Answer: 0.5m/s
This value may vary depending on the initial velocity value.
Explanation:
The question is incomplete due to absence of the initial velocity (U) of the rock. Taking the initial velocity as any value say 10m/s at 30° above the horizontal.
Time it will take to reach maximum height (Tmax) will be Usin(theta)/g
Where U is the initial velocity = 10m/s theta = 30° g is the acceleration due to gravity = 10m/s²
Substituting the values in the formula we have;
Tmax = 10sin30°/10
Tmax = sin 30°
Tmax = 0.5second
Therefore, it will take for the rock 0.5s to reach the maximum height of its trajectory if the velocity is 10m/s.
The time varies though depending on the value of the initial velocity.
A laser dazzles the audience in a rock concert by emitting green light with a wavelength of 515 nm . Calculate the frequency of the light. Express your answer with the appropriate units.
Answer:
Explanation:
This question is quite tricky. Not all parameters were provided.
λ = 515nm(nm = namo meters), 515 nm becomes 515 x 10^-9 which in turn becomes 5.15 x 10^-7.
The velocity for the green light wasn't provided but the velocity of greenlight is a constant; velocity of green light(v) = 3.00 × 108 m/s.
frequency of the light =[tex]\frac{velocity of green light}{wavelength of light}\\[/tex]
f =[tex]\frac{3.00 X 108 m/s}{5.15 X 10^{-7} }[/tex]
f=20.971 Hertz
The frequency of light will be "20.971 Hz".
According to the question,
Wavelength,
[tex]\lambda = 515 \ nm[/tex]or,
[tex]= 5.15\times 10^{-7} \ m[/tex]
Velocity of green,
[tex]v = 3.00\times 108 \ m/s[/tex]The frequency of light (f) will be:
= [tex]\frac{Velocity}{Wavelength}[/tex]
= [tex]\frac{3.00\times 108}{5.15\times 10^{-7}}[/tex]
= [tex]20.971 \ Hz[/tex]
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Which of the following must ALWAYS be equal to the buoyant force on an object?
A.the force of gravity
B.the weight of the object
C.the weight of the liquid displaced by the objec
D.the force of friction as the object moves in the water
Answer:
D.
Explanation:
It just makes sense. It is a simple answer
Colossal Cave was formed years ago by underground running water. Today, it is the home to many animals, like bats. These interactions are an example of _____. a. biosphere, atmosphere, and lithosphere b. interaction hydrosphere, lithosphere, and biosphere c. interaction hydrosphere, lithosphere, and atmosphere d. interaction atmosphere and biosphere interaction
Answer:
d. interaction atmosphere and biosphere interaction
Explanation:
hydrosphere, lithosphere, and biosphere interaction.
If A, B, and C are three activities connected with SS and FF (combination) relationships with 2-day lag on SS relationship only. Their durations are 5, 10, and 3 respectively. What is the duration of the project?a. 5 days. b. 10 days. c. 3 days. d 18 days.
Answer:
B. 10 days
Explanation:
Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface and push them together until they touch. Each ball has a mass of 4 kg. Think carefully about what to use for the distance between the two balls. And to keep things simpler, let's assume they slide without rolling.Now consider the case where both balls are free to move once you release them. Neglecting friction, calculate the final speed of one of the balls.?
Explanation:
First, we will calculate the electric potential energy of two charges at a distance R as follows.
R = 2r
= [tex]2 \times 0.1 m[/tex]
= 0.2 m
where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.
U = [tex]\frac{k \times Q \times Q}{R}[/tex]
= [tex]\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}[/tex]
= 0.081 J
As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.
Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, [tex]\frac{U}{2}[/tex].
Hence, K.E = [tex]\frac{U}{2}[/tex]
= [tex]\frac{0.081}{2}[/tex]
= 0.0405 J
Now, we will calculate the speed of balls as follows.
V = [tex]\sqrt{\sqrt{\frac{2 \times K.E}{m}}[/tex]
= [tex]\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}[/tex]
= 0.142 m/s
Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.
2 Grade 11 Physics Questions. Will Mark BRAINLIEST!!!
Which machine would be an alternative to lifting a crate with a pulley?
Ramp
Screw
Doorstop
Ax
Which of the following is an example of a simple machine?
