Answer:
1120 N
Explanation:
The velocity with which he hits the water can be found with kinematics:
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)
v = -13.3 m/s
Or it can be found with conservation of energy.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × -9.8 m/s² × -9.00 m)
v = -13.3 m/s
Sum of forces on the diver after he hits the water:
∑F = ma
F − mg = m Δv/Δt
F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)
F = 1120 N
What makes a clinical thermometer suitable for measuring small changes in body temperature? *
Answer: Because of the fine bore of the tube.
Explanation:
Temperature is the degree of hotness and coldness. And thermometer is the instrument use to measure temperature.
The two most common types of themometric fluids for thermometer are alcohol and mercury.
What makes a clinical thermometer suitable for measuring small changes in body temperature is because of the fine bore of the tube which makes it possible for small temperature changes to cause large changes in the length of mercury columns, making the thermometer very sensitive to temperature changes.
The most prominent feature of the thermometer is the kink or constriction of bore near the bulb.
Answer:
xxx
Explanation:
A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.15 s
Answer:
The magnitude of the average force on the wall during the collision is 6 N.
Explanation:
Given;
mass of snowball, m = 120 g = 0.12 kg
velocity of the snowball, v = 7.5 m/s
duration of the collision between the snowball and the wall, t = 0.15 s
Magnitude of the average force can be calculated by applying Newton's second law of motion;
F = ma
where;
a is acceleration = v / t
a = 7.5 / 0.15
a = 50 m/s²
F = ma
F = 0.12 x 50
F = 6 N
Therefore, the magnitude of the average force on the wall during the collision is 6 N.
What kind of substance can you pour from one container into another without a change in volume
Answer:
Liquids
Explanation:
Liquids take up the shape of the container it is poured into but will never change its volume.
9.
A 0.060 kg tennis ball hits the wall horizontally. The moment the ball hits the wall it accelerates at a rate of – 752 m/s2. What is the force that the ball applied to the wall?
Answer:
force = 45.12N
Explanation:
f = ma
f = 0.06 x 752 = 45.12N
What can happen if a body moves through speed of light
As waves travel into the denser medium, they slow down and wavelength decreases.
Explanation:
The denser the medium the slower the waves (speed of light) travels.
◦•●◉✿When the body approaches the speed of light, the body's length appears to contract in the direction of travel, and its mass appears to increase from the point of view of a stationary observer. Only photons move to light velocity. They don´t have length.✿◉●•◦
The skier starts from rest. The total distance travelled by the skier during the descent is 2800 m. The average resistive force on the skier is 220 N. Calculate the work done against the resistive force
Answer:
Explanation:
Force equal to resistive force will be applied for movement . So force applied
F = 220 N .
displacement = 2800 m
work done against resistive force
= force x displacement
= 220 x 2800 J
= 6.16 x 10⁵ J .
A pendulum at position A is released and swings through position B to position Con the other side.
B
1. Describe the total mechanical energy at each of the following positions. (3)
A.
B.
C
Explanation:
Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),
the conditions can be summarized bellow
point A
initial velocity= 0
final velocity=0
P.E= Max
K.E= 0
point B
initial velocity= maximum
final velocity=maximum
P.E=K.E
point C
initial velocity= min
final velocity=min
P.E= 0
K.E= max
An air-track glider of mass 0.25 kg moving at 0.60 m/s collides with and sticks to a glider of mass 0.50 kg at rest. How much kinetic energy is converted into other forms of energy as a direct result of this collision?
Answer:
