Answer:
1. SG
true
=2.689
2. The object is probably some sort of minerals and rocks such as Feldspar, Corals, Beryl, etc.
Explanation:
Given:
mass in the air= 0.0675 kg
mass in water= 0.0424 kg
The specific gravity of the object will be 2.6892. It is the ratio of the density of the given fluid and the standard fluid.
What is density?Density is specified as the mass divided by the volume. It is represented by the unit of measurement as kg/m³.
The mass of the object in air;
m=Vρ₀
m=0.0675 kg
Buoyant force on the object;
B= Vρₐg
For equilibrium;
N+B=m₀g
n=m₀g-Vρₓg
N/g=m₀-Vρₓ
N/g=0.0424 kg
[tex]\rm \frac{V\rho_0}{V\rho_x} =\frac{0.0675 }{m_0-0.0424 \ kg} \\\\ \frac{\rho_0}{\rho_x} =\frac{0.0675}{0.0675-0.0424} \\\\ \frac{\rho_0}{\rho_x} =2.6892[/tex]
Hence, the specific gravity of the object will be 2.6892.
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A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal to 3 1.15 / kg m . Find the mass of air and the overall (average) specific volume
Complete question:
A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.
Answer:
The mass of the air is 5.32 kg
The specific volume is 5.52 x 10⁻³ m³/kg
Explanation:
Given;
total volume of the container, [tex]V_t[/tex] = 5 m³
mass of granite, [tex]m_g[/tex] = 900 kg
density of granite, [tex]\rho _g[/tex] = 2,400 kg/m³
density of air, [tex]\rho_a[/tex] = 1.15 kg/m³
The volume of the granite is calculated as;
[tex]V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3[/tex]
The volume of air is calculated as;
[tex]V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3[/tex]
The mass of the air is calculated as;
[tex]m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg[/tex]
The specific volume is calculated as;
[tex]V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg[/tex]
On topographic maps, contour lines that are farther apart indicate what ?
Answer:
if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.
Answer:
gentle slopes
Explanation:
Each year 500 runners run up the stairs to the 86th floor of the Empire State Building in New York City. There are 1576 steps and each step is 0.241 m high. In 2003, Australian Paul Crake (20-29 age group) set the overall record by reaching the 86th floor in 9:33. His mass was 70.0 kg. Question 2 HomeworkUnanswered What was Paul Crake's power output during this climb
Answer:
The power is 465.44 W.
Explanation:
mass, m = 70 kg
number of steps, n = 1576
height of each step, h = 0.241 m
time taken, t = 9.33 min= 9.33 x 60 s
The power is given by the rate of doing work.
W = n m g h
W = 1576 x 70 x 9.8 x 0.241
W = 260553.776 J
The power is given by
[tex]P = \frac{W}{t}\\\\P = \frac{260553.776}{9.33\times 60}\\\\P = 465.44 W[/tex]
what is entrapersonal environment
Answer:
The interpersonal environment is considered to be a subset of the organizational environment – defined as the employee’s perception of the practices, policies, and processes of an organization
Explanation:
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
what is measured by the ammeter
Answer:
amperes
Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.
Explanation:
hope it helps
A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What
is the magnitude of the impulse?
The magnitude of the impulse of the truck is equal to 4000 Kg.m/s.
What is impulse?Impulse can be described as the integral of a force over the time interval for which it acts. Impulse is also a vector quantity since force is a vector quantity. Impulse can be applied to an object that generates an equivalent vector change in its linear momentum.
The S.I. unit of impulse is N⋅s and the dimensionally equivalent unit of momentum is kg⋅m/s. A resultant force gives acceleration and changes the velocity of an object for as long as it acts.
Given the mass of the truck, m= 2000 Kg
The initial speed of the truck, u = 6 m/s
The final speed of the truck, v = 4 m/s
The change in the linear momentum is equal to the impulse.
I = ΔP = mv - mu
I = 2000 ×4 - 2000 × 6
I = 8000 - 12000
I = - 4000 Kg.m/s²
Therefore, the magnitude of the impulse is 4000 Kg.m/s².
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Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if the system it is rotated about (a) one end of the rod and (b) the center of the rod?
Explanation:
a) [tex]I=\displaystyle \sum_{i}m_ir_i^2[/tex]
where [tex]r_i[/tex] is the distance of the mass [tex]m_i[/tex] from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,
[tex]I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2[/tex]
b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,
[tex]\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2[/tex]
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
Lauren pushed a crate of mass 28.9 kg a distance of 2.7 meters along a horizontal surface. On that part of the surface, the crate could slide with negligible friction. Lauren exerted a constant force of 121 newtons for 2.7 meters. The crate then slid down an inclined plane of height 1.8 meters, also with negligible friction. As the crate slid down the plane, the only significant forces on it were the normal force from the plane and gravity (with g = 9.81 meters per second squared). At the bottom of the incline, the crate began sliding along a horizontal surface with ordinary kinetic friction. The coefficient of friction between this surface and the crate was 0.41. This surface ended at a vertical wall after a distance of d2 = 5.2 meters but the crate did not slide that far. Attached to the vertical wall was a long ideal spring with length dy = 3.4 meters and a spring constant of 154 newtons per meter.
