Answer:
a) La magnitud del desplazamiento es de 5 m
La magnitud de la fuerza es 20 N
b) El trabajo realizado por la fuerza es de 100 J
c) El ángulo entre la fuerza y el plano es 0 °
Explanation:
a) La magnitud del desplazamiento se encuentra por la relación;
[tex]\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}[/tex]
Lo que da;
[tex]\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m[/tex]
De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;
[tex]\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]
Lo que da;
[tex]\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N[/tex]
b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X
∴ Ancho = 20 N × 5 m = 100 N · m = 100 J
c) La dirección de la fuerza viene dada por la siguiente fórmula;
[tex]tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}[/tex]
La dirección del plano viene dada por la siguiente fórmula;
[tex]tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}[/tex]
Por tanto, el ángulo entre la fuerza y el plano = 0 °
La fuerza actúa a lo largo del plano.
why do objects in equilibrium have a net force of 0?
Answer:
Because there is no net force acting o an object in equilibrium.
Explanation:
When object is in equilibrium (either at rest or moving with constant velocity, the net force acting o it is zero.A vector can only have zero magnitude fall of its components are zero this means that all forces acting on the object are balanced.
The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the sine of the angle of incidence is?
Pls help
Answer:
0.75
Explanation:
[tex]refractive \: index \: = \frac{3.0 \times {10}^{2} }{2.0 \times {10}^{2} } [/tex]
= 1.5
[tex]refractive \: index = \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) } [/tex]
[tex]1.5 = \ \frac{ \sin(i) }{ \sin(30) } [/tex]
1.5 × ½ = sin(i)
[tex] \sin(i) = 0.75[/tex]
IMPORTANT 3 QUESTIONS!
Answer:
7. 20,000,000 mL.
8. 8,000 m.
9. 120,000 secs.
10. 4
Explanation:
7. Determination of the volume in millilitres (mL)
Volume in litre (L) = 20,000 L
Volume in millilitres (mL) =..?
The volume in mL can be obtained as follow:
1 L = 1,000 mL
Therefore,
20,000 L = 20,000 x 1,000 = 20,000,000 mL.
Therefore, 20,000 L is equivalent to 20,000,000 mL.
8. Determination of the distance in metre (m)
Distance in mile = 5 mile
Distance in metre =?
First, we shall convert from mile to kilometre.
This can be done as follow:
1 mile = 1.6 km
Therefore,
5 mile = 5 x 1.6 = 8 km
Finally, we shall convert 8 km to metre (m).
This is illustrated below:
1 km = 1,000 m
Therefore,
8 km = 8 x 1,000 = 8,000 m
Therefore, 5 miles is equivalent to 8,000 m.
9. Determination of the time in seconds.
Time = 400 minutes for 5 days.
First, we shall convert 400 mins to hour.
This is illustrated below:
60 minutes = 1 hour
Therefore,
400 mins = 400/60 = 20/3 hours
The time (hours) is 20/3 hours in 1 day.
Therefore, the time (hours) in 5 days will be = 20/3 x 5 = 100/3 hours.
Next, we shall convert 100/3 hours to minutes.
This is illustrated below:
1 hour = 60 minutes
Therefore,
100/3 hours = 100/3 x 60 = 2000 mins
Finally, we shall convert 2000 mins to seconds.
This is illustrated below:
1 mins = 60 secs
2000 mins = 2000 x 60 = 120,000 secs.
Therefore, the time is 120,000 secs.
10. Determination of the number of significant figures.
To obtain the significant figures of a number, we simply count all the numbers available.
Therefore, the number of significant figures for 9876 is 4.
Un pez llamado PARGO ROJO vive a grandes profundidades. Si se pesca, al salir a la superficie puede tomar el aspecto de la foto porque su vejiga natatoria aumenta de volumen al disminuir la presión. La ley que lo explica es...
Answer:
Hay diversas leyes que podemos usar acá.
Acá sabemos que la vejiga aumenta su tamaño al reducir la presión, esto tiene sentido, pues al haber menos presión, hay menos fuerza que comprime la vejiga, lo que le permite aumentar su volumen.
