Answer:
v = 19 m/s
Explanation:
Since both rocks hit the water at the same time, and we know the total height of the bridge, we can get the time that it takes to the first rock to travel this height, due it is said that it was dropped, using the following kinematic equation:[tex]\Delta h = \frac{1}{2} * g * t^{2} (1)[/tex]
Replacing by the givens (Δh =21 m and g = 9.8 m/s2), and solving for t we get:[tex]t = \sqrt{\frac{2*21m}{9.8m/s2}} = 2.1 s (2)[/tex]
Now we need the time traveled by the first rock when it fell 7 m, that when the second rock was thrown, because the difference between the total time and this one will be the flight time for the second rock.We can use the same equation (1) replacing Δh by 7 m, and solving for t, in (2) as follows:[tex]t = \sqrt{\frac{2*7m}{9.8m/s2}} = 1.2 s (3)[/tex]
So, the total flight time for the second rock it was just the difference between (2) and (3):[tex]t_{2} = t_{tot} - t_{1} = 2.1 s - 1.2 s = 0.9 s (4)[/tex]
Since we know the total distance traveled, and the time of flight, we can use the same kinematic equation than in (1) but now taking into account the initial speed for the second rock, as follows:[tex]\Delta h = v_{o}* t_{2} + \frac{1}{2}* g*t_{2} ^{2} (5)[/tex] Since we know the value of Δh = 21 m and t₂ =0.9s, replacing in (5) and solving for v₀, we get:[tex]v_{o} =\frac{(\Delta h -\frac{1}{2}*g*t^{2})}{t_{2} } = \frac{(21m-(\frac{1}{2}*9.8m/s2*(0.9s)^{2}) ) }{(0.9m)} = 19.0 m/s (6)[/tex]
which was the initial velocity of Jill's rock taking as positive the down ward direction.Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of 25 cm. The focal length of the magnifying glass is +5.0 cm. Assume that Keesha's eye is close to the magnifying glass.
(a) What should be the distance between the magnifying glass and the beetle?
(b) What is the angular magnification?
Answer:
a) p = 4.167 cm, b) m = + 6
Explanation:
a) For this exercise we must use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm
Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.
[tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]
[tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]
[tex]\frac{1}{ p}[/tex] = 024
p = 4.167 cm
b) angular magnification
m = h ’/ h = - q / p
m = - (-25) /4.167
m = + 6
the positive sign indicates that the image is straight and enlarged
The electric field of a negative infinite line of charge: Group of answer choices Points perpendicularly away from the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and increases in strength at larger distances from the line charge Points perpendicularly away from the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Answer:
Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Explanation:
The electric field for a uniform line of charge is given by E = λ/2πε₀r where λ = charge density and r = distance from line of charge.
If λ is negative, E is negative so it points in the negative direction towards the line of charge.
Also, since for negative charges, electric field lines end up in them, the electric field for an infinitely long negative line of charge points towards the charge perpendicular to it.
Also as r increases, E decreases since E ∝ 1/r
So, the electric field decreases at larger distances from the line of charge.
So, the electric field of a negative infinite line of charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge.
3. A skater moves across the ice a distance of 12 m before a constant frictional force of 15 N cause him to stop. His initial speed is 2.2 m/s . Calculate the skater's mass.
Explanation:
thankyou fo
r the poimts
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give
Answer:
The answer is " because it is in the opposite direction of the bullet".
Explanation:
Given:
[tex]m_1= 70 \ kg\\\\m_2 = 0.01\ kg\\\\v_2 = 500\ \frac{m}{s}\\\\v_1 = ?[/tex]
Using formula:
It recoils the velocity of the rifle [tex]= -0.07142 \ \ \frac{m}{s}[/tex]
Its negative sign displays the opposite direction of the rifle.
Please helppppppp!!!!
Answer:
I think the particles transferred between brandi's body and the door knob causing the shock is the electrons.
. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass will be
Answer:
10J
Explanation:
KE = (1/2)mv²
100J = (.5)(40kg)v²
v²=(100J)/(20kg)
v²= 5
KE = 5(.5)(4kg)
KE = 10J
If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be Group of answer choices 1.0 m/s2. We cannot tell from the information given. 2.2 m/s2. 3.0 m/s2.
