Answer:
Same as before.
Explanation:
we can use momentum conservation to solve the problem.
Let mass of child be m and of the sled be M. Also let both have initial velocity u.
since, child fells off on the sled it will still have same velocity of ''u''.
By momentum conservation we have,
(M + m) x u = m x u + M x v
⇒ Mu=Mv
⇒ u=v
Hence, velocity of the sled remains the same.
Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader
Answer:
7.55 g
Explanation:
Using the relation :
Δt = temperature change = (6° - 0°) = 6°
Q = quantity of heat
C = specific heat capacity = 4190 j/kg/k
1000 J = 1kJ
333 KJ = 333000 j
The quantity of ice that will melt ;
= 0.419 * 6 * 100 / 333000
= 2514000 / 333000
= 7.549 g
The mass of ice that will melt :
2.514 / 0.333
= 7.549 g
A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string, standing motionless on the ground. Assume the kite is flying away from the child. At what rate is the child releasing the string when (a) 50 ft of the string is out
Answer:
v = 27.28 m /s, θ = 63.9º
Explanation:
For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.
* We must find the horizontal speed, for this we will find the descent time and the horizontal distance
* We look for the vertical speed
At the highest point the speed is horizontal
Let's find the time it takes for the kite to reach the ground
y = y₀ + v_{oy} t - ½ g t²
0 =y₀ + 0 -1/2 gt²
t = [tex]\sqrt{ \frac{2y_o}{g} }[/tex]
t = √(2 40/32)
t = 2.5 s
to find the horizontal velocity we must know the horizontal distance, let's use trigonometry
sin θ = y / l
θ = sin⁻¹1 y / l
θ = sin⁻¹ 40/50
θ = 53.1º
therefore the horizontal distance is
x = l cos 53.1
x = 50cos 53.1
x = 30 m
let's use the equation
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 30 / 2.5
v₀ₓ = 12 m / s
we look for the vertical component of the velocity
v_y = v_{oy} - g t
v_y = 0 - g t
v_y = - 9.8 2.5
v_y = -24.5 m / s
the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign
We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus
v = [tex]\sqrt{v_x^2 + v_y^2}[/tex]
v = [tex]\sqrt{12^2 + 24.5^2}[/tex]
v = 27.28 m /s
we use trigonometry for the angle
tan θ = v_y / vₓ
θ = tan⁻¹ v_y / vₓ
θ = tan⁻¹ 24.5 / 12
θ = 63.9º
1) A man leans against the wall and...
a) The man's shoulder pushes against the wall
b) The wall pushes against the man
c) Bricks in the wall push against each other
d) The ground pushes against the man, supporting him.
Answer:
I thinks its a, but its really about gravity im not sure
Explanation:
:)
Car X is travelling at 30m/s north. Its driver looks at car Y approaching on another road and he estimates it is moving at 15m/s south-west relative to his car. Calculate the velocity of car Y relative to the ground.
Answer: 22.1 m/s
Explanation:
The velocity of Car traveling 30 m/s towards the north
In vector form it is
[tex]v_x=30\hat{j}[/tex]
The velocity of car Y w.r.t X is
[tex]\Rightarrow v_{yx}=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}][/tex]
Solving this
[tex]\Rightarrow v_{yx}=v_y-v_x\\\Rightarrow v_y=v_{yx}+v_x[/tex]
putting values
[tex]\Rightarrow v_y=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}]+30\hat{j}[/tex]
[tex]\Rightarrow v_y=-10.606\hat{i}+19.39\hat{j}[/tex]
absolute velocity relative to ground is
[tex]\left | v_y\right |=\sqrt{(-10.606)^2+(19.39)^2}\\\left | v_y\right |=22.101\ m/s[/tex]
Suppose you are standing in front of a flat mirror which is mounted to a vertical wall. For this problem you may suppose that your height is 1.70 m and your eyes are 12 cm below the top of your head. What is the smallest mirror that will still allow you to see the full length of your body
Answer:
The right approach is "0.85 m".
Explanation:
According to the question, the diagram will is provided below.
