Answer:
Let's define t = 0s (the initial time) as the moment when Car A starts moving.
Let's find the movement equations of each car.
A:
We know that Car A accelerations with a constant acceleration of 5m/s^2
Then the acceleration equation is:
[tex]A_a(t) = 5m/s^2[/tex]
To get the velocity, we integrate over time:
[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]
Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:
[tex]V_a(t) = (5m/s^2)*t[/tex]
To get the position equation we integrate again over time:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
[tex]V_b(t) =20m/s[/tex]
To get the position equation, we can integrate:
[tex]P_b(t) = (20m/s)*t + P_0[/tex]
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
[tex]P_b(t) = (20m/s)*t + 100m[/tex]
Now we can answer this:
1) The two cars will meet when their position equations are equal, so we must have:
[tex]P_a(t) = P_b(t)[/tex]
We can solve this for t.
[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]
We only care for the positive solution, which is:
[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]
Car A reaches Car B after 11.48 seconds.
2) How far does car A travel before the two cars meet?
Here we only need to evaluate the position equation for Car A in t = 11.48s:
[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s
[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]
What would the separation between two identical objects, one carrying 4 C of positive charge and the other 4 C of negative charge, have to be if the electrical force on each was precisely 8 N
Answer:
7.46×10⁻⁶ m
Explanation:
Applying,
F = kqq'/r²............ Equation 1
make r the subject of the equation
r = √(F/kqq').......... Equation 2
From the question,
Given: F = 8 N, q' = q= 4 C
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 2
r = √[8/(4×4×8.98×10⁹)]
r = √(55.7×10⁻¹²)
r = 7.46×10⁻⁶ m
A small ball of uniform density equal to 1/2 the density of water is dropped into a pool from a height of 5m above the surface. Calculate the maximum depth the ball reaches before it is returned due to its bouyancy. (Omit the air and water drag forces).
Answer:
1.67 m
Explanation:
The potential energy change of the small ball ΔU equals the work done by the buoyant force, W
ΔU = -W
Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5
ΔU = mgΔh
ΔU = ρVgΔh
Now, W = ρ'VgΔh' where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h
So, ΔU = -W
ρVgΔh = -ρ'VgΔh'
ρVg(h - 5) = -ρ'Vgh
ρ(h - 5) = -ρ'h
Since the density of the small ball equals 1/2 the density of water,
ρ = ρ'/2
ρ(h - 5) = -ρ'h
(ρ'/2)(h - 5) = -ρ'h
ρ'(h - 5)/2 = -ρ'h
(h - 5)/2 = -h
h - 5 = -2h
h + 2h = 5
3h = 5
h = 5/3
h = 1.67 m
So, the maximum depth the ball reaches is 1.67 m.
Two metal spheres are made of the same material and have the same diameter, but one is solid and the other is hollow. If their temperature is increased by the same amount:_______.
A) the solid sphere becomes heavier and the hollow one lighter.
B) the solid sphere becomes bigger than the hollow one.
C) the hollow sphere becomes bigger than the solid one.
D) the two spheres remain of equal size.
E) the solid sphere becomes lighter and the hollow one heavier.
Answer:
D) the two spheres remain of equal size.
Explanation:
Since the body of the sphere is made up of both the same material. Thus the orientation will not affect the expansion. That is solid upon solid and hollow upon the hollow sphere. Hence it can be said that both the sphere expands and is due to the material used for making both of them is the same.If there are no other changes, explain what effect reducing the mass of the car will have on its acceleration when starting to move.
Answer:
when the mass of an object is decreased, the acceleration will increase
when mass is increased, acceleration decreases
Receptor elétrico 5 pontos Dispositivo que converte energia elétrica em outra forma de energia, não exclusivamente térmica. Exemplos: motores elétricos, ventiladores, liquidificadores, geladeiras, aparelhos de sons, vídeos, celulares, computadores?
Answer:
Electromechanical transducer and Electrical receiver.
Explanation:
Electromechanical transducer is the part of a communication system which converted electrical waves or electrical energy into sound waves. The most common example loudspeaker while on the other hand, Electrical receiver is a device that converts electrical energy into another form of energy, except thermal. Examples are cell phones, computers and television.
A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find
(i) Final velocity
(ii)The time taken
Answer:
(I)
[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]
The diagram here shows an image being formed by a convex lens. Compared to the object at right, the image at left is-
larger and upright.
smaller and upright.
smaller and upside down.
larger and upside down.
Answer:
Smaller and upside down
Explanation:
To answer the question, we must recognise that the characteristics of the image formed by a convex lens depends on the position of the object from the lens.
From the diagram given in the question above, the following data were obtained:
1. The image is smaller than the object.
2. The image is inverted i.e upside down.
3. The image is closer to the lens
4. The image between 2f and f
Now, considering the options given in question above, the correct answer to the question is:
The image is smaller and upside down.
Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant force is directed vertically upward.
Answer:
how to solve this problem ???????
The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.
Resultant of the two forces
The resultant of the two forces is determined by resolving the force into x and y component as shown below;
[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]
where;
F1 = 500 NF2 = 600 NValue of Angle θThe value of Angle θ is determined from equation (1)
-500sinθ + 600sin(30) = 0
500sinθ = 600sin(30)
500sinθ = 300
sinθ = 3/5
θ = 36.87⁰
Resultant of the two forcesThe resultant of the forces is determined using the second equation;
500cosθ + 600cos(30) = R
500 x cos(36.87) + 600 x cos(30) = R
919.6 N = R
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Why is the melting of ice a physical change?
A. It changes the chemical composition of water.
B. It does not change the chemical composition of water.
C. It creates new chemical bonds.
D. It forms new products.
E. It is an irreversible change that forms new products.
It does not change the chemical composition of water.
If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.
Answer:
The impact force is 98000 N.
Explanation:
mass = 10 tons
The impact force is the weight of the object.
Weight =mass x gravity
W = 10 x 1000 x 9.8
W = 98000 N
The impact force is 98000 N.
What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?
Answer:
The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.
1
An astronaut weighs 202 lb. What is his weight in newtons?
Answer:
978.6084 Newton
Explanation:
Given the following data;
Weight = 220 lbTo find the weight in Newtown;
Conversion:
1 lb = 4.448220 N
220 lb = 220 * 4.448220 = 978.6084 Newton
220 lb = 978.6084 Newton
Therefore, the weight of the astronaut in Newton is 978.6084.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, the weight of an object is given by the formula;
Weight = mg
Where;
m is the mass of the object.g is the acceleration due to gravity.Note:
lb is the symbol for pounds.N is the symbol for Newton.~~~NEED HELP ASAP~~~
Please solve each section and show all work for each section.
Explanation:
Forces on Block A:
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as
[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]
Substituting (2) into (1), we get
[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]
where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]
Forces on Block B:
Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as
[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]
From (5), we can solve for N as
[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]
Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by
[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]
Substituting (7) into (4) we get
[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]
Collecting similar terms together, we get
[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]
or
[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]
Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]
Find the coefficient of kinetic friction between a 3.80-kg block and the horizontal surface on which it rests if an 87.0-N/m spring must be stretched by 6.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.
Answer:
μ = 0.15
Explanation:
Let's start by using Hooke's law to find the force applied to the block
F = k x
F = 87.0 0.065
F = 5.655 N
Now we use the translational equilibrium relation since the block has no acceleration
∑ F = 0
F -fr = 0
F = fr
the expression for the friction force is
fr = μ N
if we write Newton's second law for the y-axis
N -W = 0
N = W = mg
we substitute
F = μ mg
μ = F / mg
μ = [tex]\frac{5.655}{3.8 \ 9.8}[/tex]
μ = 0.15
A source emits sound at a fixed constant frequency f. If you run towards the source, the frequency you hear is
Answer:
increased because as you run into each sound wave the time between each sound decreases meaning the period of each wave decreases to your years and since f=1/T and T is decreasing by greater than 0, f must increase.
Explanation:
If an object travels on a circular path is an acceleration? What is changing to cause an acceleration?
A 100-m long transmission cable is suspended between two towers. If the mass density is 18.2 g/cm and the tension in the cable is 6543 N, what is the speed (m/s2) of transverse waves on the cable
Una pelota se lanza verticalmente hacia arriba desde la azotea de un edificio con una velocidad inicial de 35 m/s. Si se detiene en el aire a 200 m del suelo, ¿Cuál es la altura del edificio?
a. 138,8 m
b. 51.2 m
c. 71,2 m
d. 45,0 m
Which image illustrates reflection?
A
B
с
D
Answer: I beleive A
Explanation:
Answer:
A
Explanation:
We can see the light being reflected off the mirror.
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory? A. 3 seconds B. 4 seconds C. 2 seconds D. 6 seconds
Answer:
A (3 seconds)
Explanation:
Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.
Lets identify our givens.
Givens:
Horizontal speed= 30m/s
Vertical Speed= 30 m/s
Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0
The ball is launched from the ground so y0=0
Final vertical velocity= 0
This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use
vy=vy0+ayt
Plug in our givens
0=30-10t
solve for t
t=3 seconds
suppose you have a block resting on a horizontal smooth surface. th block with a mass m is attached to a horizontal spring which is fixed at one end. the spring can be compressed and stretched. the mass is pulled to one side then released what is the formula required
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
What is meant by spring constant ?The spring constant of a spring is defined as the measurement of ratio of the force that is exerted on the spring to the displacement caused by it.
Here,
The mass of the block = m
Let F be the applied force on the spring and k be the spring constant.
When the mass attached to the spring is pulled to one side then released, it executes SHM.
Therefore we can write that, the applied force,
F = kx
Restoring force = -kx
According to Newton's law, we know that,
F = ma
So,
ma = -kx
Therefore, the acceleration,
a = (-k/m) x
For an SHM, the acceleration is given as,
a = -ω²x
Therefore, we can write that,
-ω²x = (-k/m) x
ω² = k/m
So, the time period of the spring,
T = 2[tex]\pi[/tex]/ω
T = 2[tex]\pi[/tex][√(m/k)]
Hence,
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
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How are Newton’s 1 and 2 law related?
true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields
Answer:
True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigger than the object and the screen is 5 m away from the mirror as shown in fig 5.2, calculate the focal length of the mirror.
Answer:
f = 1 m
Explanation:
The magnification of the lens is given by the formula:
[tex]M = \frac{q}{p}[/tex]
where,
M = Magnification = 4
q = image distance = 5 m
p = object distance = ?
Therefore,
[tex]4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m[/tex]
Now using thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\[/tex]
f = 1 m
If the sum of the external forces on an object is zero, then the sum of the external torques on it
must also be zero.
A) True
B) False
Answer:
True.
Explanation:
If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.
The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.
Hence, the given statement is true.
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g
Answer:
w = 1,066 rad / s
Explanation:
For this exercise we use Newton's second law
F = m a
the centripetal acceleration is
a = w² r
indicate that the force is the mass of the body times the acceleration
F = m 0.58g = m 0.58 9.8
F = 5.684 m
we substitute
5.684 m = m w² r
w = [tex]\sqrt{5.684/r}[/tex]
To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m
w = [tex]\sqrt{ 5.684/5}[/tex]
w = 1,066 rad / s
10 A turning pork creates sound cares
with
Frequency of 170Hz: To the
speed of sound in is in 340mls
calculate the wave
wave length
of
in air is
the sound wales.
Answer:
2m
Explanation:
wavelength=speed/frequency
=340/170
=2m
A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.30 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged
Answer:
The velocities of the skaters are [tex]v_{1} = 3.280\,\frac{m}{s}[/tex] and [tex]v_{2} = 0.024\,\frac{m}{s}[/tex], respectively.
Explanation:
Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:
First skater
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex] (1)
Second skater
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex] (2)
Where:
[tex]m_{1}[/tex] - Mass of the first skater, in kilograms.
[tex]m_{2}[/tex] - Mass of the second skater, in kilograms.
[tex]v_{1,o}[/tex] - Initial velocity of the first skater, in meters per second.
[tex]v_{1}[/tex] - Final velocity of the first skater, in meters per second.
[tex]v_{b}[/tex] - Launch velocity of the meter, in meters per second.
[tex]v_{2}[/tex] - Final velocity of the second skater, in meters per second.
If we know that [tex]m_{1} = 70\,kg[/tex], [tex]m_{b} = 0.043\,kg[/tex], [tex]v_{b} = 32\,\frac{m}{s}[/tex], [tex]m_{2} = 58.5\,kg[/tex] and [tex]v_{1,o} = 3.30\,\frac{m}{s}[/tex], then the velocities of the two people after the snowball is exchanged is:
By (1):
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex]
[tex]m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}[/tex]
[tex]v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}[/tex]
[tex]v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)[/tex]
[tex]v_{1} = 3.280\,\frac{m}{s}[/tex]
By (2):
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex]
[tex]v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}[/tex]
[tex]v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}[/tex]
[tex]v_{2} = 0.024\,\frac{m}{s}[/tex]
find the weight of a body of mass 200kg on the earth at a latitude 30°.(R=6400 km ,g=9.8m/s²,ω=7.27×10⁻⁵ rad/sec)
Answer:
................ftf6x
For waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths.
a) true
b) false
Answer:
false.
Explanation:
We know that for a wave that moves with velocity V, with a wavelength λ, and a frequency f, we have the relation:
V = λ*f
So, if the velocity is constant and we increase the frequency to:
f' > f
we will have a new wavelength λ'
Such that:
V = f'*λ'
And V = f*λ
Then we have:
f'*λ' = f*λ
Solvinf for λ', we get:
λ' =(f/f')*λ
And because:
f' > f
then:
(f/f') < 1
Then:
λ' =(f/f')*λ < λ
So, if we increase the frequency, we need to decrease the wavelength.
So, for higher frequency waves, we must have proportionally shorter wavelengths.
Then we can conclude that the given statement:
"or waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths"
is false.