Bicycle
Car
Steering wheel system of a boat
Ramp
Answer:
For the first one it would be screw and for the second one it is a ramp
Explanation:
Hope this helps
A 0.10-kilogram ball dropped vertically from a height of 1.0 meter above the floor bounces back to a height of 0.80 meter. The mechanical energy lost by the ball as it bounces is approximately
A) 0.20 J
B) 0.080 J
C) 0.78 J
D) 0.30 J
Answer:
A) 0.20Joules
Explanation:
The type mechanical energy acting on the body is potential energy since the body is covering a particular height.
Potential Energy = mass×acceleration due to gravity × height
Mass of the body = 0.1kg
acceleration due to gravity = 10m/s²
h is the height of the object when dropping = 1.0meters
Substituting this values in the formula to get the energy of the body on dropping, we have;
PE = 0.1×10×1.0
PE = 1.0Joules
On bouncing back to height of 0.8m, the potential energy becomes
PE = 0.1×10×0.8
PE = 0.8Joules
The mechanical energy lost by the ball as it bounces will approximately be the difference in its potential energy when dropping and when it bounces back i.e 1.0Joules - 0.8Joules = 0.20Joules
0.2 J
mgh (before) = (0.1)(9.8)(1) = 0.98
mgh (after) = (0.1)(9.8)(0.8) = 0.784
0.98-0.784=0.196
0.196 is approximately 0.2.
Which is true about rule utilitarianism and Kantianism (check all that apply) a. Both theories evaluate the moral rules in the same way b. According to both theories, right actions are those that are in line with universal moral rules c. Both are consequentialist theories d. Both are relativistic theories
Answer:
A
Explanation:
Why is that they truly evaluate both of the theories in the same way.
Who showed that our universe is heliocentric—the planets of the solar system revolve around the sun? Johannes Kepler Isaac Newton Nicolaus Copernicus Galileo Galilei
Answer:
Option (3)
Explanation:
Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.
He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.
This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.
Thus, the correct answer is option (3).
Answer:
I think it is option 'c' in the quiz.
Explanation:
I took the test and got a 100.
An object’s average density rho is defined as the ratio of its mass to its volume: rho=M/V. The earth’s mass is 5.94×1024kg, and its volume is 1.08×1012km3. What is the earth’s average density?
Final answer:
To find the Earth's average density, we divide its mass (5.94×10²⁴ kg) by its volume converted to cubic meters (1.08×10²⁴ m³), resulting in a density of approximately 5.5 kg/m³.
Explanation:
An object's average density is defined as the ratio of its mass to its volume (rho = M/V). To calculate the Earth's average density, we use the given mass of the Earth (5.94×10²⁴ kg) and the given volume of the Earth (1.08×10¹² km³).
It's important to convert the volume from cubic kilometers to cubic meters to ensure consistency in units since density is typically expressed in kilograms per cubic meter (kg/m³). There are 1,000,000,000,000 (1×10¹²) cubic meters in one cubic kilometer, so the volume in cubic meters is 1.08×10¹² km³ × 1×1012 m³/km³ = 1.08×1024 m³.
Now, dividing the mass by the volume in consistent units gives:
Density = Mass / Volume = 5.94×10²⁴ kg / 1.08×10²⁴ m³ = 5.5 kg/m³
Therefore, The average density of Earth is roughly 5.5 kg/m³.
A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b. The positive charge per unit length on the inner cylinder is λ, and there is an equal negative charge per unit length on the outer cylinder /a b, (iii) r > b Calculate the potential V(r) for (i) r ca,俪) a (Hint: The net potential is the sum of the potentials due to the individual conductors.) Take V 0 at r-b. a. 2charge per unit length lm上 b. Show that the potential of the inner cylinder with respect to the outer is -I c. Show that the electric field at any point between the cylinders has magnitude
a)
i) Potential for r < a: [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
ii) Potential for a < r < b: [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]
iii) Potential for r > b: [tex]V(r)=0[/tex]
b) Potential difference between the two cylinders: [tex]V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
c) Electric field between the two cylinders: [tex]E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]
Explanation:
a)
Here we want to calculate the potential for r < a.
Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is
[tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex]
where
[tex]\lambda[/tex] is the linear charge density
r is the distance from the wire/surface of the cylinder
By integration, we find an expression for the electric potential at a distance of r:
[tex]V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)[/tex]
Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:
[tex]E=0[/tex]
So the potential where the electric field is zero is constant:
[tex]V=const.[/tex]
iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density [tex]+\lambda[/tex] and an equal negative charge density [tex]-\lambda[/tex]. Therefore, the net charge is zero, so the electric field is zero.