Required:
Calculate the distance that the crate compressed the spring before coming to rest.
Answer:
3.034 m
Explanation:
From the law of conservation of energy, the energy at the top of the incline equals the energy at the bottom of the incline since at the top of the incline, the horizontal surface is frictionless and along the incline there is no friction.
The work done in moving the crate a distance, d = 2.7 m with a force of F = 121 N to the top of the incline is W = Fd = 121 N × 2.7 m = 326.7 J.
From work-kinetic energy principles, this work W = kinetic energy of the crate at the top of the incline, K₁.
Now, the total mechanical energy at the top of the incline, E equals the total mechanical energy at the bottom of the incline E' since there is no friction along the incline.
So, E = E'
U₁ + K₁ = U₂ + K₂ where U₁ = potential energy of crate at top of incline = mgh where m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s², h = height of incline = 1.8 m, K₁ = kinetic energy of crate at top of incline = 326.7 J, U₂ = potential energy of crate at bottom of incline = 0 J(since it is at an elevation h = 0) and K₂ = kinetic energy of crate at bottom of incline
So, substituting the values of the variables into the equation, we have
U₁ + K₁ = U₂ + K₂
mgh + K₁ = U₂ + K₂
28.9 kg × 9.8 m/s² × 1.8 m + 326.7 J = 0 J + K₂
509.796 J + 326.7 J = K₂
K₂ = 836.496 J
K₂ ≅ 836.5 J
Now since the vertical wall is a distance d2 away and the long ideal spring has a length dy = 3.4 m, let x be the compression of the spring. So, the distance moved by the crate is thus D = d2 - dy - x.
Now, the change in kinetic energy of the crate ΔK equals the work done by friction and that done by the spring W.
So ΔK = -W (from work-kinetic energy principles)
Let W' = work done by friction = μmgD where μ = coefficient of kinetic friction between surface and crate = 0.41, m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s² and D = distance moved by crate = D = d2 - dy - x = 5.2 m - 3.4 m - x = 1.8 - x
So, W' = μmgD
W' = 0.41 × 28.9 kg × 9.8 m/s² (1.8 - x)
W' = 116.12(1.8 - x)
W' = 2090.16 - 116.12x
The work done by the spring W" = 1/2k(x₀² - x²) where k = spring constant = 154 N/m, x₀ = initial spring length = dy = 3.4 m and x = final spring compression.
So, W" = 1/2k(x₀² - x²)
W" = 1/2 × 154 N/m[(3.4 m)² - x²]
W" = 77 N/m[11.56 m² - x²]
W" = 890.12 - 77x²
So, W = W' + W"
W = 2090.16 - 116.12x + 890.12 - 77x²
W = 2980.28 - 116.12x - 77x²
Since the crate stops, final kinetic energy K₃ = 0. So, ΔK = K₃ - K₂ = 0 - 836.5 J = -836.5 J
Also, ΔK = -W
-836.5 = -(2980.28 - 116.12x - 77x²)
836.5 = 2980.28 - 116.12x - 77x²
77x² + 116.12 -2980.28 + 836.5 = 0
77x² + 116.12x -2143.78 = 0
dividing through by 77, we have
x² + 1.508x -27.841 = 0
Using the quadratic formula to find x, we have
[tex]x = \frac{-1.508 +/-\sqrt{1.508^{2} - 4 X 1 X (-27.841)} }{2 X 1.508} \\x = \frac{-1.508 +/-\sqrt{2.274064 + 111.364} }{3.016} \\x = \frac{-1.508 +/-\sqrt{113.638064} }{3.016} \\x = \frac{-1.508 +/- 10.66}{3.016} \\x = \frac{-1.508 - 10.66}{3.016} or x = \frac{-1.508 + 10.66}{3.016} \\x = \frac{-12.168}{3.016} or x = \frac{9.152}{3.016} \\x = -4.03 or 3.034[/tex]
x = -4.03 or 3.034
Since the compression of the spring is positive, we choose x = 3.034
So, the crate compresses the spring 3.034 m
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
WOODHAVEN MI AND SAGINAW MI
In a separate location, take notes from the sources you have identified. This will take place over two or more days. While taking notes, consider using these reading strategies. From your notes, select one piece of evidence describing a climate similarity or difference that you discovered between these two cities. Write it in the space provided.
Answer:
LOL PUEDO HABLAR CON
Explanation:
Answer:
Whats the anwser/????