0.03 Joules have been converted into other forms of energy as the direct result of the collision.
Explanation:
Let's start studying the conservation of momentum for the system:
[tex]P_i=P_f\\(0.25\,kg)\,{0.6\,m/s)+(0.5\,kg)\,(0\,m/s)=(0.25\,kg+0.5\,kg)\, v_f \\\\\\ 0.15\,kg\,m/s=0.75\,kg\,\,v_f\\v_f=0.15/0.75\,\,m/s\\v_f=0.2\,\,m/s[/tex]
Now that we know the speed of the newly created object, we can calculate how the final kinetic energy differs from the initial one:
[tex]K_i=\frac{1}{2} (0.25)\,(0.6)^2+\frac{1}{2} (0.5)\,(0)^2=0.045\,\,J\\ \\K_f=\frac{1}{2} (0.75)\,(0.2)^2=0.015\,\,J\\[/tex]
Then, when we subtract one from the other, we can estimate how much kinetic energy has been converted into other forms of energy in the collision:
0.045 J - 0.015 J = 0.03 J
Q.) Miscellaneous conversations. a) mass=120*10^8 g (Convert this value in mg and kg Write in standard form after converting) b) length=200000*10^3 Convert this value in micrometres cm and km Write in standard form after converting
Answer:
a. Convert 120 × 10⁸ g to i mg = 1.2 × 10¹³ mg ii. to g = 1.2 × 10⁷ kg
b. Convert 200000 × 10³ m to i. micrometers = 0.2 × 10³ μm ii. to cm = 2 × 10⁶ cm iii. to km = 2 × 10⁵ km
Explanation:
a. i. To convert the mass = 120 × 10⁸ g to mg, We know that 1000 mg = 10³ mg = 1 g, Since we are converting to mg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10³ mg = 120 × 10¹¹ mg = 1.2 × 10² × 10¹¹ mg = 1.2 × 10¹³ mg
ii. To convert the mass = 120 × 10⁸ g to kg, We know that 1000 g = 10³ g = 1 kg, 1 g = 10⁻³ kg. Since we are converting to kg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10⁻³ kg = 120 × 10⁵ kg = 1.2 × 10² × 10⁵ kg = 1.2 × 10⁷ kg
b. i.To convert the length = 200000 × 10³ m to micrometers, We know that 1/1000000 μm = 10⁻⁶ mg = 1 m, Since we are converting to micrometers, μm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/1000000 μm = 200000/1000000 × 10³ μm = 0.2 × 10³ μm
ii. To convert the length = 200000 × 10³ m to cm, We know that 100 cm = 10² cm = 1 m, 1 m = 10⁻² cm = 1/100 cm. Since we are converting to cm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/100 cm = 200000/100 × 10³ cm = 2000 × 10³ cm = 2 × 10³ × 10³ cm = 2 × 10⁶ cm
iii. To convert the length = 200000 × 10³ m to km, We know that 1000 m = 10³ m = 1 km, 1 m = 10⁻³ km = 1/1000 km Since we are converting to km, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/1000 km = 200000/1000 × 10³ km = 200 × 10³ km = 2 × 10² × 10³ km = 2 × 10⁵ km
Answer part (d) please
Answer:
MARK me brainliest please and follow my page
Explanation:
All you have to do to get the average speed is to calculate the total distance covered and divide it by the total time taken
= 16/18 = 0.88m/s
Average speed = (distance covered) / (time to cover the distance)
For the full 18 seconds described by the graph . . .
Average speed = (16 meters) / (18 seconds)
Average speed = (16 / 18) m/s
Average speed = 0.89 m/s
What type of image does a concave lens form? A. real image B. magnified image C. virtual image D. reverse image
Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine the net force charge acting at q1 (+ 2.0 × 10^-5C), caused by q2 (-4.0 × 10-5 C) and q3 (-4.0 × 10^-5C). They create a right angles triangle, where q1 is at the 90° corner
Determine the net electric field acting at q1
Answer:
E≅1.2×10^7 N/C
Explanation:
First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!
Edit: I ended up adding it anyways ;P
Since we are observing the net electric field acting at q1, we need to use the formula: [tex]E=k\frac{q}{r^{2} }[/tex]
And since we are observing the effects of multiple charges at once...
E=ΣE, which just means wee need to add all the observed electric fields together:
ΣE= [tex]k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }[/tex]
Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:
q2=q3they are the same distance from q1These are actually really nice to have, because now we can simplify our expression to:
[tex]E=k\frac{2q}{r^{2} }[/tex]
Now let's plug in our values and get an answer out.
E= 2(8.99×10^9)(4×10^-5)/(0.24)
Plugging all that in, I get:
E≅1.2×10^7 N/C
If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.
Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!
Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.
1. Find the energy required to melt 255g of ice at 0°C into water at 0°C
Answer:
E = 85170 J (/ 85.2 kJ)
Explanation:
Take the latent heat of fusion of water be 334J / g.
From the equation E = ml,
E = energy required (unknown),
mass m = 255g,
latent heat of fusion l = 334J / g,
E = 255 x 334
E = 85170 J (/ 85.2 kJ)
can all alpha beta and gamma radiation treat cancer
Answer:
No
Explanation:
Only a few of alpha beta and gamma radiation can treat cancer
10 advantages of friction
Answer:
1. it helps to change the direction.
2. it helps us to walk on ground.
3. it helps the vechils to break while moving.