So that as per the diagram,
The values will be:
My height,
AO = 1.70 m
My eyes at,
AB = 12 cm
i.e.,
= 0.12 m
As we can see, the point of incidence lies between the feet as well as the eyes, then
BO = 1.58 m
Now,
⇒ [tex]O'D = \frac{1.58}{2}[/tex]
[tex]=0.79 \ m[/tex]
The point of incidence of the ray will be:
⇒ [tex]CO'=1.70-\frac{0.12}{2}[/tex]
[tex]=1.70-0.06[/tex]
[tex]=1.64 \ m[/tex]
Hence,
The smallest length of the mirror will be:
= [tex]CO'-O'D[/tex]
On substituting the values, we get
= [tex]1.64-0.79[/tex]
= [tex]0.85 \ m[/tex]
A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding
Answer:
0.6
Explanation:
Given that :
Radius, R = 7m
Period, T = 6.9s
The Coefficient of static friction, μs can be obtained using the relation :
μs = v² / 2gR
Recall, v = 2πR/T
μs becomes ;
μs = (2πR/T)² / 2gR
μs = (4π²R² / T²) ÷ 2gR
μs = (4π²R² / T²) * 1/ 2gR
μs = 4π²R / T²g
μs = 4π²*7 / 6.9^2 * 9.8
μs = 28π² / 466.578
μs = 276.34892 / 466.578
μs = 0.5922887
μs = 0.6
A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Explanation:
Given:
mass of truck M = 1370 kg
speed of truck = 12.0 m/s
mass of car m = 593 kg
collision is elastic therefore,
Applying law of momentum conservation we have
momentum before collision = momentum after collision
1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)
Also for a collision to be elastic,
velocity of approach = velocity of separation
12 -0 = v2-v1 ....(ii)
using (i) and (ii) we have
So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 140 m above the launch point. (a) Calculate the horizontal component of its velocity. (b) Calculate the vertical component of its velocity just after launch. (c) At one instant during its flight the vertical component of its velocity is found to be 65 m/s. At that time, how far is it above or below the launch point
Answer:
a). 53.78 m/s
b) 52.38 m/s
c) -75.58 m
Explanation:
See attachment for calculation
In the c part, The negative distance is telling us that the project went below the lunch point.
Part D Here is one last question as a final check on your understanding of your work for this problem, looking at this problem as an example of the Conservation of Energy. The action in this problem begins at location A , with the block resting against the uncompressed spring. The action ends at location B, with the block moving up the ramp at a measured speed of 7.35 m/s . From A to B, what has been the work done by non-conservative forces, and what has been the change in the mechanical energy of the block-Earth system (the ramp is a part of the Earth)
Answer:
The answer is "39.95 J".
Explanation:
Please find the complete question in the attached file.
[tex]\to W_{AC}=(\mu \ m \ g \ \cos \theta ) d[/tex]
[tex]=(0.45 \times 1.60 \times 9.8 \times \cos 26^{\circ}) 6.30 \\\\=(7.056 \times \cos 26^{\circ}) 6.30 \\\\=6.34189079\times 6.30\\\\=39.95 \ J\\\\[/tex]
[tex]\therefore \\\\\bold{\Delta E =39.95 \ J}[/tex]
A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun
The Sun is divided into three regions.
True оr False?
Answer:
false I think
Explanation:
hope that help
so it's not divided in 3 regions
Fat Albert (the TV show character) runs up the stairs on Monday. On Tuesday, he walks up the same set of stairs. Which day did he do more work?
Answer:
Tuesday bc instead of running he/she was walking bc he/she might not have as much energy
Explanation:
Match the following:
Answer:
iron metal :chromiummachinery part :nickel or chromiumornamentation and decoration pieces :silver and goldprocessed food :tin coated iron canbridges and automobiles :zinc metaldistilled water:bad conductorWhich one is it? Help ASAP
Answer:
extreme heat, because no physical damage can demagnetize a magnet
Explanation:
Answer:
the 3rd one
Explanation:
Which graph represents the relationship between the magnitude of the gravitational force exerted by earth on a spacecraft the distance between the center of the spacecraft the center of earth
Answer:
B as distance increase force decrease, but it is not a linear relationship.