Explanation:
Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
An irregular shape object has a mass of 19 oz. A graduated cylinder with and initial volume of 33.9 mL. After the object was dropped in the graduated cylinder, it had a volume of 92.8 mL. What is the density of object( g/mL)
Explanation:
m = 19 oz × (28.3 g/1 oz) = 537.7 g
V = 92.8 mL
[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]
Ahmed is pushing a 4 Kg box to the right and Rashid is Pushing it to the right as well with a force of 12 N , the box accelerates by 5 m/s^2. What is the Force that is applied by Ahmed
Answer:
8 N
Explanation:
Applying,
(F'+F) = ma............... Equation 1
Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.
From the question,
Given: F = 12 N, m = 4 kg, a = 5 m/s²
Substitute these values into equation 1
(F'+12) = 4×5
(F'+12) = 20
F' = 20-12
F' = 8 N.
Hence Ahmed's force is 8 N
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Answer:
A. Power generated by meteor = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Workdone = 981000 J
Power required = 19620 Watts
Note: The question is incomplete. A similar complete question is given below:
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?
Explanation:
A. Power = workdone / time taken
Workdone = Kinetic energy of the meteor
Kinetic energy = mass × velocity² / 2
Mass of meteor = 1.5 g = 0.0015 kg;
Velocity of meteor = 50 km/s = 50000 m/s
Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J
Power generated = 1875000/2.1 = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Work done by elevator against gravity = mass × acceleration due to gravity × height
Work done = 1000 kg × 9.81 m/s² × 100 m
Workdone = 981000 J
Power required = workdone / time
Power = 981000 J / 50 s
Power required = 19620 Watts
Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. We assume the upward direction to be positive, and the downward direction to be negative.
(a) How long are her feet in the air?(b) What is her highest point above the board?(c) What is her velocity when her feet hit the water?
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
[tex]2gh = v_f^2 - v_i^2[/tex]
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]
h = 0.82 m
Now, for the time in air during upward motion we use first equation of motion:
[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]
(c)
Now we will consider the downward motion and use the third equation of motion:
[tex]2gh = v_f^2-v_i^2[/tex]
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]
vf = 7.17 m/s
Now, for the time in air during downward motion we use the first equation of motion:
[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
t = 1.14 s
Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿ a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
The question is incomplete, the complete question is;
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
A) The atom moves to a state of lower energy
B) The atom is ionized
C) One of the electrons leaves the atom
D) The atom can be excited to a higher energy state
Answer:
The atom can be excited to a higher energy state
Explanation:
According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.
If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.
Rather, the atom is excited from ground state to a higher energy level.
three people are standing by maintaining social distancing i.e 6 feet from each other. A doctor checks one person first then go to the other standing towards the east and then third person towards north. what will be the resultant displacement covered the doctor?
Answer:
8.5 feet.
Explanation:
A sketch of the position of the three people gives a right angled triangle. The hypotenuse of the triangle gives the resultant displacement, while the two other sides are 6 feet each.
The resultant displacement, R, is the overall displacement covered by the doctor. This can be determined by;
R = [tex]\sqrt{d_{1} ^{2} + d_{2} ^{2} }[/tex]
where: [tex]d_{1}[/tex] = 6 feet and [tex]d_{2}[/tex] = 6 feet.
Thus,
R = [tex]\sqrt{6^{2} + 6^{2} }[/tex]
= [tex]\sqrt{72}[/tex]
R = 8.49
The resultant displacement covered by the doctor is 8.5 feet.
12. What type of circuit is the diagram below?
series circuit
parallel circuit
Answer:
parallel circuit
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;
I. Series circuit
II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.
Hence, the type of circuit that the above diagram above represents is a parallel circuit.
Answer:
parallel circuit
Explanation:
I got it right on my exam
who is the biggest man in the world
Answer:
Sultan Kösen
here is a pic
Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2 < r1) and whose turns per unit length are n1 and n2.
Answer:
M' = μ₀n₁n₂πr₂²
Explanation:
Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.
So, M = N₂Ф₂₁/i₁
substituting the values of the variables into the equation, we have
M = N₂Ф₂₁/i₁
M = N₂B₁A₂/i₁
M = n₂lμ₀n₁i₁πr₂²/i₁
M = lμ₀n₁n₂πr₂²
So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²
M' = μ₀n₁n₂πr₂²
During which radioactive decay process does a neutron change into a proton?
A. Alpha decay
B. Gamma decay
C. Beta decay (positron)
D. Beta decay (electron)
A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.
What is beta decay?A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.
A positron is an electron with a positive charge.
During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.
Learn more about beta decay at: https://brainly.com/question/25455333
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What does it mean when work is positive?
O Velocity is greater than kinetic energy.
O Kinetic energy is greater than velocity.
O The environment did work on an object.
O An object did work on the environment.
Answer:
O The environment did work on an object
Explanation:
Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