Acá tenemos una relación inversa de la forma: V = K/P
Una relación inversa donde la presión esta en el denominador y K es un termino que no depende ni del volumen ni de la presión.
Entonces, a medida que aumenta P, el denominador aumenta, por lo que el valor del volumen decrece.
Un ejemplo de una ecuación similar es la del gas ideal, por ejemplo, para un gas ideal dentro de un globo de volumen V para una dada presión P:
V = nRT/P
donde n es el numero de moles, R es la constante termodinámica y T es la temperatura, acá podemos ver que esta ecuación tiene la misma forma fundamental que la escrita arriba.
briefly explain what EMF of a cell cell is
Answer:
The emf of a cell is the sum of the electric potential differences (PDs) produced by a separation of charges (electrons or ions) that can occur at each phase boundary (or interface) in the cell. The magnitude of each PD depends on the chemical nature of the two contacting phases.
Explanation:
pls mark brainliest
Every time Dylan takes his dog Frisbee for a walk, the dog always pulls hard on the leash. This nearly pulls Dylan's arm out of its socket and Frisbee almost chokes himself! Explain these effects in terms of Newton's third law.
Answer:
See explanation below for Newton's third law of "action generates reaction" for this example
Explanation:
From Newton's third law: "For every action there is an equal and opposite reaction."
Frisbee is pulling on the leash with force F which is felt by Dylan as a tension force in the leash trying to pull his arm out of its socket, and Frisbee feels an equal and opposite tension force from the leash chocking him.
3. What two characteristics do you need to include when defining a force? (a) speed and direction (b) strength and direction (c) acceleration and time (d) distance and direction
4. When a bug traveling west hits the windshield of a car that is traveling east, what can be said about the collision? (a) The bug feels a stronger force than the car. (b) The car and the bug feel the same size force. (c) The car accelerates more than the bug. (d) The bug does not accelerate due to the force.
5. What is the reaction force to the force on a baseball from a bat? (a) The force of the bat on the ball (b) The force of the ball on the ground (c) The force of the bat on the players hand (d) The force of gravity on the
Answer:
2
yes
Explanation:
been submitted to the Theme 20 participants will be there was a lot of fun and I will be there at Home
Help please answer the two questions
Explanation:
this is it. hope u understand
The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total momentum a collision or explosion is equal to the total momentum a collision or explosion.what is momentum
Answer:
The momentum of an object is equal to the product of its mass and its velocity.
Explanation:
Consider an object of mass [tex]m[/tex] travelling at a velocity [tex]\vec{v}[/tex]. The momentum [tex]\vec{p}[/tex] of this object would be:
[tex]\vec{p} = m \cdot \vec{v}[/tex].
For the law of conservation of momentum, consider two objects: object [tex]\rm a[/tex] and object [tex]\rm b[/tex]. Assume that these two objects collided with each other.
Let [tex]m_{\rm a}[/tex] and [tex]m_{\rm b}[/tex] denote the mass of the two objects. Let [tex]\vec{v}_{\rm a}(\text{initial})[/tex] and [tex]\vec{v}_{\rm b}(\text{initial})[/tex] denote the velocity of the two object right before the interaction. Let [tex]\vec{v}_{\rm a}(\text{final})[/tex] and [tex]\vec{v}_{\rm b}(\text{final})[/tex] denote the velocity of the two objects right after the interaction. The momentum of the two objects right before the collision would be [tex]m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})[/tex] and [tex]m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})[/tex], respectively. The momentum of the two objects right after the collision would be [tex]m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})[/tex] and [tex]m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})[/tex], respectively.The sum of the momentum of the two objects would be:
[tex]m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})[/tex] right before the collision, and[tex]m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})[/tex] right after the collision.Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:
[tex]m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})[/tex].
What is the cat's displacement t=0.5s to 1.5s,
Answer:
3.75m
Explanation:
area under graph
5+2.5=7.5
7.5/2=3.75
3.75x0.5=1.875
1.875x2=3.75m
Monochromatic light of wavelength 649 nm is incident on a narrow slit. On a screen 2.25 m away, the distance between the second diffraction minimum and the central maximum is 1.99 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.
