Answer:
This means that if the voltage is high the current is high, and if the voltage is low the current is low. Likewise, if we increase the resistance, the current goes down for a given voltage and if we decrease the resistance the current goes up.
What impact could lowering the pH of ocean water have? choose all that apply
O sea level rise
O shellfish decreasing
O decreasing amount of coral
Ofish being "burned by the acid"
A person holds a gallon of milk with a mass of 3 kg away from their body so that their forearm is completely parallel with the ground. The length of the forearm is .38 meters. The biceps brachii attaches to the forearm 4 cm from the elbow at an angle of 60 deg. How much force must be exerted by the bicep in order to keep the forearm level while holding the milk
Answer:
"322.5 N" is the appropriate solution.
Explanation:
The given values are:
Mass,
m = 3 kg
Length of forearm,
= .38 meters
Angle,
= 60°
As we know,
⇒ [tex]FSin60^{\circ}\times 0.04=mg\times 0.38[/tex]
⇒ [tex]F =\frac{mg\times 0.38}{Sin60^{\circ}\times 0 .04}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{3\times 9.8\times 0.38}{0.04\times \frac{\sqrt{3}}{2}}[/tex]
⇒ [tex]=\frac{11.172}{0.0346}[/tex]
⇒ [tex]=322.5 \ N[/tex]
A still ball of mass 0.514kg is fastened to a cord 68.7cm long and is released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.63kg steel block at rest on a horizontal frictionless surface. On collision, one-half the kinetic energy is converted to internal energy. Find the final speeds.
Answer:
1.21 m/s
Explanation:
From the law of conservation of energy,
U₁ + K₁ + E₁ = U₂ + K₂ + E₂
where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.
K₁ = initial kinetic energy of system = 0
E₁ = initial internal energy of system = unknown and
U₂ = final potential energy of system = 0
K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block
E₁ = final internal energy of system = unknown
So,
U₁ + K₁ + E₁ = U₂ + K₂ + E₂
mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂
mgh = 1/2(m + M)v² + (E₂ - E₁)
Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4
Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)
mgh = 1/2(m + M)v² + (m + M)v²/4
mgh = 3(m + M)v²/4
So, making v subject of the formula, we have
v² = 4mgh/3(m + M)
taking square root of both sides, we have
v = √[4mgh/3(m + M)]
Substituting the values of the variables into the equation, we have
v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]
v = √[13.8422/{3(3.144 kg)}]
v = √[13.8422 kgm/s²/{9.432 kg)}]
v = √(1.4676 m²/s²)
v = 1.21 m/s
The final speed of the given ball law of conservation of energy. The final speed of the given ball is 1.21 m/s.
The law of conservation of energy,
U₁ + K₁ + E₁ = U₂ + K₂ + E₂
where
U₁ = initial potential energy
K₁ = initial kinetic energy
E₁ = initial internal energy
U₂ = final potential energy
K₁ = final kinetic energy
E₁ = final internal energy
So,
mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂
mgh = 1/2(m + M)v² + (E₂ - E₁)
Given that (E₂ - E₁) = change in internal energy,
ΔE = 1/2ΔK
Where
ΔK = change in kinetic energy.
So,
ΔE = 1/2ΔK = 1/2(K₂ - K₁)
ΔE = K₂/2
ΔE = 1/2(m + M)v²/2
ΔE = (m + M)v²/4
Thus,
mgh = 1/2(m + M)v² + (E₂ - E₁)
mgh = 1/2(m + M)v² + (m + M)v²/4
mgh = 3(m + M)v²/4
v² = 4mgh/3(m + M)
Take square root of both sides,
v = √[4mgh/3(m + M)
put the values in the formula,
v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/3(0.514 kg + 2.63 kg)
v = 1.21 m/s
Therefore, the final speed of the given ball is 1.21 m/s.
To know more about Kinetic Energy,
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A student measured the density of Galena to be 7.9g/cm3 however the known density of Galena is 7.6g/cm3 . Calculate the percent error of the measurements.
Answer:
~4%
Explanation:
% = |(7.6 - 7.9)|/7.9
= 0.3/7.9 ≈ 0.04 = 4%
VUTE SUNDUI Physical Science / Motion / Lesson 89 1. The graph shown below is a velocity vs. time graph. The y-value represents the velocity, and the x-value represents the time. Explain the acceleration of the object in three distinct portions of the graph. Your conclusions of the acceleration should be based on a graphical analysis. Your answer should be one paragraph and written in your own words. Be sure to include proper spelling, grammar, and punctuation.
Answer:
this stuff hard no cap
Explanation:stop cheating cuhhh
You are cooking spaghetti and you look into the pot of boiling water before you put the spaghetti in. You notice the water moving around in the pot. What type of thermal energy transfer occurs to heat the water?
Answer:
Heat transfer is an exchange of thermal energy between two objects. The rate of heat transfer depends upon the temperatures of each entity and the medium through which the thermal energy is being transferred. In cooking, heat transfer refers to heating your food items through a cooking appliance, such as a stove, fryer, microwave, or oven.
Explanation:
I hope it helps
Answer:
Convection
Convection is heat transfer by mass motion of a fluid such as air or water when the heated fluid is caused to move away from the source of heat, carrying energy with it. Convection above a hot surface occurs because hot air expands, becomes less dense, and rises (see Ideal Gas Law).
Explanation:
hope this helps:-)
Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the two currents flow in opposite directions, what is the magnitude (in N/m) and direction of the force per unit length of one wire on the other
Answer: [tex]1.298\times 10^{-5}\ N/m[/tex]
Explanation:
Given
Current in the first wire [tex]I_1=2.2\ A[/tex]
Current in the second wire [tex]I_2=5.9\ A[/tex]
wires are [tex]20\ cm[/tex] apart
Force per unit length between the current-carrying wires is
[tex]\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}[/tex]
Force exerted by the wires is the same
Put the values
[tex]\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m[/tex]
This force will be repulsive in nature as the current is flowing opposite
In terms of their location, what is the difference between subcutaneous fat and visceral fat
4.17 In an n- type semiconductor bar, there is an increase in electron concentration from left to right and an electric field pointing to the left. With a suitable sketch, indicate the directions of the electron drift and diffusion current flow and explain why. If we double the electron concentration everywhere, what happens to the diffusion current and the drift current
Answer:
i) Jn = q Dn [tex]\frac{dn}{dx}[/tex]
ii) Diffusion current and drift current will be doubled
Explanation:
i) Determine the directions of the electron drift and diffusion current flow
the electron concentration is higher on the right side of the N-type semiconductor hence any change in the concentration of electrons will cause diffusion of electrons from right to left and electron drift is from left to right and this is because the electrons in the semiconductor bar are attracted to the electric field.
Jn = q Dn [tex]\frac{dn}{dx}[/tex] ( electron diffusion current )
ii) Determine what happens to the diffusion current and drift current If we double the electron concentration everywhere
The electron diffusion current ( J'n ) = 2q Dn [tex]\frac{dn}{dx}[/tex] = 2Jn ( i.e. diffusion current doubles )
Drift current = ( Jdrift = nq*μn*ε)
when the electron concentration is double
Drift current = 2 ( nq*μn*ε )
hence the drift current will double
The toadfish makes use of resonance in a closed tube to produce very loud sounds. The tube is its swim bladder, used as an amplifier. The sound level of this creature has been measured as high as 100 dB. (a) Calculate the intensity of the sound wave emitted. W/m2 (b) What is the intensity level if three of these toadfish try to make a sound at the same time
Answer:
a) I = 10⁻² W / m², b) β = 104.8 db
Explanation:
a) For this exercise we use the definition of decibels
β = 10 log [tex]\frac{I}{I_o}[/tex]
where I and Io are the intensities produced and the sound threshold
I₀ = 10⁻¹² W / m²
log \frac{I}{I_o} = β / 10
I = Io [tex]10^{\beta /10}[/tex]
let's calculate
I = 10⁻¹² [tex]10^{100/10}[/tex]
I = 10⁻² W / m²
b) each toad produces the same sound for which the total intensity is
I_total = 3 I
I_total = 3 10⁻² W / m²
expressed in decibels
β = 10 log ([tex]\frac{3 \ 10^{-2} }{10^{-12} }[/tex])
β = 104.8 db
The toadfish make sue of the resonance on the closed system tube to make a very loud noise and fish tube is the bladder which us used for amplifying the sounds. The sound level of the creatures has been taken as 100DB.
The intensity of the sound waves emitted by the fish is I = 10⁻² W / m², The intensity of he waves and level of the three of these toadfish is make a sound at the same time is that of β = 104.8 DB.Learn more about the use of resonance in a closed tube to produce.brainly.com/question/14700223.
Cruiser A was traveling east at 63 mi/h on a police emergency call when it was hit at an intersection by car B, which was traveling south at high speed. After sliding together on the wet pavement, the two cars hit cruiser C, which was traveling north at 45 mi/h and was 63 ft from the intersection at the time of the first collision. Stuck together, the three cars hit a wall and came to a stop at D. Knowing that car B weighs 3600 lb and that each of the cruisers weighs 3000 lb, solve the problem by neglecting the forces exerted on the cars by the wet pavement and treating the cars as point masses. Knowing that the speed of car B was 76.8757 mi/h and that the time elapsed from the first collision to the stop at D was 1.5 s, determine the coordinates of D.
Answer:
R = 45.05 ft, θ = -29º
Explanation:
Let's analyze your problem, two cars collide that get stuck, then this collide with a third car and they get stuck and the whole collides with a wall stopping, this problem has several parts
We define a form system for all the cars, so that during the crashes the forces have been internal and the moment can be preserved
Let's slow down to the English system
vₐ₀ = 63 mi / h (5280 ft / 1 mile) (1 h / 3600 s) = 92.4 ft / s
v_{bo] = 76.8757 mi / h = 112.7 ft / s
v_{c0} = 45 mi / h = 66 ft / s
Let's start by looking for the velocity in the first collision, let's set a reference system with the x-axis in the eastern direction
X axis
initial instant. Before crash
p₀ = mₐ vₐ₀
final instant. Right after the crash
p_f = (mₐ + m_b) v₁ₓ
the moment will be preserved
p₀ = p_f
mₐ vₐ₀ = (mₐ + m_b) v₁ₓ
v₁ₓ = [tex]\frac{m_a}{m_a+m_b} \ v_{ao}[/tex]
let's look for speed
note that the unit of free is the weight (W = mg), but since it is in the numerator and denominator the value of gravity is eliminated, so we do not have to reduce this unit
v₁ₓ = [tex]\frac{3000}{3000+3600} \ 92.4[/tex]
v₁ₓ = 42 ft / s
Y axis
initial
p₀ = -m_b v_{bo}
the negative sign is because it travels south (negative of the y-axis)
final
p_f = (mₐ + m_b) v_{1y}
conservation moment
p₀ = p_f
-m_b v_{bo} = (mₐ + m_b) v_{1y}
v_{1y} = [tex]- \frac{m_a}{m_a+m_b} \ v_{1y}[/tex]
we look for the value
v_{1y} = [tex]- \frac{3000}{3000+3600} \ 112.7[/tex]
v_{1y} = -51.227 ft / s
Now let's solve the second clash
X axis
initial
p₀ = (mₐ + m_b) v₁ₓ
final
p_f = (mₐ + m_b + m_c) v₂ₓ
conservation moment
(mₐ + m_b) v₁ₓ = (mₐ + m_b + m_c) v₂ₓ
M = mₐ + m_b + m_c
v₂ₓ = [tex]\frac{m_a+m_b}{M} \ v_{1x}[/tex]
we calculate
M = 3000 + 3600 + 3000 = 9600
v₂ₓ = [tex]\frac{ 3000+3600}{9600} \ 42[/tex]
v₂ₓ = 26.25 ft / s
Y axis
initial
p₀ = (mₐ + m_b) v_{1y} + m_c v_{co}
final
m_f = M v_{2y}
conservation moment
(mₐ + m_b) v_[1y} + m_c v_{co} = M v_{2y}
v_{2y} = [tex]\frac{m_a+m_b}{M } \ v_{1y} + \frac{m_c}{M} \ v_{co}[/tex]
we calculate
v_{2y} = [tex]- \frac{3000+3600}{9600} \ 51.227 + \frac{3000}{9600} \ 66[/tex]
v_{2y} = - 35.22 + 20.625
v_{2y} = - 14.594 ft / s