How much heat is needed to boil 5.30 kg of water at its boiling point?
Use Q = mass x latent heat of vaporization.
A. 1760 kJ B. 22.2 kJ C. 530 kJ D. 12,000kJ
Answer:
Required heat Q = 11,978 KJ
Explanation:
Given:
Mass = 5.3 kg
Latent heat of vaporization of water = 2,260 KJ / KG
Find:
Required heat Q
Computation:
Required heat Q = Mass x Latent heat of vaporization of water
Required heat Q = 5.3 x 2260
Required heat Q = 11,978 KJ
Required heat Q = 12,000 KJ (Approx.)
Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed
Answer:
The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg
Explanation:
The given percentage by weight of protein solids in raspberries = 10 weight%
The ratio of sugar to raspberries in ja-m = 45:55
The mass of the mixture after boiling = 0.4 weight fraction water
Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry
The mass of raspberry, r = 1 kg
The percentage by weight of water in raspberry = 90 weight %
The mass of water in 1 kg of raspberry = 90/100 × 1 kg = 0.9 kg
The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55
∴ s = 1 kg × 55/45 = 11/9 kg
The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg
The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction
Let 'w' represent the mass of water boiled off, we have;
(0.9 - w)/(20/9 - w) = 0.4
(0.9 - w) = 0.4 × (20/9 - w)
0.9 - w = 8/9 - 0.4·w
9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w
(81 - 80)/(90) = (6/10)·w
1/90 = (6/10)·w
w = ((10/6) × 1/90) = 1/54
w = 1/54
The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg
In an effort to be the star of the half-time show, the majorette twirls a highly unusual baton made up of four mases fastened to the ends of light rods. Each rod is 1.0 m lone. Find the moment of inertia of the system about an axis perpendicular to the page and passing through the point where the rods cross.
Answer:
"0.25 kg-m²" is the appropriate answer.
Explanation:
The diagram of the question is missing. Find the attachment of the diagram below.
According to the diagram, the values are:
m₁ = 0.2
m₂ = 0.3
m₃ = 0.3
m₄ = 0.2
d₁ = d₂ = d₃ = d₄ = 0.5 m
As we know,
The moment of inertia is:
⇒ [tex]I=\Sigma M_id_i^2[/tex]
then,
⇒ [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]
⇒ [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]
On substituting the values, we get
⇒ [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]
⇒ [tex]=0.25\times 1[/tex]
⇒ [tex]=0.25 \ Kg-m^2[/tex]
Suppose you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 50 mi/hr with a constant 5-second time headway between them. All vehicles in lane 2 are traveling at 55 mi/hr with a constant 6-second time headway between them. All vehicles in lane 3 are traveling at 60 mi/hr with a constant 10-second time headway between them. You collected spot speed data for all vehicles as they crossed your observation point for 75 minutes. After 10 minutes, vehicles in lane 1 stopped arriving, and after 30 minutes, vehicles in lane 2 stopped arriving. What is the flow and density of the observed traffic stream
Answer:
lane 3 Ф = 450 vehicles, ρ = 0.1 vehicle / s
lane 2 Ф_{average} = 300 vehicles, ρ _{average} = 6.66 10⁻² vehicles/s
lane 1 Ф_{average} = 300 vehicles, ρ_{average} = 2.66 10⁻² vehicle/s
Explanation:
Before solving this exercise we must clarify the concepts the flow is defined as the occurrence of an event in a time interval, in this case the passage of a car through time
Flux Density is the flux between unit area or unit time
Let's start by calculating the calculation for lane 3
the flow.
Let's use a direct rule of proportions (rule three) if the number of vehicles per unit of time (t₀ = 10s), for the observation time how many vehicles passed in the observation time (t_total = 75 * 60 = 4500 s)
Ф = 4500 s (1 vehicle / 10 s)
Ф = 450 vehicles
The flux density is the flux per unit area, in this case the area is not indicated, so we can define the flux density as the flux per unit of time.
ρ = 450/4500
ρ = 0.1 vehicle / s
Lane 2
we look for the flow
we can have separates the interval into two parts
* for the first t₁1 = 30 * 60 = 1800 s
Ф₁ = 1800 s (1 vehicle / 6s)
Ф₁ = 300 vehicles during t₁
* for the rest of the time t₂ = 4500-1800 = 2700 s
Ф₂ = 0
the average density is the total number of vehicles between the total time
#_ {vehicle} = 300 +0
Ф_{average) = # _vehicle
Ф_{average} = 300 vehicles in all time
The density is
ρ 1 = fi1 / t1
ρ1 = 300/1800
ρ1 = 1.66 10-1 vehicles / s
the average density is
ρ_{average} = [tex]\frac{\phi_1 + \phi_2}{ t_{total}}[/tex]
ρ _{average} = (300 +0) / 4500
ρ _{average} = 6.66 10⁻² vehicles / s
Lane 1
flow
* first time interval t₁ = 10 * 60 = 600 s
Ф₁ = 600 s (1 vehicle / 5s)
Ф₁ = 120 vehicles in interval t₁
* second interval t₂ = 4500-600 = 3900 s
Ф2 = 0
average flow
Ф = Ф1 + Ф2
Ф = 120 vehicles at all time
Density
* first interval
ρ₁ = 120/600
ρ₁ = 0.2 vehicles / s
* second interval
ρ₂ = 0
average density
ρ+{average} = 120/4500
ρ_{average} = 2.66 10⁻² vehicle/s
A ball is dropped off the side of a bridge,
After 1.55 S, how far has it fallen?
(Unit=m)
Answer:
Distance S = 11.77 m (Approx.)
Explanation:
Given:
Time t = 1.55 Second
Gravity acceleration = 9.8 m/s²
Find:
Distance S
Computation:
S = ut + (1/2)(g)(t)²
S = (0)(1.55) + (1/2)(9.8)(1.55)²
S = (0)(1.55) + (1/2)(9.8)(1.55)²
Distance S = 11.77 m (Approx.)
Activities:
1. Name the instrument that is used to measure Air Pressure.
2.Explain what is Cyclone and Anticyclone
Answer: barometer.
A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.
A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating over one period of the waveform. Hint: the equation for a square wave will be a piecewise function and it will be convenient to start the integration where the voltage changes; for example in this problem we could define that during the first half of the period the voltage is 0 V and for the second half of the period the voltage is 4 V
Answer:
V_{average} = [tex]\frac{1}{2} V_o[/tex] , V_{average} = 2 V
Explanation:
he average or effective voltage of a wave is the value of the wave in a period
V_average = ∫ V dt
in this case the given volage is a square wave that can be described by the function
V (t) = [tex]\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.[/tex]
to substitute in the equation let us separate the into two pairs
V_average = [tex]\int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt[/tex]
V_average = [tex]V_o \ \int\limits^{1/2}_0 {} \, dt[/tex]
V_{average} = [tex]\frac{1}{2} V_o[/tex]
we evaluate V₀ = 4 V
V_{average} = 4 / 2)
V_{average} = 2 V
PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE
Which best explains why species living in Australia are found nowhere else on Earth? This is an example of Geologic Evolution.
A.
Australia has an ecosystem different from any other area on Earth.
B.
Humans have genetically altered many Australian species in laboratories.
C.
Australian species were genetically altered after a comet hit the landmass.
D.
Australia separated from other continents and species there evolved independently.
If the mass of an object increases, then its kinetic energy will increase proportionally because mass and kinetic energy have a linear relationship when graphed.
My Response:
If the mass of an object increases, then its kinetic energy will increase proportionally because mass and kinetic energy have a linear relationship when graphed.
Sample Response:
If the speed of an object increases, then its kinetic energy will increase proportionally because speed and kinetic energy have a linear relationship when graphed.
Fossil clues are one of the _____________ clues that support the theory of continental drift.
A. crust B. resource C. climate D. rock
Answer:
a
Explanation:
I think don't get mad if I'm wrong
A 0.6 kg basketball is 3.0 high calculate its potential Energy PE=mgh
plz help me with my career!!!
part one...