Answer:
a)0.51°
b)1.47×10^-4m
Explanation:
a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.
bsin(θ)= mλ.........................(*)
Where b= width of the slit
The distance on the screen from Central angle can be expressed as
Sin(θ)= y/d............. (**)
d and y is the horizontal distance between slit and screen
If we input eqn(**) into equation (*) we have
y= mλd/b................(z)
In order to find angle (θ) we have
(θ)= sin-(1.99×10^-2)/2.25
= 0.51°
Therefore, angle of diffraction θ of the second minimum is 0.51°
b)to find the width of the sloth using eqn(z) by substitute the values, we have
b= (2)(649×10^-9)(2.25)/1.99×10^-2
b= 1.47×10^-4m
Therefore, the width of the slit is 1.47×10^-4m
Which statement correctly compares the number of protons and electrons in atoms of rubidium (Rb) and calcium (Ca)?
(1 point)
Rubidium has fewer protons and fewer electrons than calcium.
Rubidium has fewer protons and more electrons than calcium.
Rubidium has more protons and fewer electrons than calcium.
Rubidium has more protons and more electrons than calcium.
Answer
#1. number of valence electrons
#2. oxygen
#3. Rubidium has more protons and more electrons than calcium.
#4. 20
#5. helium (he)
Explanation:
Answer:
Rubidium has more protons and more electrons than calcium.
Explanation:
for connexus honors chemistry unit 3 lesson 3 quick check :)
6. What are the methods to control noise pollution?
Answer: Some of the ways to control noise pollution are as follows: (1) Control at Receiver's End (2) Suppression of Noise at Source (3) Acoustic Zoning (4) Sound Insulation at Construction Stages (5) Planting of Trees (6) Legislative Measures.
Answer:
You have to:
a) Improve your insulation.
b) Install a fence
c) Use modern Acoustic wall panels
d) Plant trees
e) Reduce electronic volumes,e.t.c.
Explanation:
okay.
The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive
Answer:
Positive
Explanation:
In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive
what is the scientific notation for 1,500,000
Answer:
[tex]1.5 * 10^6[/tex]
Explanation:
Given
Digit: 1,500,000
Required
Represent using scientific notation
We start by expressing the given digit as a product of 1.5 and a multiple of 10
[tex]Digit = 1.5 * 1,000,000[/tex]
1,000,000 can be represented as 10⁶
So; the digit becomes
[tex]Digit = 1.5 * 10^6[/tex]
Hence, the scientific notation of 1,500,000 is 1.5 * 10⁶
The Doppler effect a. can be applied only to sound waves. b. is the basic explanation for the blue shift of light in our Universe. c. occurs when the frequency of sound waves received is lower if the wave source is moving toward you than if it's moving away. d. occurs when the pitch of a sound gets lower if the source is receding.
Answer:
a. can be applied only to sound waves.
c. occurs when the frequency of sound waves received is lower if the wave source is moving toward you than if it's moving away
d. occurs when the pitch of a sound gets lower if the source is receding.
Explanation:
[tex].[/tex]
the gravitational force between two masses at distance of 2.5×10^6 metre is 250 Newton . what should be distance between them to reduce the gravitational force by half?
Answer:
distance between should be 1.25 × 10^6 m
Explanation:
Force is proportional to separation distance.
[tex]250 \: newtons = 2.5 \times {10}^{6} \: metres \\ ( \frac{1}{2} \times 250) \: newtons = d \\ \\ d = \frac{125 \times 2.5 \times {10}^{6} }{250} \\ d = 1.25 \times {10}^{6} \: metres[/tex]
Answer:
distance between should be 1.25×10^6.
Explanation:
force Is proportional to separation .
mrk above answer branilest .
Velocity may be positive,negative or zero.Explain this statement with diagram.
Answer:
/
Explanation:
Write any three importance of Measurement.
Stay Safe,Stay Healthy and Stay Happy
Please answer me this question fast, PLEASE
Explanation:
The three importance of measurement are as follows :-
measurement is used to compare the items when barter system takes place .measurement helps in weighing the foods , groceries etc .Measurement are important in laboratory when there experiments is being taken for different purposes .Hope it is helpful to you
✌️✌️✌️✌️✌️✌️✌️
1. Si tengo medio kilo de fruta y te doy un cuarto y tú me das tres cuartos de kilo, ¿cuánto tengo? 2. Si en una carrera te queda por recorrer la mitad de la mitad de 1 km, ¿cuánto te falta? 3. ¿Qué pesa mas, un kilo y medio de hierro o tres medios kilos de paja? porfavor es urgente.
Answer:
1. Tienes 1 kg de fruta.
2. Queda por recorrer 1/4 km.
3. Ambos pesan lo mismo.
Explanation:
1. Tienes 1/2 kg y cuando te doy 1/4 te queda:
[tex] m = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} [/tex]
Ahora cuando te doy 3/4 kg te queda en total:
[tex] m_{T} = \frac{1}{4} + \frac{3}{4} = 1 kg [/tex]
Por lo tanto, tienes 1 kg de fruta al final.
2. Si falta por recorrer la mitad de la mitad, tenemos:
[tex] d = \frac{1/2}{2} = \frac{1}{4} [/tex]
Entonces, queda por recorrer 1/4 km.
3. El peso (P) del hierro es:
[tex] P = m*g [/tex]
[tex] P = (1 + 1/2)kg*9.81 m/s^{2} = 14.72 N [/tex]
Y el peso de la paja es:
[tex] P = 3/2 kg*9.81 m/s^{2} = 14.72 N [/tex]
Por lo tanto, ambos pesan lo mismo.
Espero que te sea de utilidad!
If, on a particular day, atmospheric pressure is 730 mm
of mercury, what is this
a) in pascals
b) in atmospheres
c) in millibars?
Answer:
b) in atmostphere
Explanation:
I made lots of research and it should be b.
a. On a particular day, atmospheric pressure is 730 mm of mercury In pascals 97392.2896252, b. On a particular day, atmospheric pressure is 730 mm of mercury in atmospheres 0.960526316 atm, c. on a particular day, atmospheric pressure is 730 mm of mercury in millibars 973.70355547 mb.
a) To convert atmospheric pressure from mm of mercury to pascals, we can use the conversion factor: 1 mm Hg = 133.322387415 pascals.
Pressure in pascals = 730 mm Hg * 133.322387415 = 97392.2896252 pascals.
b) To convert atmospheric pressure from mm of mercury to atmospheres, we can use the conversion factor: 1 atm = 760 mm Hg.
Pressure in atmospheres = 730 mm Hg / 760 = 0.960526316 atm.
c) Atmospheric pressure is often expressed in millibars (mb), where 1 atmosphere is approximately equal to 1013.25 millibars.
Pressure in millibars = 0.960526316 atm * 1013.25 mb/atm = 973.70355547 mb.
Atmospheric pressure is the force exerted by the weight of air molecules in the Earth's atmosphere. It can be measured in various units, including millimeters of mercury, pascals, atmospheres, and millibars.
Converting between these units involves using conversion factors based on established relationships between them.
In this case, the provided pressure of 730 mm Hg was converted to pascals, atmospheres, and millibars using the appropriate conversion factors.
Conversions help express pressure measurements in units that are more suitable for different contexts or calculations.
Know more about atmospheric pressure:
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A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end is allowed to fall. What is the angular acceleration as it is released
Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
A body decelerates uniformly to a constant speed and after some time it accelerates uniformly Draw the shape of speed time graph for such a motion . Label the three sections of this graph. What is the quantity which is measured by the area occupied under speed time graph ?
Answer:
here I am just giving an idea of how the graph will be like ...
In the pic..
Hope it helped u if yes mark me BRAINLIEST!
Tysm!
:)
A girl walks 55 m downhill in 3.2 s. What is her average velocity?
Answer:
V = 17.2
Explanation:
Speed = distance(m) / time(s)
= 55m / 3.2s
=17.2 m/s
Explain what happens to the sound waves when a singer hits the high pitched notes during the National Anthem. Be sure to use the terms amplitude and frequency in your answer.