We already have the velocity of the set of vehicles, now we can use the kinematics relations to find the diastase at point D
For this last part we must make some assumptions
* we despise collision times
* we find the distance that the given time advances (t = 1.5 s) using the equations of uniform motion
X axis
x = v₂ₓ t
x = 26.25 1.5
x = 39.375 ft
Axis y
y = v_{2y} t
y = -14.594 1.5
y = -21.89 ft
We can give the answer in two ways
* the position is 39.375 ft to the East and 21.89 ft to the South
* In the form of module and angle
Let's use the Pythagorean theorem
R = [tex]\sqrt{x^2+ y^2}[/tex]
R = [tex]\sqrt{39.375^2 + 21.89^2}[/tex]
R = 45.05 ft
let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (-21.89 / 39.375)
θ = -29º
this angle is measured clockwise from the x axis
A physics class performs an experiment to determine the winner of a race. An empty can, a solid cylindrical battery and a marble roll, all with the same radius) without slipping down an inclined plane of vertical height H (they are all released at the same height). A box slides without friction on another inclined plane of the same height, H. Which object reaches the bottom of the incline first
Answer:
a) SPHERE must be the winner
b) BOX is most fastest
Explanation:
For this exercise we must use conservation of energy
starting point. Highest part of the plane
Em₀ = U = m g h
final point. Lowest part of the plane
Em_f = K = ½ m v² + ½ I w²
Note that as the objects roll, the kinetic energy of rotation is included. Energy is conserved
Em₀ = Em_f
mg h = ½ m v² + ½ I w²
the linear and rotational variables are related
v = w r
w = v / r
the moment of inertia of the bodies is tabulated
ring (empty can) I = m r²
cylinder I = ½ m r²
sphere (marbles) I = 2/5 m r²
mgh = ½ m v² + ½ I v²/r² = ½ mv² (+ I / mr²)
2gh / (1 + I / mr²) = v² (1)
Let's analyze the value of I / mr2
can I / mr² = mr² / mr² = 1
cylinder I / mr² = ½ mr² / mr² = ½
sphere I / mr² = 2/5 mr² / mr² = 2/5
we substitute in equation 1
v = [tex]\sqrt{ \frac{2gh}{ (1+ \frac{I}{m r^2}) } }[/tex]
can v = √(2gh / 2) = √ gh
v = √gh
cylinder v= [tex]\sqrt{ \frac{2gh}{ \frac{3}{2} } }[/tex] = √(4gh/3)
v = 1,155 √gh
sphere v = [tex]\sqrt{ \frac{2gh}{ \frac{7}{5} } }[/tex] = √(10gh/7)
v = 1.20 √gh
therefore the object with the highest speed is the one that takes less time, consequently the SPHERE must be the cattle.
b) for a box on a frictionless surface, there is no rotational kinetic energy
mgh = ½ m v²
v = √2gh
v = 1.41 √(gh)
When comparing with the latter, this would be the one that arrives first
Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block
Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil, [tex]\rho _o[/tex] = 926 kg/m³
density of the wood, [tex]\rho _{wood}[/tex] = 974 kg/m³
density of water, [tex]\rho _w[/tex] = 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
[tex]F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm[/tex]
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.
Is this a balanced equation? 6O2+ C6H12O6---> 6 CO2 + 6H2O
]
Yes, this equation is balanced.
You can check it using the law of conservation of mass, i.e., the total mass of the products formed must be equal to the total mass of the reactants.
104in rotational kinetic energy is equals to 1 by 2 I w2 that is equals to 2 by 2 I what does w represent
Answer:
I is the moment of inertia which is a scalar
w is the angular velocity
Explanation:
Kinetic energy has the Formula
K = ½ m v²
if the system is rotating, the linear velocity is related to the angular velocity
v = wr
we substitute
K = ½ m w² r²
K = ½ (m r²) w²
If we consider the body as punctual
I = m r²
we substitute
K = ½ I w²
we can see that
I is the moment of inertia which is a scalar
w is the angular velocity
K is called the rotational kinetic energy
You are tasked with calibrating the springs for a pinball machine. Your method of testing the springs is by attaching masses onto the end of the springs and measuring the stretch from initial position to final position. You hang a 14kg mass on a spring and notice that it stretches from 50 to 78cm. What is the spring constant for that spring
Answer:
[tex]490.5\ \text{N/m}[/tex]
Explanation:
m = Mass attached to spring = 14 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
x = Displacement of spring = [tex]78-50=28\ \text{cm}[/tex]
k = Spring constant
The force balance of the system is given by
[tex]kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{14\times 9.81}{0.28}\\\Rightarrow k=490.5\ \text{N/m}[/tex]
The spring constant for that spring is [tex]490.5\ \text{N/m}[/tex].
Help me with both questions please?
Answer:
1. They all accelerate at the same rate.
2.The object travels at a constant velocity throughout the fall.
Explanation:
Earths gravitational pull is at a constant 9.08 m/s^2. so when objects are free falling, the objects in question can only fall so fast before it would break gravity so to speak.
Select the correct answer.
In general, how does an increase in distance from the Sun affect a planet?
ОА. .
The amount of sunlight the planet receives decreases.
OB.
The gravitational force exerted by the Sun increases.
ОС.
The thickness of a planet's atmosphere increases.
OD. The planet's atmosphere absorbs more heat.
Answer:
a.
gravity decreases by distance
thickness doesn't change
Less heat.
Which materials do you wrap the egg in to prevent it from breaking at 5m?
Site: (https://www.healthinteractives.studio/bydesign/egg-drop/)
Answer:
If you do not have bubble wrap but do have other packing materials, like packing peanuts, inflated plastic packing packets, packing paper, cotton balls, or crumpled newspaper, you can use these materials to cushion the egg, as well.
A closed pipe creates a fifth
harmonic frequency of 125 Hz.
What is the next lower frequency
that will create a standing wave in
the pipe?
(Speed of sound = 343 m/s)
(Unit = Hz)
PLEASE HELP QUICKLY, WILL GIVE BRAINLIEST
Answer:
75 hz
Explanation:
Trust
The next frequency will be 75.0 Hertz
Wait, how do you know if it is correct?!
Well, it is correct for Acellus:)
Try it in other schools/academies:)
HELP PLEASEEEEE ASAP
Answer:
point D, hope this helps!
Which of the following situations would cause the greatest decrease in the
motion of molecules in a system?
A
Explanation:
hindi ako sure kung tama ba yan
A 40.0 kg block of lead is heated from -25°C to 200°C.
How much heat is absorbed by the lead block?
A. 2,354,000 J
B. 1,170,000 J
C. 56,891 J
D. 10,650 J
Answer:
B. 1,170,000 J
Explanation:
Given;
mass of lead block, m = 40 kg
initial temperature, t₁ = -25 ⁰C
final temperature, t₂ = 200 ⁰C
The heat absorbed the lead block is calculated as;
H = mcΔt
where;
c is the specific heat capacity of lead = 130 J/kg⁰C
H = 40 x 130 x (200 - (-25))
H = 40 x 130 x (200 + 25)
H = 40 x 130 x 225
H = 1,170,000 J
Therefore, the heat absorbed the lead block is 1,170,000 J
The graph shows the range of frequencies commonly heard by some animals. A graph with frequency in Hertz on the axis from 0 to 100, 000. On the y axis are porpoise, bat, moth, cat, human, bird, fish, cricket and snake. Porpoise is 60 to 105,000 Hertz. Bat is 55 to 100,000 Hertz. Moth is 250 to 100,000 Hertz. Cat is 60 to 80, 000 Hertz. Human is 20 to 12,000 Hertz. Bird is 65 to 9000 Hertz. Fish is 80 to 2000 Hertz. Cricket is 55 to 1,000 Hertz. Lastly snake is 100 to 1000 Hertz. Which animals can hear a sound wave that has 18,500 cycles in 0.75 seconds? Check all that apply. bats moths cats humans birds fish crickets
Answer:A,B,C or you could say bats, moths, cats
Explanation:
Trust me i got you, i promise u gonna get the answer right
The animals that can hear a sound wave that has 18,500 cycles in 0.75 seconds are bats, moths and cats.
What is frequency?It is the number of oscillations per second of the sinusoidal wave.
The graph shows the range of frequencies commonly heard by some animals. A graph with frequency in Hertz on the axis from 0 to 100, 000. On the y axis are porpoise, bat, moth, cat, human, bird, fish, cricket and snake.
Porpoise is 60 to 105,000 Hertz.
Bat is 55 to 100,000 Hertz.
Moth is 250 to 100,000 Hertz.
Cat is 60 to 80, 000 Hertz.
Human is 20 to 12,000 Hertz.
Bird is 65 to 9000 Hertz.
Fish is 80 to 2000 Hertz.
Cricket is 55 to 1,000 Hertz.
Snake is 100 to 1000 Hertz.
Thus, the animals that can hear a sound wave that has 18,500 cycles in 0.75 seconds are bats, moths and cats.
Learn more about frequency.
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A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same direction at 1.40 m/s. All cars have a mass of mass 1.6 104 kg. Determine the following.(a) speed of the three coupled cars after the collision (Give your answer to at least 2 decimal places.) m/s(b) kinetic energy lost in the collision J
Answer:
[tex]2.09\ \text{m/s}[/tex]
[tex]22298.4\ \text{J}[/tex]
Explanation:
m = Mass of each the cars = [tex]1.6\times 10^4\ \text{kg}[/tex]
[tex]u_1[/tex] = Initial velocity of first car = 3.46 m/s
[tex]u_2[/tex] = Initial velocity of the other two cars = 1.4 m/s
v = Velocity of combined mass
As the momentum is conserved in the system we have
[tex]mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}[/tex]
Speed of the three coupled cars after the collision is [tex]2.09\ \text{m/s}[/tex].
As energy in the system is conserved we have
[tex]K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}[/tex]
The kinetic energy lost during the collision is [tex]22298.4\ \text{J}[/tex].
A 70 kg bicyclist rides his 9.8 kg bicycle with a speed
of 16 m/s.
How much work must be done by the brakes to bring the bike and rider to a stop?
Hello,
QUESTION)✔ We have Ek = 1/2m x v²
Ek = 1/2 x 79.8 x 16²Ek = 10 214.4 JIn order to come to a complete stop, the cyclist must convert all his kinetic energy into thermal energy. Given that the braking force opposes movement, the work is therefore resistant, i.e. W = -10 214.4 J.
Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Strings of holiday lights can be designed in one of two ways. In some strings of lights, each light is connected to the others along a single wire (in series). In others, each light is attached to its own wire (in parallel). Suppose a single light bulb burns out. How do you think this will affect lights that are strung along a single wire
Answer:
They would go out
Explanation:
This is because, in a series connection, the same current passes through each light. Since the current is the same, if one light burns out, it cuts off the rest of the other lights and thus, no current flows in the string again.
Whereas, in a parallel connection, each light is attached to its own wire and thus has a different current flowing through it than the rest of the other wires. If one of the lights goes out, current stops flowing through it but, it doesn't affect the other lights.
Sound waves rely on matter to transmit their energy. They cannot ravel in a vacuum. True or false
Answer:
false
Explanation:
Hello,
QUESTION)True,Sound travels in a material medium, in space, there is no matter, so sound cannot propagate.
when the mass of an object increases, the forcé of gravity
Answer:
increace
Explanation:
they are both going up
how does speed measure
Answer:
Speed has the dimensions of distance divided by time. The SI unit of speed is the metre per second, but the most common unit of speed in everyday usage is the kilometre per hour or, in the US and the UK, miles per hour. For air and marine travel the knot is commonly used.
Explanation:
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