Answer:
We cannot tell from the information given
Explanation:
Given;
mass of the box, m = 5 kg
first force, F₁ = 10 N
second force, F₂ = 5 N
(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;
∑Fx = 10 N - 5 N
= 5 N
Apply Newton's second law of motion;
∑Fx = ma
a = ∑Fx/m
a = 5 / 5
a = 1 m/s² in the direction of the 10 N force.
(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;
∑Fx = 10 N + 5 N
∑Fx = 15 N
a = 15 / 5
a = 3 m/s²
Therefore, the information given is not enough to determine the acceleration of the box.
tell me about Orion nebula
In Case 1, a mass M hangs from a vertical spring having spring constant k and is at rest in its equilibrium position. In Case 2 the mass has been lifted a distance D vertically upward. If we define the potential energy in Case 1 to be zero, what is the potential energy of Case 2
Answer: hello your question is incomplete attached below is the complete question
answer : 1/2 KD^2 ( option A )
Explanation:
P.E ( potential energy ) = mgd
In case 1 P.E = 0 i.e. mgd = 0
Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring
The potential energy of the M in case 2
= P.E of M at rest + P.E of the spring
= 0 + 1/2 KD^2
Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Answer:
39.40 MeV
Explanation:
Determine the minimum possible Kinetic energy
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
MCQs. How does the heat from the Sun reach the Earth.(options)a: conduction b: convertion c: radiation d: combustion.
which one of the following is not a simple machine
1 ladder
2 wheel Barrow
3 pulley
4 electric pole
Discuss the chemical bond exist in silicon crystal?
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
a). How fast is the plane moving at takeoff?
b). How long does ot take the plane to travel down the runway?
Which statements are true of noble gases?
Check all that apply.
A. They are metalloids.
B. Their valence shells are full of electrons.
C. They are not very reactive.
D. All of the noble gases have at least two electron shielding layers.
A man who works for a moving company is loading a box onto a moving van. He pushes a 200N box up a 5m long ramp. If he pushes with a force of 60N and the ramp is 1m high, what is the efficiency of the inclined plane
Answer:
η = 0.667 = 66.7%
Explanation:
The efficiency of the man can be given by the following formula:
η = output/input
where,
η = efficiency of man = ?
output = potential energy gain of the box = Wh
input = work done by man = Fd
Therefore,
[tex]\eta = \frac{Wh}{Fd}[/tex]
where,
W = weight of box = 200 N
h = height gained by box = 1 m
F = force exerted by man = 60 N
d = length of ramp = 5 m
Therefore,
[tex]\eta = \frac{(200\ N)(1\ m)}{(60\ N)(5\ m)}[/tex]
η = 0.667 = 66.7%
What is the primary evidence used to determine how the Moon formed? O A. Moon craters and Earth craters were caused by the same asteroid
strike.
B. Moon rocks and Earth rocks are made up of many of the same
materials.
O C. The Moon and Earth are exactly the same age.
D. The Moon and Earth have similar atmospheres.
What is the primary evidence used to determine how the Moon formed?
[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]
A. Moon craters and Earth craters were caused by the same asteroid strike. ✅
[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]
the organelle involved in cell secretion is
Answer:
Golgi apparatus (bodies).
Explanation:
A cell can be defined as the fundamental or basic functional, structural and smallest unit of life for all living organisms. Some living organisms are unicellular while others are multicellular in nature.
Generally, cells have the ability to independently replicate themselves. In a cell, the "workers" that perform various functions or tasks for the survival of the living organism are referred to as organelles. Some examples of cell organelles found in all living organisms such as trees, birds, and bacteria include; nucleus, cytoplasm, cell membrane, golgi apparatus, mitochondria, lysosomes, ribosomes, chromosomes, endoplasmic reticulum, vesicles, etc.
Golgi apparatus is also referred to as Golgi bodies and it functions as a packaging unit in living organisms, especially eukaryotic cells because it prepares protein and lipid molecules for export by chemically tagging them.
Hence, the Golgi apparatus (bodies) is an organelle involved in cell secretion through the transportation (export) of proteins and lipids out of a cell.
In an experiment, a student brings up the rotational speed of a piece of laboratory apparatus to 24 rpm. She then allows the apparatus to slow down uniformly on its own, and counts 236 revolutions before the apparatus comes to a stop. The moment of inertia of the apparatus is known to be 0.076 kg m2. What is the magnitude of the torque on the apparatus
Answer:
T = 6.43 x 10⁻⁵ N.m
Explanation:
First, we will calculate the deceleration of the apparatus by using the third equation of motion:
[tex]2\alpha \theta = \omega_f^2-\omega_i^2[/tex]
where,
α = angular decelration = ?
θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad
ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s
ωf = final angular speed = 0 rad/s
Therefore,
[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]
negative sign shows deceleration
Now, for torque:
T = Iα
where,
T = Torque = ?
I = moment of inertia = 0.076 kg.m²
Therefore,
T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)
T = 6.43 x 10⁻⁵ N.m
A pair of glasses uses a nonreflective coating of index of refraction 1.4 to minimize reflection of light with wavelength 500nm. If the index of refraction of the glass is 1.5, what is the minimum non-zero thickness of the coating
Answer:
d = 178.57 10⁻⁹ m
Explanation:
For this exercise we must find the thickness to minimize the reflection, so the interference for the reflection must be destructive.
To find the expression we must take into account, two things:
* When the light goes from an index mordant medium to one with a higher refractive incoe, it undergoes a phase change of 180 (pi radians)
* within the film the wavelength of light is modulated by the index of refraction
λₙ = λ₀/ n
In this case the light passes from the air to the reflective layer and undergoes a phase change of ∫π rad, then it is reflected in the film-glass layer where it undergoes another phase change of π rad, therefore the total change of phase is 2π radians, this change is the or changes its value
period of the trigonometric functions, therefore its value does not change
the expression for destructive interference is
d sin θ = (me + ½) λₙ
d sin θ = (m + ½) λ₀ / n
the minimum thickness occurs for m = 0 and if we take perpendicular incidence the sine = 1
d = λ₀ /2 n
l
et's calculate
d = 500 10⁻⁹ /( 2 1.4)
d = 178.57 10⁻⁹ m
Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, the box (initially at rest) is traveling at a speed of 6 m/s. How much work (in Joules) did friction do in this process
Answer: 321 J
Explanation:
Given
Mass of the box [tex]m=3\ kg[/tex]
Force applied is [tex]F=25\ N[/tex]
Displacement of the box is [tex]s=15\ m[/tex]
Velocity acquired by the box is [tex]v=6\ m/s[/tex]
acceleration associated with it is [tex]a=\dfrac{F}{m}[/tex]
[tex]\Rightarrow a=\dfrac{25}{3}\ m/s^2[/tex]
Work done by force is [tex]W=F\cdot s[/tex]
[tex]W=25\times 15\\W=375\ J[/tex]
change in kinetic energy is [tex]\Delta K[/tex]
[tex]\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J[/tex]
According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
[tex]\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J[/tex]
Therefore, the magnitude of work done by friction is [tex]321\ J[/tex]
Which statement is true?
a particle of violet light has less energy than a particle of red light
a particle of violet light has more energy than a particle of red light
a particle of violet light has exactly the same energy as a particle of red light
particles of light do not have any energy, regardless of what color the light is
a particle of violet light has exactly the same energy as a particle of red light
What is the period of revolution of a satellite with mass m that orbits the earth in a circular path of radius 7880 km (about 1500 km above the surface of the earth)
Answer:
Explanation:
For time period of revolution , the expression is as follows .
T² = 4π² R³ /GM , M is mass of the earth.
Putting the values
T² = 4π² (7880 x 10³)³ /(6.67 x 10⁻¹¹ )( 5.97 x 10²⁴ )
T² = 4.846 x 10⁷ s
T = 6.961 x 10³ s
= 6961 s
= 116 minutes .
An atom has 20 protons and 22 neutrons and 18 electrons. The charge of this atom is: ________
Answer:
the number of electrons should equal to the the number of protons in a neutral atom
if there is a inequality between the numbers it means the atom has a + or - charge
The charge of this atom=+(20-18)=+2HELP ME PLEASE !!!!!!!!!!!!
Answer:
D
Explanation:
Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0
A plane mirror produces images of objects that have an orientation that is _____, a size that is _______ (compared to that of the object) and a type that is _____
Answer:
RIGHT, SAME SIZE, VIRTUAL
Explanation:
Plane mirrors comply with the law of reflection where the angle of incidence is equal to the angle of reflection
therefore to complete the sentences:
A plane mirror produces images of objects that have an orientation that is RIGHT __, a size that is _SAME SIZE____ (compared to that of the object) and a type that is VIRTUAL_____
A certain 20-A circuit breaker trips when the current in it equals 20 A. What is the maximum number of 100-W light bulbs you can connect in parallel in an ideal 120-V dc circuit without tripping this circuit breaker
Answer: 28
Explanation:
Given
Circuit breaker current is [tex]I=20\ A[/tex]
Power of the light bulb is [tex]P=100\ W[/tex]
Voltage of the DC-circuit is [tex]V=120\ V[/tex]
If the resistance are connected in parallel, they must have same voltage i.e. 120 V
So, Resistance is given by
[tex]\Rightarrow R=\dfrac{V^2}{P}\\\\\Rightarrow R=\dfrac{120^2}{100}\\\\\Rightarrow R=144\ \Omega[/tex]
For the 20 A current and 120 V battery, net resistance is
[tex]\Rightarrow R_{net}=\dfrac{120}{20}\\\\\Rightarrow R_{net}=6\ \Omega[/tex]
Suppose there are n resistance in the circuit connected in parallel.
[tex]\Rightarrow \dfrac{144}{n}=R_{net}\\\\\Rightarrow n=\dfrac{144}{6}\\\\\Rightarrow n=28.8\approx 28\ \text{for current to be less than 20A}[/tex]
Thus, there can maximum of 28 bulbs.
why doping method is used to design a diode circuit
Answer:
To increase the conductivity of the material.
Explanation:
Generally , the group 4 elements are non conductor but in certain conditions, such as doping or the increase in temperature, they becomes conductor.
The doping is the process of mixing of pentavalent or the trivalent material into tetra valent material in the very small amount, so that the material becomes conductor.
In making a diode we need two types of the materials, n type semiconductor and p type semi conductor.
When the trivalent impurity is added in the tetra valent element, the semiconductor becomes n type because an electron is left for the conduction.
When the pentavalent impurity is added in the tetra valent element, the semiconductor becomes p type because a hole is left for the conduction.
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what location are the kinetic energy and the potential energy the same
Given :
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.
The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
To Find :
At what location are the kinetic energy and the potential energy the same.
Solution :
Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.
So,
[tex]P.E = K.E\\\\\dfrac{kx^2}{2} = \dfrac{mv^2}{2}\\\\x = v\sqrt{\dfrac{m}{k}}\\\\x = 1.5 \times \sqrt{\dfrac{0.5}{20}}\ m\\\\x = 0.238 \ m[/tex]
Hence, this is the required solution.
Engineers are working on a design for a cylindrical space habitation with a diameter of 7.50 km and length of 29.0 km. The habitation will simulate gravity by rotating along its axis. With what speed (in rad/s) should the habitation rotate so that the acceleration on its inner curved walls equals 8 times Earth's gravity
Answer:
The speed will be "0.144 rad/s".
Explanation:
Given that,
Diameter,
d = 7.50 km
Radius,
R = [tex]\frac{7.5}{2} \ Km[/tex]
Acceleration on inner curve,
= 8 times
Now,
As we know,
⇒ [tex]\omega^2R=8g[/tex]
or,
⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]
⇒ [tex]=\sqrt{\frac{78.4}{3750} }[/tex]
⇒ [tex]=\sqrt{0.0209}[/tex]
⇒ [tex]=0.144 \ rad/s[/tex]