fall when jumping across to another building. We know that he's attempting to jump across a gap that is 2 meters wide, and the other building is 2.5 meters lower. Given that the acceleration due to gravity is approximated as 10 m/s² and assuming no air resistance, we can calculate the time it will take for the policeman to travel horizontally the 2 meters distance. Since the policeman's horizontal velocity is 10 m/s, the time to cross 2 meters is 2 meters / 10 m/s = 0.2 seconds.
Next, we need to calculate the vertical displacement during this time. Using the formula for vertical displacement s under constant acceleration a due to gravity, with initial vertical velocity u = 0 (since the policeman is only moving horizontally initially), s = u⋅t + 0.5⋅a⋅t². Substituting the values, we get s = 0⋅(0.2 s) + 0.5⋅(10 m/s²)⋅(0.2 s)². Simplifying this, we find s = 0.5⋅(10)⋅(0.04) = 0.2 meters.
Therefore, the policeman will fall a distance of 0.2 meters during the time it takes to jump across to the other rooftop.
The moment an object in freefall hits the ground, its final velocity will be: a. Zero
b. Greater than the initial
c. Less than the initial
d. Constant
Answer:
Option b
Explanation:
An object is said to fall freely when there is no force acting on the object other than the gravitational force. Thus the acceleration of the object is solely due to gravity and no other acceleration acts on the body.
Also the initial velocity of the body in free fall is zero and hence less than the final velocity.
As the body falls down closer to earth, it experiences more gravitational pull and the velocity increases as it falls down and the moment it touches the ground the period of free fall ends at that instant.
Thus the final velocity of an object in free fall is not zero because the final velocity is the velocity before coming in contact with the ground.
You mount a ball launcher to a car. It points toward the back of the car 39 degrees above the horizontal and launches balloons at 9.7 m/s. When you drive the car straight ahead at 2.2 m/s, what is the initial angle in degrees of the ball's trajectory as seen by someone standing on the ground?
Answer:[tex]\theta =32.08 ^{\circ}[/tex]
Explanation:
Given
Ball launcher is mounted on car at angle of [tex]39^{\circ}[/tex]
launching velocity is u=9.7 m/s
Speed of car =2.2 m/s
So in horizontal component of balloon speed of car will be added
Thus [tex]u_x=9.7cos39+2.2=7.53+2.2=9.73 m/s[/tex]
[tex]u_y=9.7sin39=6.10 m/s[/tex]
therefore Appeared trajectory angle is
[tex]tan\theta =\frac{6.10}{9.73}=0.627[/tex]
[tex]\theta =32.08 ^{\circ}[/tex]
A ball of charge containing 0.47 C has radius 0.26 m. a) What is the electric field strength at a radius of 0.18 m?
b) What is E at 0.50 m?
c) What is E at infinity?
d) What is the volume charge density?
Answer:
a) 4.325*10^10 V/m
b) 1.689*10^10 V/m
c) 0
d) 6.384 C/m^3
Explanation:
Hello!
The electric field of a sphere of uniform charge is a piecewise function, let a be the raius of the sphere
For r<a:
[tex]E = kQ\frac{r}{a^{3}}[/tex]
For r>a:
[tex]E=kQ/r^{2}[/tex]
Since a=0.26m and k= 8.987×10⁹ N·m²/C²
a)
[tex]E= 8.987\times10^{9}\frac{0.47C \times0.18m}{(0.26m)^{3}}[/tex]
E=4.325*10^10 V/m
b)
[tex]E= 8.987\times10^{9}\frac{0.47C}{(0.5m)^{2}}[/tex]
E=1.689*10^10 V/m
c)
Since r --> ∞ 1/r^2 --> 0
E(∞)=0
d)
The charge density may be obtained dividing the charge by the volume of the sphere:
[tex]\rho = \frac{Q}{V} =\frac{0.47C}{\frac{4}{3} \pi (0.26m)^{3}}=6.384 C/m^{3}[/tex]
Represent 8953 ms with Sl units having an appropriate prefix Express your answer to four significant figures and include the appropriate units. PÅ ROS? 8953 ms = 8.953 . 10 ms
Answer:
8.953 s
Explanation:
Here a time of 8953 ms is given.
Some of the prefixes of the SI units are
mili = 10⁻³
micro = 10⁻⁶
nano = 10⁻⁹
kilo = 10³
The number is 8953.0
Here, the only solution where the number of significant figures is mili
1 milisecond = 1000 second
[tex]1\ second=\frac{1}{1000}\ milisecond[/tex]
[tex]\\\Rightarrow 8953\ milisecond=\frac{8953}{1000}\ second\\ =8.953\ second[/tex]
So 8953 ms = 8.953 s
An object carries a charge of -6.1 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from the first to the second object so that both objects have the same charge?
To equalize the charges of two objects, the number of electrons transferred from the first object to the second is calculated by dividing the total charge needed (-4.1 x 10^-6 C) by the charge per electron (-1.602 x 10^-19 C/e).
Explanation:To balance the charges of the two objects, we must first calculate how many electrons represent the disparity in charge between the two objects. This disparity is of 4.1 µC (-6.1 µC - (-2.0 µC) = -4.1 µC). When we convert this to the fundamental unit of charge (Coulombs), we get -4.1 x 10-6 C.
The charge on an electron is -1.602 x 10-19 C. Therefore, the number of electrons that must be transferred to balance the charges can be found by dividing the total charge needed by the charge per electron. That gives us -4.1 x 10-6 C ÷ -1.602 x 10-19 C/e
The result of this calculation denotes how many electrons must be transferred from the first object to the second to make their charges equal.
Learn more about Charge Balance here:https://brainly.com/question/13344069
#SPJ12
The cubit is an ancient unit of length based on the distance between the elbow and the tip of the middle finger of the measurer. Assume that the distance ranged from 43 to 53 cm, and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 5.0 cubits and a diameter of 3.0 cubits. For the stated range, what are the lower values for (a) the cylinder's length in meters, (b) the cylinder's length in millimeters, and (c) the cylinder's volume in cubic meters? What are the upper values for (d) the cylinder's length in meters, (e) the cylinder's length in millimeters, and (f) the cylinder's volume in cubic meters?
Answer:
a) 215 cm
b) 1290 mm
c) 2.81 m³
d) 265 cm
e) 1590 mm
f) 5.26 m³
Explanation:
Lower value
1 cubit = 43 cm
Length of cylinder = 5 cubits
So,
a) Length of cylinder = 5×43 = 215 cm
Diameter of cylinder = 3 cubit
1 cubit = 43 cm = 430 mm
b) Diameter of cylinder = 3×430 = 1290 mm
Radius of cylinder = 129/2 = 64.5 cm
Volume of cylinder
[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.645^2\times 2.15\\\Rightarrow v=2.81 m^3[/tex]
c) Volume of cylinder = 2.81 m³
Upper value = 53 cm
Length of cylinder = 5 cubits
So,
d) Length of cylinder = 5×53 = 265 cm
Diameter of cylinder = 3 cubit
1 cubit = 53 cm = 530 mm
e) Diameter of cylinder = 3×530 = 1590 mm
Radius of cylinder = 159/2 = 79.5 cm
Volume of cylinder
[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.795^2\times 2.65\\\Rightarrow v=5.26 m^3[/tex]
f) Volume of cylinder = 5.26 m³
Because Earth's rotation is gradually slowing, the length of each day increases: The day at the end of 1.0 century is 1.0 ms longer than the day at the start of the century. In 43 centuries, what is the total of the daily increases in time (that is, the sum of the gain on the first day, the gain on the second day, etc.)?
Final answer:
To find the total increase in the length of each day over 43 centuries, we use the sum of an arithmetic series: S = n/2(2a1 + (n-1)d). With 4300 days in 43 centuries, an initial increase of 0 ms, and a daily increase of 0.002 ms, the total increase is 18471.9 milliseconds.
Explanation:
The length of the day increases because Earth's rotation is gradually slowing, primarily due to the friction of the tides. To calculate the total increase in the length of each day over 43 centuries, given that the length of a day at the end of a century is 1 ms longer than the day at the start of the century, we use the arithmetic progression formula where the total sum (S) is equal to n/2 times the first term (a1) plus the last term (an). In this case, the difference between each day (common difference, d) is constant at 0.002 ms.
The first term (a1) is 0 ms, as there is no increase on the first day of the first century, and the last term (an) after 43 centuries will be 43 ms, since we're told that each century contributes an additional 1 ms to the length of a day. So, for 43 centuries, we have n=4300 days (number of days in 43 centuries), a1=0 ms, an=43 ms, and d=0.002 ms/day.
Using the formula for the sum of an arithmetic series, S = n/2(2a1 + (n-1)d), we get the following:
S = 4300/2 [2(0) + (4300-1)(0.002 ms)]
S = 2150 [0 + 4299(0.002 ms)]
S = 2150 x 8.598 ms
S = 18471.9 ms
Therefore, over 43 centuries, the total increase in the length of each day adds up to 18471.9 milliseconds.
A simple pendulum of mass 0.50 kg and length 0.75 m is held still and then released from an angle of 10° at t = 0. At what time does the pendulum first reach its maximum kinetic energy? (a) 0.43 s, (b) 0.53 s, (c) 1.1 s, (d) 1.7 s, (e) 3.4 s. AD string on a guitar is 648 mm long, has a mass of 1.92 g and a fundamental frequency of 147 Hz. How far from the end of the string is the fret associated with the G note, which has a frequency of 110 Hz? (a) 81.5 mm, (b) 163 mm, (c) 245 mm, (d) 326 mm, (e) 408 mm.
Answer:
Explanation:
The motion of the pendulum will be be SHM with time period equal to
T = [tex]2\pi\sqrt{\frac{l}{g} }[/tex]
l = .75 m , g = 9.8
T = [tex]2\pi\sqrt{\frac{.75}{9.8} }[/tex]
T = 1.73 s .
Time to reach the point of maximum velocity or maximum kinetic energy
= T /4
= 1.73 /4
= 0.43 s
For notes on Guitar , The formula is
n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]
n is fequency of the note , T is tension of string , m is mass per unit length of string , l is length of string.
For fundamental note , l = .648 m and f = 147 Hz
147 = [tex]\frac{1}{2\times.648} \sqrt{\frac{T}{m} }[/tex]
For G note
110 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{m} }[/tex]
[tex]\frac{147}{110} =\frac{l}{.648}[/tex]
l = 866 mm
Distance from the end of string
866 - 648 = 218 mm or 245 mm
Option c ) is correct .
A 10000 kg rocket blasts off vertically from the launch pad with a constant upward of 2.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 525 m, its engine suddenly fails so that only gravity acts on it after that. What is the maximum height that the rocket reaches?
Answer:
The maximum height that the rocket reaches is 645.5 m.
Explanation:
Given that,
Mass = 10000 kg
Acceleration = 2.25 m/s²
Distance = 525 m
We need to calculate the velocity
Using equation of motion
[tex]v^2=u^2+2as[/tex]
Put the value in the equation
[tex]v^2=0+2\time2.25\times525[/tex]
[tex]v=\sqrt{2\times2.25\times525}[/tex]
[tex]v=48.60\ m/s[/tex]
We need to calculate the maximum height with initial velocity
Using equation of motion
[tex]v^2=u^2-2gh[/tex]
[tex]h=\dfrac{v^2-u^2}{-2g}[/tex]
Put the value in the equation
[tex]h=\dfrac{0-(48.60)^2}{-2\times9.8}[/tex]
[tex]h=120.50\ m[/tex]
The total height reached by the rocket is
[tex]h'=s+h[/tex]
[tex]h'=525+120.50[/tex]
[tex]h'=645.5\ m[/tex]
Hence, The maximum height that the rocket reaches is 645.5 m.
If my mass is 196 lbm and I tackle one of my teammates - while decelerating from a velocity of 6.7 m/s to 0 m/s in 0.5 s, how much force (kN) do I impart to the teammate's body?
Answer:
the force acting on the team mate is 1.19 kN.
Explanation:
given,
mass = 196 lbm
while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s
time taken for deceleration = 0.5 sec
F = mass × acceleration
acceleration = [tex]\dfrac{0-6.7}{0.5}[/tex]
= -13.4 m/s²
1 lbs = 0.453 kg
196 lbs = 196 × 0.453 = 88.79 kg
F = 88.79 × 13.4
F = 1189.786 N = 1.19 kN
hence, the force acting on the team mate is 1.19 kN.
A 25 kg gazelle runs 3 km up a slope at a constant speed, increasing their elevation by 40 m. While running, they experience a constant drag force of 20 N. It then jumps a height of 2 m onto a ledge. What is the minimum take off speed required to make the jump?
Answer:
v = 75 m/s
Explanation:
given data:
mass of gazelle is 25 kg
length of slope is 3 km
height of slope is 40 m
drag force is 20 N
jump height is 2 m
By work energy theorem we have
[tex]\frac{1}{2} mv^2 = F*L + mg(h_1 +h_2)[/tex]
[tex]\frac{1}{2}*25*v^2 = 20*3000 + 25 *9.8(40 + 2)[/tex]
solving for v
[tex]v = \sqrt{\frac{70290}{12.5}}[/tex]
v = 75 m/s
Given light with frequencies of: 5.5 x 1014 Hz, 7 x 1014 Hz and 8 x 1014 Hz, calculate the wavelengths in free space. What color are these waves?
Explanation:
The relation between frequency and wavelength is shown below as:
c = frequency × Wavelength
Where, c is the speed of light having value = 3×10⁸ m/s
So, Wavelength is:
Wavelength = c / Frequency
Given:
Frequency = 5.5 × 10¹⁴ Hz
So,
Wavelength = 3×10⁸ m/s / 5.5 × 10¹⁴ Hz = 545 × 10⁻⁹ m = 545 nm (1 nm = 10⁻⁹ m )
This wavelength corresponds to green color.
Frequency = 7 × 10¹⁴ Hz
So,
Wavelength = 3×10⁸ m/s / 7 × 10¹⁴ Hz = 428 × 10⁻⁹ m = 428 nm (1 nm = 10⁻⁹ m )
This wavelength corresponds to Violet color.
Frequency = 8 × 10¹⁴ Hz
So,
Wavelength = 3×10⁸ m/s / 8 × 10¹⁴ Hz = 375 × 10⁻⁹ m = 375 nm (1 nm = 10⁻⁹ m )
It does not fall in visible region. So no color. It is in UV region.
A baseball (m=140g) traveling 32m/s moves a fielders
glovebackward 25cm when the ball is caught. What was the average
forceexerted by the ball on the glove?
Answer:
Average force, F = 286.72 N
Explanation:
Given that,
Mass of the baseball, m = 140 g = 0.14 kg
Speed of the ball, v = 32 m/s
Distance, h = 25 cm = 0.25 m
We need to find the average force exerted by the ball on the glove. It is solved using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
F = mg
[tex]\dfrac{1}{2}mv^2=Fh[/tex]
[tex]F=\dfrac{mv^2}{2h}[/tex]
[tex]F=\dfrac{0.14\times (32)^2}{2\times 0.25}[/tex]
F = 286.72 N
So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.
A 2000 kg car rounds a circular turn of radius 20 m. If
theroad is flat and the coefficient of friction between tires and
roadis 0.70, how fast can the car go without skidding?
Answer:
The velocity of car is 11.71 m/s.
Explanation:
Given that,
Mass of car = 2000 kg
Radius = 20 m
Coefficient of friction = 0.70
We need to calculate the velocity of car
Using relation centripetal force and frictional force
[tex]F= \dfrac{mv^2}{r}[/tex]...(I)
[tex]F=\mu mg[/tex]...(II)
From equation (I) and (II)
[tex]\dfrac{mv^2}{r}=\mu mg[/tex]
[tex]v=\sqrt{\mu\times r\times g}[/tex]
Put the value into the formula
[tex]v=\sqrt{0.70\times20\times9.8}[/tex]
[tex]v=11.71\ m/s[/tex]
Hence, The velocity of car is 11.71 m/s.
Answer:
car can move at 11 m/s without skidding.
Explanation:
given,
mass of car = 2000 kg
radius of turn = 20 m
μ = 0.7
using centripetal force
F = [tex]\dfrac{mV^2}{R}[/tex].....................(1)
and we know
F = μ N = μ m g........................(2)
equating both the equation (1) and (2)
μ m g = [tex]\dfrac{mV^2}{R}[/tex]
v = [tex]\sqrt{\mu R g}[/tex]
v = [tex]\sqrt{0.7 \times 20 \times 9.81}[/tex]
v = 11.71 m/s
hence, car can move at 11 m/s without skidding.
Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?
Answer: Your code returns a number of 99.123456789 +0.00455679
Ok, you must see where the error starts to affect your number.
In this case, is in the third decimal.
So you will write 99.123 +- 0.004 da da da.
But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.
in the error, after the 4 comes a 5, so it rounds up.
So the final presentation will be 99.123 +- 0.005
you are discarding all the other decimals because the error "domains" them.
Create a mathematical model for the pressure variation as a function of position and time for a sound wave, given that the wavelength of the wave is λ = 0.190 m and the maximum pressure variation is ΔPmax = 0.270 N/m2. Assume the sound wave is sinusoidal. (Assume the speed of sound is 343 m/s. Use the following as necessary: x and t. Assume ΔP is in Pa and x and t are in m and s, respectively. Do not include units in your answer.)
Answer:
The equation of position and time for a sound wave is [tex]\Delta p=0.270(33.06 x-11342.40 t)[/tex].
Explanation:
Given that,
Wavelength = 0.190 m
Maximum pressure [tex]\Delta P_{max}= 0.270 N/m^2[/tex]
We know that,
The function of position and time for a sound wave,
[tex]\Delta p=\Delta p_{max}(kx-\omega t)[/tex]....(I)
We need to calculate the frequency
Using formula of frequency
[tex]f=\dfrac{v}{\lambda}[/tex]
Put the value into the formula
[tex]f=\dfrac{343}{0.190}[/tex]
[tex]f=1805.2\ Hz[/tex]
We need to calculate the angular frequency
Using formula of angular frequency
[tex]\omega =2\pi f[/tex]
Put the value into the formula
[tex]\omega=2\pi\times1805.2[/tex]
[tex]\omega=11342.40\ rad/s[/tex]
We need to calculate the wave number
Using formula of wave number
[tex]k = \dfrac{2\pi}{\lambda}[/tex]
Put the value into the formula
[tex]k=\dfrac{2\pi}{0.190}[/tex]
[tex]k=33.06[/tex]
Now, put the value of k and ω in the equation (I)
[tex]\Delta p=0.270(33.06 x-11342.40 t)[/tex]
Hence, The equation of position and time for a sound wave is [tex]\Delta p=0.270(33.06 x-11342.40 t)[/tex].
Final answer:
The mathematical model for the pressure variation of a sound wave involves sine functions with specific parameters. To find the frequency and speed of the sound wave, formulas relating wavelength, speed, and frequency are utilized.
Explanation:
The mathematical model for the pressure variation of a sound wave is given by the equation: ΔP = ΔPmax * sin((2π / λ) * x - (2πf) * t). In this equation, ΔPmax is the maximum pressure variation, λ is the wavelength, x is the position, t is the time, and f is the frequency.
To find the frequency of the sound wave, you can use the formula: f = v / λ, where v is the speed of sound. Substituting the given values allows you to calculate the frequency.
The speed of the sound wave can be determined using the formula: v = f * λ. By substituting the frequency and wavelength values, you can find the speed of the sound wave.
When a field goal kicker kicks a football as hard as he can at 45° to the horizontal, the ball just clears the 3-m-high crossbar of the goalposts 45.7 m away. (a) What is the maximum speed the kicker can impart to the football? (b) In addition to clearing the crossbar, the football must be high enough in the air early during its flight to clear the reach of the onrushing defensive lineman. If the lineman is 4.6 m away and has a vertical reach of 2.5 m, can he block the 45.7-m field goal attempt? (c) What if the lineman is 1.0 m away?
Answer:
Part a)
[tex]v = 21.9 m/s[/tex]
Part b)
[tex]y = 4.17 m[/tex]
So he will not able to block the goal
Part c)
[tex]y = 0.98 m[/tex]
yes he can stop the goal
Explanation:
As we know by the equation of trajectory of the ball
[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]
[tex]y = 3 m[/tex]
[tex]x = 45.7 m[/tex]
[tex]\theta = 45 degree[/tex]
now from above equation we have
[tex]y = 45.7 tan 45 - \frac{9.81(45.7)^2}{2v^2cos^245}[/tex]
[tex]3 = 45.7 - \frac{20488}{v^2}[/tex]
[tex]v^2 = 479.81[/tex]
[tex]v = 21.9 m/s[/tex]
Part b)
If lineman is 4.6 m from the football
[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]
[tex]y = 4.6tan45 - \frac{9.81(4.6^2)}{2(21.9^2)cos^245}[/tex]
[tex]y = 4.17 m[/tex]
So he will not able to block the goal
Part c)
If lineman is 1 m from the football
[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]
[tex]y = 1tan45 - \frac{9.81(1^2)}{2(21.9^2)cos^245}[/tex]
[tex]y = 0.98 m[/tex]
yes he can stop the goal
You throw a ball vertically from an apartment balcony to the ground 16.0 m below. Find the ball's initial velocity if it hits the ground 1.02 s after you release it.
Answer:
u(intial velocty)= 10.688 m/s apprx
Explanation:
given data:
height of apartment = 16 m
time of hiitng = 1.02 s
Using equation of motion we have
h = ut + 0.5gt2 putting all value to get inital velocity value
16 = u(1.02)+ 0.5(9.8)(1.02)^2
SOLVING FOR u
16 - 5.09 = U1.02
u(intial velocty)= 10.688 m/s apprx
Direction will be towards ground
To find the ball's initial velocity, use the equation of motion for free fall and solve for v0. The initial velocity of the ball is approximately 10.78 m/s.
Explanation:To find the ball's initial velocity, we can use the equation of motion for free fall: d = v0t + 0.5gt2, where d is the distance, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity. In this case, d = 16.0 m, t = 1.02 s, and g = 9.8 m/s2. Plugging in these values, we can solve for v0.
16.0 = v0(1.02) + 0.5(9.8)(1.02)2
Simplifying the equation gives us:
16.0 = 1.02v0 + 5.01
Subtracting 5.01 from both sides gives:
10.99 = 1.02v0
Finally, dividing both sides by 1.02 gives:
v0 ≈ 10.78 m/s
A 4.55 nF parallel-plate capacitor contains 27.5 μJ of stored energy. By how many volts would you have to increase this potential difference in order for the capacitor to store 55.0 μJ of potential energy?
Express your answer in volts as an integer.
Answer:
[tex] \Delta V=V_{2}-V_{1}=45.4V[/tex]
Explanation:
The energy, E, from a capacitor, with capacitance, C, and voltage V is:
[tex]E=\frac{1}{2} CV^{2}[/tex]
[tex]V=\sqrt{2E/C}[/tex]
If we increase the Voltage, the Energy increase also:
[tex]V_{1}=\sqrt{2E_{1}/C}[/tex]
[tex]V_{2}=\sqrt{2E_{2}/C}[/tex]
The voltage difference:
[tex]V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}[/tex]
[tex]V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V[/tex]