This means that the electric potential is constant, so we can write:
[tex]\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)[/tex]
However, we know that the potential at b is zero, so
[tex]V(r)=V(b)=0[/tex]
ii) The electric field in the region a < r < b instead it is given only by the positive charge [tex]+\lambda[/tex] distributed over the surface of the inner cylinder of radius a, therefore it is
[tex]E=\frac{\lambda}{2\pi r \epsilon_0}[/tex]
And so the potential in this region is given by:
[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex] (1)
i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):
E = 0
This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):
[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
And so, for r<a,
[tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
b)
Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.
We have:
- Potential at the surface of the inner cylinder:
[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
- Potential at the surface of the outer cylinder:
[tex]V(b)=0[/tex]
Therefore, the potential difference is simply equal to
[tex]V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
c)
Here we want to find the magnitude of the electric field between the two cylinders.
The expression for the electric potential between the cylinders is
[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]
The electric field is just the derivative of the electric potential:
[tex]E=-\frac{dV}{dr}[/tex]
so we can find it by integrating the expression for the electric potential. We find:
[tex]E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]
So, this is the expression of the electric field between the two cylinders.
Learn more about electric fields:
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Two small, identical particles have charges Q1 = +5.5 μC and Q2 = -17μC. The particles are conducting and are brought together so that they touch. Charge then moves between the two particles so as to make the excess charge on the two particles equal. If the particles are then separated by a distance of 60 mm, what is the magnitude of the electric force between them?
Answer:
Explanation:
Given
Charge [tex]Q_1=+5.5\mu C[/tex]
[tex]Q_2=-17\ mu C[/tex]
When the two charges are touch then the net charge on the two sphere will be same after some time
[tex]Q=\frac{Q_1+Q_2}{2}[/tex]
[tex]Q=\frac{5.5-17}{2}=-5.75\ \mu C[/tex]
Now the force between this tow charges are when they are separated with [tex]d=60\ mm[/tex]
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
[tex]F=\frac{9\times 10^9\times (-5.75)^2}{(60\times 10^{-3})^2}[/tex]
[tex]F=82.65\ N[/tex]
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far along the length of the ladder can a 69.1-kg painter climb before the ladder begins to slip?
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=[tex]\frac{N2}{mg}*l[/tex] * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= [tex]\frac{241.75}{69.1*9.8} *l[/tex] * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
Answer:
The weight of the ladder is given as well as the length ; l = 11.9 m long and weighing W = 46.1 N rests against a smooth vertical wall.
how far along the length of the ladder = x = 4.14m
Explanation:
The detailed step and calculation is as shown in the attached file.
The acceleration of a particle varies with time according to the equation a(t) = pt^2 - qt^3. Initially, the velocity and position are zero, (a) What is the velocity as a function of time? (b) What is the position as a function of time?
Answer:
Explanation:
Given
acceleration of a particle is given by [tex]a(t)=pt^2-qt^3[/tex]
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=pt^2-qt^3[/tex]
[tex]dv=(pt^2-qt^3) dt[/tex]
Integrating we get
[tex]\int dv=\int \left ( pt^2-qt^3\right )dt[/tex]
[tex]v=\frac{pt^3}{3}-\frac{qt^4}{4}+c_1[/tex]
at [tex]t=0\ v=0[/tex]
therefore [tex]c_1=0[/tex]
We know velocity is given by
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]vdt=dx[/tex]
integrating
[tex]\int dx=\int \left ( \frac{pt^3}{3}-\frac{qt^4}{4}+\right )dt[/tex]
[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}+c_2[/tex]
using conditions
at [tex]t=0\ x=0[/tex]
[tex]c_2=0[/tex]
[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}[/tex]
(a) The velocity of the particle as function of time is pt³/3 - qt⁴/4 + C.
(b) The position of the particle as function of time is pt⁴/12 - qt⁵/20 + Ct + C.
Velocity of the particleThe velocity of the particle is the integral of the acceleration of the particle.
The velocity of the particle is calculated as follows;
v = ∫a(t)
v = ∫(pt² - qt³)dt
v = pt³/3 - qt⁴/4 + C
Position of the particleThe position of the particle as function of time is the integral of velocity of the particle.
x = ∫v
x = ∫(pt³/3 - qt⁴/4 + C)dt
x = pt⁴/12 - qt⁵/20 + Ct + C
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The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of 39 m/s (88 mi/h). To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.
Answer:
44.1613858478 m/s
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement = 99.4
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s[/tex]
If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s