The wheelbarrow's final kinetic energy at the bottom of the ramp is 4500 J, found by calculating the work done on the wheelbarrow by gravity and friction and applying the work-energy theorem.
Explanation:To determine the wheelbarrow's kinetic energy at the bottom of the ramp, we need to calculate the work done on the wheelbarrow by both gravity and friction as it slides down the ramp, and then use this to find the change in kinetic energy.
The work done by gravity is simply the component of the wheelbarrow's weight (500 N) that acts parallel to the ramp (i.e., 500 N * sin(20) * 50 m = 8500 J).
The work done by friction is the frictional force (80 N) times the distance (50 m), but since friction opposes the motion it’s negative, so -4000 J.
By the work-energy theorem, the total work done on the wheelbarrow equals its change in kinetic energy. The wheelbarrow starts from rest, so its initial kinetic energy is zero. Hence, its final kinetic energy (at the bottom of the ramp) is the total work done on it: 8500 J - 4000 J = 4500 J.
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Suppose you exert a 400-N force on a wall but the wall does not move. The work you are doing on the wall is : unknown, because the mass of the wall is not given
zero, because the wall is not moving
3920 J
400 J
Answer:
W = 0 :The work done on the wall is zero,because the wall is not moving
Explanation:
Work theory
Work is the product of a force applied to a body and the displacement of the body in the direction of this force.
W= F*d Formula (1)
W: Work (Joules) (J)
F: force applied (N)
d=displacement of the body (m)
The work is positive (W+) if the force goes in the same direction of movement.
The work is negative (W-)if the force goes in the opposite direction to the movement
Data
F= 400-N
d= 0
Problem development
We apply formula (1) to calculate the work done on the wall:
W= 400*0
W=0
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.
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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.
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 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.
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°.
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.
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
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.
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.
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 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]
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.
Explain why two equipotential lines cannot cross each other.
Answer:
Explained
Explanation:
Equipotential lines cannot cross each other because. The equipotential at a given point in space is has single value of potential throughout. If two equipotential lines intersect with each other, that would mean two values of potential, that would not mean equipotential line. Hence two equipotential lines cannot cross each other.
Equipotential lines are imaginary lines that connect points with the same electric potential. Two equipotential lines cannot cross each other because that would mean two different points on the same line have the same electric potential, which is not possible.
Explanation:Equipotential lines are imaginary lines that connect points with the same electric potential. Two equipotential lines cannot cross each other because that would mean two different points on the same line have the same electric potential, which is not possible.
For example, imagine a hill with contour lines representing lines of equal height. If two contour lines were to cross each other, it would mean that two different points on the hill have the same height, which is not possible.
In electricity, the electric potential difference between two points on an equipotential line is zero. If two equipotential lines were to cross, it would create a contradiction, as the potential difference between the two points of intersection would be both zero and non-zero.
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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 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]
Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements are true. It can always be expressed as the difference between the initial and final values of a potential energy function. None of these statements are true. It is independent of the path of the body and depends only on the starting and ending points. When the starting and ending points are the same, the total work is zero.
The false statement among the provided ones about conservative forces is 'None of these statements are true'. Conservative forces rely on the starting and ending points, not the path taken, and the work done on any closed path is always zero. These forces also have a close relationship with potential energy.
Explanation:The statements provided about conservative forces generally hold true however the false statement is 'None of these statements are true', as all other statements presented are indeed accurate. A conservative force is a type of force where the work done is independent of the path. That is, the work it does only depends on the starting and ending points, not the route taken. This implies that during any closed path or loop, the total work done by a conservative force is zero.
Consider a simple example, such as the gravitational force. If you lift an object to a certain height and then back to its original position, the total work done by the gravitational force is zero as the starting and ending points are the same.
An important concept related to the conservative force is potential energy. When work is done against a conservative force, potential energy is accumulated. Conversely, the component of a conservative force, in a particular direction, equals the negative of the derivative of the potential energy for that force, with respect to a displacement in that direction.
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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.
Two spheres are cut from a certain uniform rock. One has radius 4.10 cm. The mass of the other is eight times greater.
Final answer:
The volume of the larger sphere is eight times greater than that of the smaller sphere, and since their masses are proportional to their volumes for objects of uniform density, we can find the radius of the larger sphere to be twice that of the smaller sphere with a radius of 4.10 cm, resulting in a radius of 8.20 cm for the larger sphere.
Explanation:
The question concerns a comparison of the volumes of two spheres made from the same uniform rock. Since the mass of one sphere is eight times greater than the other, and since mass is proportional to volume for objects with uniform density, we can deduce that the volume of the larger sphere is also eight times greater than the smaller sphere. Given that the volume of a sphere (V) is calculated as V = (4/3)πR³, where R is the radius, we understand that because the volumes are proportional to the cube of the radii, we can calculate the radius of the larger sphere if we know the radius of the smaller one. Specifically, if the radius of the smaller sphere is 4.10 cm and its volume is V, then the volume of the larger sphere is 8V, and its radius is 2 times that of the smaller sphere since 2³ equals 8. Therefore, the radius of the larger sphere is 2 × 4.10 cm = 8.20 cm.
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
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.
If a wave vibrates up and down twice each second and travels a distance of 20 m each second and travels a distance of 20 m each second, what is its frequency? Its wave speed?
Answer:
Frequency is 0.5 Hz and the wave speed is 10 m/s.
Explanation:
As we know that frequency is defined as the how many times the no of cycles repeat in one second so if the wave is vibrating up and down twice during 1 second then the frequency in 1 second is
[tex]f=\frac{1}{2} hz\\F=0.5hz[/tex]
Therefore frequency is 0.5 Hz.
Now the distance of wawe in each second is,
[tex]d=20m[/tex]
Now the wave velocity is,
[tex]v=fd[/tex]
Here, f is frequency, d is the distance, v is the wave velocity.
Substitute all the variables
[tex]v=0.5\times 20\\v=10m/s[/tex]
Therefore the wave speed is 10 m/s.
In the given question, the wave's frequency is 2 Hertz (Hz), which means it oscillates twice per second. The wave speed, or distance covered by the wave per second, is identified as 20 m/s. These are fundamental concepts in the Physics of wave mechanics.
Explanation:In this question, we're dealing with the topics of wave frequency and wave speed. The frequency of a wave relates to how many cycles of the wave occur per unit of time - in this case, the wave is oscillating twice per every second, meaning that its frequency is 2 Hertz (Hz).
The wave speed is the speed at which the wave is travelling. Given that the wave travels a distance of 20 m each second, the speed of the wave is 20 m/s (meters per second).
It's important to note that these two properties are interconnected. In general, the speed of a wave (v) is calculated by multiplying its wavelength (λ) by its frequency (f). So, using the relationship v = λf, the properties of a wave can be determined if the other two are known.
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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
A glider moves along an air track with constant acceleration a. It is projected from the start of the track (x = 0 m) with an initial velocity of v0. At time t = 8s, it is at x = 100 cm and is moving along the track at velocity vt = − 0.15 m/s. Find the initial velocity v0 and the acceleration a.
Answer:
vo = 0.175m/s
a = -0.040625 m/s^2
Explanation:
To solve this problem, you will need to use the equations for constant acceleration motion:
[tex]x = \frac{1}{2}at^2 +v_ot+x_o \\v_f^2 - v_o^2 = 2(x-x_o)a[/tex]
In the first equation you relate final position with the time elapsed, in the second one, you relate final velocity at any given position. In both equations, you will have both the acceleration a and the initial velocity vo as variables. We can simplify with the information we have:
1. [tex]x = \frac{1}{2}at^2 +v_ot+x_o\\0.1m = \frac{1}{2}a(8s)^2 +v_o(8s)+0m \\0.1 = 32a + 8v_o[/tex]
2. [tex]v_f^2 - v_o^2 = 2(x-x_o)a\\(-0.15m/s)^2 - v_o^2 = 2(0.1m-0m)a\\0.0225 - v_o^2 = 0.2a\\a = \frac{0.0225 - v_o^2}{0.2} = 0.1125 - 5v_o^2[/tex]
Replacing in the first equation:
[tex]0.1 = 32(0.1125 - 5v_o^2) + 8v_o\\0.1 = 3.6 - 160v_o^2 + 8v_o\\160v_o^2 - 8v_o - 3.5 = 0[/tex]
[tex]v_0 = \frac{-(-8) +- \sqrt{(-8)^2 - 4(160)(-3.5)}}{2(160)} \\ v_o = 0.175 m/s | -0.125 m/s[/tex]
But as you are told that the ball was projected om the air track, it only makes sense for the velocity to be positive, otherwise it would have started moving outside the air track, so the real solution is 0.175m/s. Then, the acceleration would be:
[tex]a = 0.1125 - 5v_o^2\\a = -0.040625 m/s^2[/tex]
A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is the smallest distance the student could possibly be from the starting point?
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
[tex]A\pm \Delta A[/tex] and [tex]B\pm \Delta B[/tex]
The sum is represented as
[tex]Sum=(A+B)\pm (\Delta A+\Delta B)[/tex]
For the the values given to us the sum is calculated as
[tex]Sum=(2.9+3.9)\pm (0.1+0.2)[/tex]
[tex]Sum=6.8\pm 0.3[/tex]
Now the since the uncertainity inthe sum is [tex]\pm 0.3[/tex]
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals [tex]6.8-0.3=6.5[/tex]meters
A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to provide a 2-V voltage. Sketch the circuit. Assuming exact-valued resistors, what output voltage (measured to ground) and equivalent output resistance result? If the resistors used are not ideal but have a ±5% manufacturing tolerance, what are the extreme output voltages and resistances that can result?
Answer:
circuit sketched in first attached image.
Second attached image is for calculating the equivalent output resistance
Explanation:
For calculating the output voltage with regarding the first image.
[tex]Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}[/tex]
[tex]Vout = 5 \frac{2000}{5000}[/[tex]
[tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V[/tex]
For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.
so.
[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200[/tex]
Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.
if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:
[tex]Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V[/tex]
[tex]Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V[/tex]
[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140[/tex]
[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260[/tex]
so.
[tex]V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}[/tex]
At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building that is 300 m tall. How much earlier does the thrown rock strike the ground? Neglect air resistance. Please show all work and formulas used thanks
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
[tex]x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration[/tex]
So now we have an equation and unkown value.
for the thrown rock
[tex]\frac{1}{2}(9.8)*t^2+29*t-300=0[/tex]
for the dropped rock
[tex]\frac{1}{2}(9.8)*t^2+0*t-300=0[/tex]
solving both equation with the quadratic formula:
[tex]\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}[/tex]
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
The thrown rock hits the ground approximately 1.89 seconds earlier than the dropped rock when released simultaneously from a height of 300 meters.
Explanation:To determine how much earlier the thrown rock strikes the ground compared to the dropped rock, we need to use the kinematic equation Projectile motion that relates displacement (Δx), initial velocity (vi), acceleration (a), and time (t): Δx = vi * t + 0.5 * a * t₂.
Since we're dealing with gravity, acceleration due to gravity (a) is 9.8 m/s². For the rock that is dropped, the initial velocity (vi) is 0 m/s, while the rock thrown downward has an initial velocity of 29 m/s.
First, we'll find the time it takes for the dropped rock to hit the ground:
0 = vi * t + 0.5 * a * t₂ - Δx0 = 0 * t + 0.5 * 9.8 * t₂ - 3000 = 4.9 * t₂ - 300t₂ = 61.22t ≈ √61.22 ≈ 7.83 secondsNext, we use the same equation to find the time for the thrown rock:
0 = vi * t + 0.5 * a * t₂ - Δx0 = 29 * t + 0.5 * 9.8 * t₂ - 3000 = 29 * t + 4.9 * t₂ - 300This forms a quadratic equation in the form of at₂ + bt + c = 0. To solve for t, we use the quadratic formula. We only need the positive root since time cannot be negative:
t = (-b ± √(b2 - 4ac))/(2a)t ≈ (5.94 seconds)Finally, the difference in time between when the two rocks hit the ground is:
Time difference = Time for dropped rock - Time for thrown rockTime difference ≈ 7.83 s - 5.94 s ≈ 1.89 secondsTherefore, the thrown rock hits the ground approximately 1.89 seconds earlier than the dropped rock.
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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