4. helps in changing one form of enegry to another form. eg when we rub our hands we feel heat energy.
5. it opposites the force.
6. it helps us to change shape of objects.eg we roll the dough to make it roti.
7. it changes the state of body from rest motion.eg when we push any obj from inclined plane it moves.
i all know is just 7..
Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de sección por el que sale el agua con un chorro continuo. A) ¿Qué cantidad de líquido saldrá del depósito expresada en m3/s?
Answer:
a) El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].
Explanation:
a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:
[tex]\Delta z = \frac{v_{out}^{2}}{2\cdot g}[/tex]
Donde:
[tex]\Delta z[/tex] - Diferencia de altura, medida en metros.
[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.
[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.
Se despeja la rapidez de salida del chorro:
[tex]v_{out} = \sqrt{2\cdot g \cdot \Delta z}[/tex]
Si [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]\Delta z = 0.3\,m[/tex], entonces la rapidez de salida del chorro es:
[tex]v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}[/tex]
[tex]v_{out} \approx 2.426\,\frac{m}{s}[/tex]
Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:
[tex]\dot V_{out} = v_{out}\cdot A_{t}[/tex]
Donde:
[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.
[tex]A_{t}[/tex] - Área transversal del orificio, medido en metros cuadrados.
[tex]\dot V_{out}[/tex] - Caudal de salida del chorro, medido en metros cúbicos por segundo.
Dado que [tex]v_{out} = 2.426\,\frac{m}{s}[/tex] y [tex]A_{t} = 5\,cm^{2}[/tex], el caudal de salida del chorro es:
[tex]\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)[/tex]
[tex]\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex]
El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].
What is the last step in creating an argumentative essay?
outlining
prewriting
researching
revising
Answer:
The answer is Revising
Answer:
D
Explanation:
Revising
Which measurement is a potential difference?
O A. 115 N
O B. 115 C
O C. 115 J
O D. 115 V
Answer:
option d is answer because pd is measured in volt.
How many atoms of oxygen in the chemical formula 2 Ca(ClO2)2?
Answer:
8
Explanation:
Ca(ClO2)2 - 2*2 = 4 Oxygen atoms
2 Ca(ClO2)2 - 2*4 = 8 Oxygen atoms
Una bola de 1 kg gira alrededor de un circulovrtical en el extremo de un cuerda. El otro extremo de la cuerda esta fijo en el centro del circulo. Calcular la diferencia entre las tensiones (de la cuerda) maxima y minima
Answer:
La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.
Explanation:
Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:
[tex]T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]
En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:
[tex]T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]
Donde:
[tex]T_{min}[/tex], [tex]T_{max}[/tex] - Tensiones mínima y máxima, medidas en newtons.
[tex]m[/tex] - Masa de la bola, medida en kilogramos.
[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.
[tex]L[/tex] - Distancia con respecto al eje de rotación, medida en metros.
[tex]v[/tex] - Rapidez tangencial, medido en metros por segundo.
Se elimina la aceleración centrípeta de ambas expresiones por igualación:
[tex]T_{min} + m\cdot g = T_{max} - m\cdot g[/tex]
Ahora, la diferencia entre las tensiones máxima y mínima es:
[tex]T_{max} - T_{min} = 2\cdot m \cdot g[/tex]
Si [tex]m = 1\,kg[/tex] y [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], entonces:
[tex]T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]T_{max}-T_{min} = 19.614\,N[/tex]
La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.
a 0.350 kg block at -27.5 °C is added to 0.217 kg of water at 25.0 °C. they come to equilibrium at 16.4 °C. what is the specific heat of the block?
Answer:
[tex]C_{pb}=0.501\ kJ/kg.K[/tex]
Explanation:
Given that
[tex]m_1=0.35 kg[/tex]
[tex]T_1=-27.5^oC[/tex]
[tex]m_2=0.214 kg[/tex]
[tex]T_2=25^oC[/tex]
[tex]T=16.4^oC[/tex]
We know that
[tex]C_{pw}=4.187 kJ/kg.K[/tex]
By using energy conservation
Heat lost by water = Heat gain by block
[tex]m_2\times C_{pw}\times (T_2-T)=m_1\times C_{pb}\times (T-T_1)[/tex]
[tex]0.214\times 4.187\times (25-16.4)=0.35\times C_{pb}\times (16.4+27.5)[/tex]
[tex]C_{pb}=0.501\ kJ/kg.K[/tex]
Therefore the specific heat of the block will be 0.501 kJ/kg.K
A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of Ï3 ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s.