Answer:
#1 Yes
Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.
Question 1: Crops.
Question 2: Diagnostic Services.
Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.
Question 4: A bachelor's degree in energy research.
Question 5: Environmental Resources.
If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.
A trough is 10 meters long, 1 meters wide, and 2 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 2 meters, and base, on top, of length 1 meters). The trough is full of water (density 1000kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g
Answer:
The amount of work required to empty the trough by pumping the water over the top is approximately 98,000 J
Explanation:
The length of the trough = 10 meters
The width of the through = 1 meter
The depth of the trough = 2 meters
The vertical cross section of the through = An isosceles triangle
The density of water in the through = 1000 kg/m³
Let 'x' represent the width of the water at a depth
x/y = 1/2
∴x = y/2
The volume of a layer of water, dV, is given as follows;
dV = 10 × y/2 × dy = 5·y·dy
The mass of the layer of water, m = ρ × dV
∴ m = 1000 kg/m³ × 5·y·dy m³ = 5,000·y·dy kg
The work done, W = m·g·h
Where;
h = The the depth of the trough from which water is pumped
g = The acceleration due to gravity ≈ 9.8 m/s²
[tex]\therefore \, W \approx \int\limits^2_0 {5,000 \times y \times 9.8 \, dy} = \left[24,500\cdot y^2 \right]^2_0 = 98,000[/tex]
The work done by the pump to pump all the water in the trough, over the top W ≈ 98,000 J
Jim and Sally both do identical jobs. Jim works quickly while Sally works slowly. Which of the following is true?
A) Sally uses more energy.
B) Jim uses more energy.
C) Jim uses more power.
D) Sally uses more power.
Suppose that white light strikes a flat piece of flint glass in air, coming in at an angle of 60 degrees to the surface (30 degrees from the normal or perpendicular). The index of refraction of this dense glass for red light is 1.710, for green light is 1.723, and for blue light is 1.735. What is the order of colors you would see in the refracted light inside the glass as they leave the surface
Answer:
The order of the light starting from the light closest to the normal line is
Blue light, followed by green ight and then lastly red light
Explanation:
White light travels from one medium to another such as from air to glass is refracted according to Snell's law as follows
n₁·sin(θ₁) = n₂·sin(θ₂)
The given parameters of the white light are;
The angle the incident (incoming) light makes with the surface = 90°
The angle of incidence of the light, θ₁ = 30°
The index of refraction of red light for the glass, n₂ = 1.710
The index of refraction of green light for the glass, n₃ = 1.723
The index of reaction of blue light for the glass, n₄ = 1.735
The refractive index of air, n₁ = 1
The angle of refraction of the red light, θ₂ is given as follows;
1 × sin(30°) = 1.710 × sin(θ₂)
sin(θ₂) = 1 × sin(30°)/1.710
θ₂ = sin⁻¹(1 × sin(30°)/1.710) ≈ 17°
The angle of refraction (to the surface's normal line) of the red light, θ₂ ≈ 17°
The angle of refraction of the green light, θ₃ is given as follows;
1 × sin(30°) = 1.723 × sin(θ₃)
sin(θ₃) = 1 × sin(30°)/1.723
θ₃ = sin⁻¹(1 × sin(30°)/1.723) ≈ 16.869°
The angle of refraction of the green light, θ₃ ≈ 16.869°
The angle of refraction of the green light, θ₄ is given as follows;
1 × sin(30°) = 1.723 × sin(θ₄)
sin(θ₄) = 1 × sin(30°)/1.735
θ₄ = sin⁻¹(1 × sin(30°)/1.735) ≈ 16.749°
The angle of refraction of the green light, θ₄ ≈ 16.749°
The order of colors we see as the in the refracted light inside the glass as the light leave the surface are;
The red light, with an angle of refraction of approximately 17° will be furthest from the normal
The green light which has an angle of refraction of 16.869° will follow and will be intermediate between the red and the blue light
The blue light which has an angle of refraction of 16.749° will follow next and it will be closest to the normal
The order of the light from the normal line will be blue, followed by green and then red light