Answer: The frequency increases as the pitch increases, and the amplitude increases as the volume increases
Explanation:
Waves have the property of:
v = f*λ
where v is the speed of the wave (which is almost constant for soundwaves, v = 340 m/s)
f is the frequency of the wave, and λ is the wavelength.
Now, we know that when the pitch of a note increases, also does the frequency of the soundwave (so the wave oscillates faster).
Now, we also want to include the amplitude of the soundwave in this.
The amplitude is related to the volume of the soundwave (actually is related to the energy, and as higher is the energy, more "loud" is the sound).
As the high pitch part is usually "louder", we can assume that we have an amplitude increase.
Then the answer would be:
"The frequency increases as the pitch increases, and the amplitude increases as the volume increases"
An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,000 Hz as it approaches and then passes by the observer without slowing down. Assuming the speed of sound is 340 m/s, how much of a frequency change did the observer hear?
7351.35Hz
f0= v-Vo/v-Vs × FSA
= 340-0 /340+30 ×8000
= 340/370× 8000
= 7351.35hz
The length and breadth of rectangular sheet of metal are 4.234m, 1.005,
and 2.01cm respectively.
Give the area and volume of sheet to correct significant figures..
Answer:
The area of the sheet of metal = 4.255 m²
The volume of the sheet of metal = 0.08553 m³
Explanation:
The area of a rectangle is given by the following relation;
Area, A = Length × Breadth
Therefore, the area of the rectangular sheet of metal, A[tex]_m[/tex], is given as follows;
A[tex]_m[/tex] = 4.234 m × 1.005 m = 4.255 m²
The area of the sheet of metal = 4.255 m²
The volume of the metal sheet id given by the relation, volume V[tex]_m[/tex]= area × Thickness
The thickness of the sheet of metal = 2.01 cm = 0.0201 m
Therefore,
V[tex]_m[/tex] = 4.255 m × 0.0201 m = 0.08553 m³
The volume of the sheet of metal = 0.08553 m³.
Answer:
Are yarr mei bhi Indian hu.....
Explanation:
Tumse milkar accha laga
Write a formula by showing the relation among density mass and volume.
Lets do
[tex]\boxed{\sf P=\dfrac{m}{v}}[/tex]
[tex]\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}[/tex]
[tex]\\ \sf\longmapsto Mass=Density\times volume[/tex]
[tex]\\ \sf\longmapsto Volume=\dfrac{Mass}{Density}[/tex]
SI unit of density is kg/m^3SI unit of mass is kgSI unit of Volume is m^3A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 76.8 m/s2 for 1.99 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?
Answer:
Explanation:
Take note that when the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be moving up until all the forces of gravity would dominate and change the direction of the rocket. Hence, there will be a need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.
Given:
a = 76.8 m/s^2
t = 1.99 s
Solution:
d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.99) + 0.5 (76.8) (1.99)^2
d = 0+38.4×3.9601
d = 152.068m
vf = vi + at
vf = 0 + 76.8 (1.99)
vf = 152.83 m/s (velocity when the fuel is consumed completely)
Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 152.83+(-9.8) (t)
0 = 152.83 + (-9.8) (t)
-152.83 = -9.8t
t = 152.83/9.8 s
t = 15.59s
Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 152.83 (15.59) + 0.5 (-9.8) (15.59^2)
d = 2382.6197- 1190.93
d = 1191.68m
Then, we determine the maximum altitude:
d1 + d2 = 152.068 m + 1191.68m = 1343.748m
When light travels from air into water, Group of answer choices its wavelength changes, but its velocity and frequency does not change its velocity remains constant, but its frequency and wavelength changes its velocity, wavelength and frequency all change its velocity and wavelength changes, but its frequency does not change
Answer:
its velocity, wavelength and frequency all change
Explanation:
Whenever a wave crosses the boundary between two media of different densities, its velocity, frequency and wavelength changes. This appears as a bending of the wave as it crosses the boundary from one medium to another.
Refraction is a fundamental property of waves. Hence when light is travelling from air into water, it wavelength, speed and frequency all changes at the interface between the two media.