Answer:
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
Explanation:
To solve the problem/ we first write the differential equation governing the motion. So,
[tex]m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0[/tex]
with m = 1 slug and k = 9 lb/ft, the equation becomes
[tex]\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0[/tex]
The characteristic equation is
D² + 9 = 0
D = ±√-9 = ±3i
The general solution of the above equation is thus
x(t) = c₁cos3t + c₂sin3t
Now, our initial conditions are
x(0) = -1 ft and x'(0) = -√3 ft/s
differentiating x(t), we have
x'(t) = -3c₁sin3t + 3c₂cos3t
So,
x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)
x(0) = c₁cos(0) + c₂sin(0)
x(0) = c₁ × (1) + c₂ × 0
x(0) = c₁ + 0
x(0) = c₁ = -1
Also,
x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)
x'(0) = -3c₁sin(0) + 3c₂cos(0)
x'(0) = -3c₁ × 0 + 3c₂ × 1
x'(0) = 0 + 3c₂
x'(0) = 3c₂ = -√3
c₂ = -√3/3
So,
x(t) = -cos3t - (√3/3)sin3t
Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)
where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.
Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3
x(t) = (2√3/3)sin(3t + 4π/3)
So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)
We now find the times when v(t) = 3 ft/s
So (2√3)cos(3t + 4π/3) = 3
cos(3t + 4π/3) = 3/2√3
cos(3t + 4π/3) = √3/2
(3t + 4π/3) = cos⁻¹(√3/2)
3t + 4π/3 = ±π/6 + 2kπ where k is an integer
3t = ±π/6 + 2kπ - 4π/3
t = ±π/18 + 2kπ/3 - 4π/9
t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9
t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3
t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3
Since t is not less than 0, the values of k ≤ 0 are not included
So when k = 1,
t = 5π/18 and π/6. So,
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
The time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.
Given data:
The mass suspended from spring is, m = 1 slug.
The spring constant is, k = 9 lb/ft.
The magnitude of upward velocity is, v = 3 ft/s.
The magnitude of downward velocity is, v' = 3 ft/s.
The given problem can be resolved by framing a differential equation that governs the motion of spring. The differential equation governing the motion of spring is,
[tex]m \dfrac{d^{2}x}{dt^{2}}=-kx\\\\\\\dfrac{d^{2}x}{dt^{2}}+\dfrac{k}{m}x=0[/tex]
With m = 1 slug and k = 9 lb/ft, the equation becomes
[tex]\dfrac{d^{2}x}{dt^{2}}+\dfrac{9}{1}x=0\\\\\\\dfrac{d^{2}x}{dt^{2}}+9x=0[/tex]
Now, the characteristic equation is,
D² + 9 = 0
D = ±√-9 = ±3i
And the general solution of the above equation is,
x(t) = c₁cos3t + c₂sin3t
Now, our initial conditions are
x(0) = -1 ft and x'(0) = -√3 ft/s
differentiating x(t), we have
x'(t) = -3c₁sin3t + 3c₂cos3t
So,
x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)
x(0) = c₁cos(0) + c₂sin(0)
x(0) = c₁ × (1) + c₂ × 0
x(0) = c₁ + 0
x(0) = c₁ = -1
Also,
x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)
x'(0) = -3c₁sin(0) + 3c₂cos(0)
x'(0) = -3c₁ × 0 + 3c₂ × 1
x'(0) = 0 + 3c₂
x'(0) = 3c₂ = -√3
c₂ = -√3/3
So,
x(t) = -cos3t - (√3/3)sin3t
Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)
where A = √c₁² + c₂²
= √[(-1)² + (-√3/3)²]
= √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3
and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.
Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3
x(t) = (2√3/3)sin(3t + 4π/3)
So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)
We now find the times when v(t) = 3 ft/s
So (2√3)cos(3t + 4π/3) = 3
cos(3t + 4π/3) = 3/2√3
cos(3t + 4π/3) = √3/2
(3t + 4π/3) = cos⁻¹(√3/2)
3t + 4π/3 = ±π/6 + 2kπ
where k is an integer
3t = ±π/6 + 2kπ - 4π/3
t = ±π/18 + 2kπ/3 - 4π/9
t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9
t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3
t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3
Since t is not less than 0, the values of k ≤ 0 are not included
So when k = 1,
t = 5π/18 and π/6.
t = 5π/18 + 2nπ/3
here, n is a natural number.
