Answer:
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
Explanation:
The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:
[tex](m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}[/tex] (1)
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex] - Masses of the first, second and third fragments, in kilograms.
[tex]\vec v_{o}[/tex] - Initial velocity of the object, in meters per second.
[tex]\vec v_{1}[/tex], [tex]\vec v_{2}[/tex], [tex]\vec v_{3}[/tex] - Velocities of the first, second and third fragments, in meters per second.
If we know that [tex]m_{1} = 0.5\,kg[/tex], [tex]m_{2} = 1.3\,kg[/tex], [tex]m_{3} = 1.2\,kg[/tex], [tex]\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right][/tex], the velocity of the third fragment is:
[tex](-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)[/tex]
[tex]1.2\cdot \vec v_{3} = (1.4,1.95)[/tex]
[tex]\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right][/tex]
The speed of the third fragment is the magnitude of the result found above:
[tex]v_{3} = 2\,\frac{m}{s}[/tex]
And the direction of the third fragment is:
[tex]\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)[/tex]
[tex]\theta_{3} \approx 54.316^{\circ}[/tex]
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
A 50kg crate is being push on a horizontal floor at constant velocity. Given that the coefficient of kenitic friction between crate and floor is hk=0.1 . What is the push force?
Answer:
F = 49 N
Explanation:
For this exercise we must use Newton's second law. Let's set a reference frame with the x axis parallel to the floor.
As they indicate that the box is going at constant speed, its acceleration is zero
Y axis y
N-W = 0
N = mg
X axis
F-fr = 0
F = fr
the friction force has the formula
fr = μ N
fr = μ mg
we substitute
F = μ m g
let's calculate
F = 0.1 50 9.8
F = 49 N
g 4.86 Separators are used to separate liquids of diff erent densities, such as cream from skim milk, by rotating the mixture at high speeds. In a cream separator, the skim milk goes to the outside while the cream migrates toward the middle. A factor of merit for the centrifuge is the centrifugal acceleration force (RCF), which is the radial acceleration divided by the acceleration due to gravity. A cream separator can operate at 9000 rpm (rev/min). If the bowl of the separator is 20 cm in diameter, what is the centripetal acceleration if the liquid rotates as a solid body, and what is the RCF
Answer:
Centripetal Acceleration = 88826.44 m/s²
RCF = 9054.7
Explanation:
First, we will find the value of the centripetal acceleration by using the following formula:
[tex]Centripetal\ Acceleration = \frac{v^2}{r}\\[/tex]
where,
v = linear speed of liquid or separator = rω
ω = angular speed of liquid or separator = (9000 rpm)(2π rad/rev)(1 min/60 s) = 942.48 rad/s
r = radius of seperator = diameter/2 = 20 cm/2 = 10 cm = 0.1 m
Therefore,
[tex]Centripetal\ Acceleration = \frac{(r\omega)^2}{r}\\Centripetal\ Acceleration = r\omega^2\\Centripetal\ Acceleration = (0.1\ m)(942.48\ rad/s)^2\\[/tex]
Centripetal Acceleration = 88826.44 m/s²
Now, for the RCF:
[tex]RCF = \frac{Centripetal\ Acceleration}{g}\\RCF = \frac{88826.44\ m/s^2}{9.81\ m/s^2}\\[/tex]
RCF = 9054.7
Calculate the force that would be required to enact 50 joules of work on an object that was displaced by 2 meters.
Answer:
Force = 25 N
Explanation:
Given:
Work done = 50 J
Displacement = 2 m
Find:
Force
Computation:
Work done = Force x displacement
50 = Force x 2
Force = 25 N
In this experiment, you will use a track and a toy car to explore the concept of movement. You will measure the time it takes the car to travel certain distances, and then complete some calculations. In the space below, write a scientific question that you will answer by doing this experiment.
Answer: if weight affects how fast they go?
Explanation:
Answer:
How can we change the speed of a toy car on a racetrack to describe the car’s motion?
Explanation:
thats the sample respond
A planet with a mass one-half that of Earth has a radius that is 3 times that of Earth's radius. What is the gravitational field strength of the planet?
The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²)
of the force that you feel on the surface of the Earth.
That's (1/18) as much as on Earth.
The acceleration of gravity there would be about 0.545 m/s².
This is about 12% less than the gravity on Pluto.
I don’t understand this
Answer:
true
Explanation:
force or powerbecause he pushes a disk
A hair dryer draws a current of 12.8 A.
(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.
(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.
Answer:
(a) 8.85×10⁻³ minutes
(b) 4.24×10¹⁹ electrons
Explanation:
(a) Using,
Q = it............................. Equation 1
Where Q = quantity of charge, i = current, t = time.
Make t the subject of the equation
t = Q/i............................. Equation 2
Given: Q = 6.8×10⁰ C, i = 12.8 A
Substitute these values into equation 2
t = 6.8×10⁰/12.8
t = 8.85×10⁻³ minutes
(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3
Where n = number of electrons.
Given: Q = 6.8×10⁰ C
Substitute into equation 2
n = 6.8×10⁰/1.602×10⁻¹⁹
n = 4.24×10¹⁹ electrons
(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b) Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
What will be the time of the charge and number of the electrons in the charge ?As we know Q = IT
Where Q = quantity of charge, i = current, T = time.
From the above equation
T= Q/I.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A
Substitute these values
T= [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8
T = [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes
Now the number of the electrons present in the charge will be
n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])
Where n = number of electrons.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C
Substitute Value of Q
n = [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]
n = [tex]4.24\times\d10^{19}[/tex] electrons
Thus
(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b)Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
To know more about electric charge follow
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calculate the resistance if a potential difference of 3v exists when a current of 2A flows through the conductor
Answer:
Resistance = 1.5 ohms
Explanation:
Given:
Potential difference = 3v
Current flow = 2 A
Find:
Resistance
Computation:
Resistance = Potential difference / Current flow
Resistance = 3 / 2
Resistance = 1.5 ohms
A strong electromagnet produces a uniform magnetic field of 1.60 T and a cross-sectional area of 0.200 m2. If we place a coil having 240 turns and a total resistance of 21.0 around the electromagnet and then we then smoothly reduced the current in the electromagnet until it reaches zero in 20.0 ms, what is the current induced in the coil
Answer:
the magnitude of the induced current is 182.86 A.
Explanation:
Given;
number of turns, N = 240 turns
cross sectional area of the loop, A = 0.2 m²
uniform magnetic field strength, B = 1.6 T
resistance of the loop, R = 21 ohms
time, Δt = 20.0 ms
The magnitude of the induced emf is calculated as;
[tex]emf = \frac{NA B}{t} \\\\emf = \frac{240 \times 0.2 \times 1.6}{20 \times 10^{-3}} \\\\emf = 3,840\ V[/tex]
The induced current in the loop is calculated as;
[tex]I = \frac{emf}{R} \\\\I = \frac{3840}{21} \\\\I= 182.86 \ A[/tex]
Therefore, the magnitude of the induced current is 182.86 A.
A motorcycle traveling due east at a constant speed covers 75 kilometers in
1.5 hours. What is its velocity in km/h?
Answer:
The velocity of the motorcycle is 50km/hr due east.
Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion
what colors of light are absorbed when white light falls on a green object?
what determines the magnification of an imagev
measure:what the current values of
Answer:
The magnification of an image is equal to the ratio of the image height to the object height.
help me plz!!!!!!. what difference between center of gravity and center of mass
Answer:
I think that the centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field
35.15 .. Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fringe is at 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen
Answer:
[tex]1.199\ \mu\text{m}[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength = 600 nm
D = Distance of the light source from screen = 3 m
y = Distance of first order bright fringe from center = 4.84 mm
d = Distance between slits
m = Order = 1
We have the relation
[tex]y=\dfrac{D\lambda}{d}\\\Rightarrow d=\dfrac{D\lambda}{y}\\\Rightarrow d=\dfrac{3\times 600\times 10^{-9}}{4.84\times 10^{-3}}\\\Rightarrow d=0.0003719\ \text{m}[/tex]
From the question we have
[tex]y=\dfrac{\dfrac{1}{2}3\lambda}{d}\\\Rightarrow \lambda=\dfrac{2}{3}yd\\\Rightarrow \lambda=\dfrac{2}{3}\times 4.84\times 10^{-3}\times 0.0003719\\\Rightarrow \lambda=0.000001199\ \text{m}=1.199\ \mu\text{m}[/tex]
The required wavelength of light is [tex]1.199\ \mu\text{m}[/tex].
(will give brainliest to whoever answers first and explains reasoning) A 10kg object is spun around in a circle with a centripetal acceleration of 3.5m/s^2. What is the centripetal force acting on the object?
Answer:
35 N
Explanation:
F = ma
centripetal force = 10(3.5) = 35 N
A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
the cat's displacement during this motion?
A. 6m
B. 2.5 m
C. 8.5 m
D. 4.5 m
20
А
B
D
Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
If an 800 N force causes an 80kg fullback to change his velocity by 10 m/s, what impulse was experienced?
Answer:
7
Explanation:
7
n
w
t
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 640 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 12th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source
Answer:
533.33 nm
Explanation:
Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.
Since the fringes coincide,
m'λ = m"λ'
λ' = m'λ/m"
= 10 × 640 nm/12
= 6400 nm/12
= 533.33 nm
starting from rest, your bicycle can reach a speed of 4.0 m/s in 50 s. Assuming that your bicycle accelerates at a constant rate, what is its acceleration?
Answer:
0.08 ms^-2
Explanation:
by using v= u + at
initial velocity is zero as it is starting from rest
4= 0 + a x 50
4/50 = a = 0.08 ms^-2
A car's initial speed of 15m/s is running with the acceleration of 32m/
s2 in 8 seconds. What is the car's final velocity?
Explanation:
15m/s
acceleration= (+)
so, 15m/s +32m/s=47m/s
42m/s. X 8 = 336
which energy conservation takes place when a toaster is switched on?
Answer:
A toaster usually takes in electrical energy.
Two types of energy are created.
In the first step, all of the input electrical energy is transformed into heat energy. That heat goes first into the coils or heating elements.
The heating elements get hot and glow.
The hot elements then transfer thermal energy (heat energy) into the air inside the toaster.
Explanation:
Answer: electrical energy
Explanation: An electric toaster takes in electrical energy from the power outlet and converts it into heat, very efficiently. If you want your toast to cook quickly, you need a toaster that radiates as much heat as possible each second onto your bread. hope this helps. Can u pls give me brainliest
Hello everyone I have a question for you today. So you know we have discovered black holes and have theorized about white holes so let me ask you without you going to the internet and looking it up what would happen if a white hole and a black hole collided?
Answer:
White hole is an impossible object in universe. ... This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.15 mm apart and position your screen 3.93 m from the slits. Although Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 647 nm . How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe
Answer:
a) y = 2.21 10⁻³ m, b) y = 5.528 10⁻³ m
Explanation:
In the double-slit experiment the interferences occur at the positions
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference
let's use trigonometry for the angle
tan θ = y / L
as in this experiment the angles are very small
tan θ = sin θ/cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ constructive interference
d y / L = (m + ½) λ destructive interference
Let's answer the questions
a) first line of constructive interference me = 1
y = m λ L / d
y = 1 647 10⁻⁹ 3.93 /1.15 10⁻³
y = 2.21 10⁻³ m
b) second dark band m = 2
y = (m + ½) λ L / d
y = (2 + ½) 647 10⁻⁹ 3.93 /1.15 10⁻³
y = 5.528 10⁻³ m
If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?
___Hz
Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
The second harmonic will be three times the first harmonic. The answer is 750 Hz
VIBRATION OF WAVES IN PIPESClosed pipes have odd multiples of frequencies or harmonics. That is,
If [tex]F_{0}[/tex] = fundamental frequency = first harmonic
[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic
[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic
[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic
Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.
By using the formula above,
second harmonic will be 3 x 250 = 750Hz
Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz
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someone please help I can mark brainless
A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They use rope with a mass per length of 1.50 kg/m under 6000 N tension. If they are limited to a wave amplitude of 0.500 m, what must be the frequency of waves necessary to transmit power at the average rate of 2.00 kW
Answer:
the required frequency of waves is 2.066 Hz
Explanation:
Given the data in the question;
μ = 1.50 kg/m
T = 6000 N
Amplitude A = 0.500 m
P = 2.00 kW = 2000 W
we know that, the average power transmit through the rope can be expressed as;
p = [tex]\frac{1}{2}[/tex]vμω²A²
p = [tex]\frac{1}{2}[/tex]√(T/μ)μω²A²
so we solve for ω
ω² = 2P / √(T/μ)μA²
we substitute
ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²
ω² = 4000 / 23.71708
ω² = 168.65
(2πf)² = ω²
so
(2πf)² = 168.65
4π²f² = 168.65
f² = 168.65 / 4π²
f² = 4.27195
f = √4.27195
f = 2.066 Hz
Therefore, the required frequency of waves is 2.066 Hz
Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t[tex]_{min[/tex] = ( m + 1/2) × λ/2n
we substitute
t[tex]_{min[/tex] = ( 0 + 1/2) × 711 /2(1.21)
t[tex]_{min[/tex] = 0.5 × 711/2.42
t[tex]_{min[/tex] = 0.5 × 293.80165
t[tex]_{min[/tex] = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
What is happening in the graph shown below?
A.
The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
B.
The object moves toward the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 2 m/s.
C.
The object moves toward the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 8 m/s.
D.
The object moves away from the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 8 m/s.
Answer:
D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
Explanation:
I just got it right lol
Scientific work is currently under way to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that driv- ers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long expo- sure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 3 1023 T, oscillating sinusoidally at 60.0 Hz. If the diameter of a red blood cell is 8.00 mm, determine the maximum emf that can be generated around the perimeter of a cell in this field.
Answer:
fem = 7.58 10⁻⁵ V
Explanation:
For this exercise we use Faraday's law
fem = [tex]- \frac{d \Phi _B}{dt}[/tex]
the magnetic flux is
Ф_B = B. A = B A cos θ
Tje bold are vectros. Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1
The red blood cell area is
A =π r²
indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m
the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T, therefore we can write it
B = B₀ sin (wt) = B₀ sin( 2π f t)
we substitute
fem = - A dB / dt
fem = - A B₀ [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]
fem = - π r² Bo (2πf cos 2πft)
the maximum electromotive force occurs when the function is ±1
fem = 2 π² r² B₀ f
let's calculate
fem = 2π² (8.00 10⁻³)² 1.00 10⁻³ 60
fem = 7.58 10⁻⁵ V