Thus, we can conclude that the time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.
Learn more about the differential equation here:
https://brainly.com/question/14620493
During or after exercise,it is normal for a student to comment that his/her heart feels like it is "beating out of my chest"?
Answer:
Yes it is normal
Explanation:
When you exercise, your heart beat goes up, resulting in people saying that their heart feels like it is "beating out of their chests".
coma Narrows Bridge across the Puget Sound collapsed. The bridge was a suspension bridge. The wind blowing through the narrows matched the natural frequency of the bridge. This resulted in a large movement of roadway, which eventually caused the bridge to fail. What characteristic of waves caused the bridge to collapse
Answer:
amplitude
Explanation:
Amplitude is the characteristic of waves which caused the bridge to collapse. Amplitude of a wave is the maximum amount of displacement of a particle occurs in the medium from its rest position. When the frequency of a wave reaches the natural frequency of the bridge, the oscillation of the bridge produce an amplitude where it causing the destruction of the bridge which is called Resonance. So we can say that amplitude is the characteristic of waves which is responsible for the collapse of the bridge.
Answer: C.interference, because constructive interference occurred when the wind frequency matched the natural frequency of the bridge
Explanation:
Which statement accurately describes this atomic spectrum? There is a horizontal rectangle with the colors of the rainbow from violet to red. There are lines in purple, blue, green and orange. The black lines represent the energy emitted by the electrons. The black lines represent the energy absorbed by the electrons. The colored lines represent the energy emitted by the electrons. The colored lines represent the energy absorbed by the electrons.
Answer:
The black lines represent the energy absorbed by the electrons.
Explanation:
Atoms emit lights when they are excited. These lights are of particular wavelengths that match with different colors. A series of colored lines appear along with spaces in the middle of the two colors. The middle of the colors is filled with dark spaces. Each spectral line of an element represents a specific characteristic of the element. These colored lines appearing in the form of series are termed to be the atomic spectrum of the element. Identification of the elements is done through the line of the spectrum they possess.
Answer:
(B) The black lines represent the energy absorbed by the electrons.
Explanation:
Magnetic field lines exit out of the . Magnetic field lines enter into the . Magnetic field lines travel around a bar magnet in
Answer:
Magnetic field lines exit out of the North pole . Magnetic field lines enter into the South pole. Magnetic field lines travel around a bar magnet in closed loops.
Explanation:
Magnetic field lines shows the direction of a magnetic force and how it acts, it gives the direction of the magnetic field at that point in time.
For a bar magnetic, the magnetic field lines runs from the north pole to the south pole, i.e. it exits the north pole and enters into the south pole. This magnetic field lines also go through the magnet forming closed loops without ends.
Answer:
Magnetic field lines exit out of the
✔ north pole
.
Magnetic field lines enter into the
✔ south pole
.
Magnetic field lines travel around a bar magnet in
✔ a closed loop
.
Explanation:
A ball has a mass of 0.25 kg and is moving to the right at 1.0 m/s. It hits a ball of mass 0.15 kg that is initially at rest. After the collision, the 0.15 kg ball moves off to the right with a velocity of 0.75 m/s. What is the final velocity of the 0.25 kg ball? 0.42 m/s to the right 0.42 m/s to the left 0.55 m/s to the right 0.55 m/s to the left
Answer:
C-0.55 m/s to the right
Explanation:
edg
Answer:
0.55 to the right
Explanation:
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).
Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
[tex]C=2\pi r[/tex] (1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
[tex]r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m[/tex]
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
[tex]L=Lo[1+\alpha \Delta T][/tex] (2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
[tex]L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m[/tex]
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
[tex]r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m[/tex]
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m
Explanation:
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).
Choose the friction which opposes the relative motion between surfaces in motion a.Static friction b.Kinetic friction c.Sliding friction d.Both kinetic and sliding friction
Answer:
d. Both kinetic and sliding friction
Explanation:
Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.
The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU
Answer:
Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;
The closest (minimum) distance of Neptune from the Earth is 29 AU
The farthest (maximum) distance of Neptune fro the Earth is 31 AU
Explanation:
The following parameters are given;
The distance from the Earth to the Sun = 1 AU
The distance of Neptune from the Earth = 30 AU
We have;
When the Sun is between the Earth and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 + 1 = 31 AU
When the Earth is between the Sun and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 - 1 = 29 AU
Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU
The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.
Answer:
29 AU
Explanation: