Answer:
I don't know but you got this!!
A 0.6 kg basketball is 3.0 high calculate its potential Energy PE=mgh
Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision
Answer:
v = 28.98 ft / s
Explanation:
For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision
In the exercise they indicate the weight of each car
Wₐ = 1500 lb
W_b = 1125 lb
Car B's velocity from v_b = 42.0 mph westward, car A travels east
let's find the mass of the vehicles
W = mg
m = W / g
mₐ = Wₐ / g
m_b = W_b / g
mₐ = 1500/32 = 46.875 slug
m_b = 125/32 = 35,156 slug
Let's reduce to the english system
v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s
We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved
we assume the direction to the east (right) positive
initial instant. Before the crash
p₀ = mₐ v₀ₐ - m_b v_{ob}
final instant. Right after the crash
p_f = (mₐ + m_b) v
the moment is preserved
p₀ = p_f
mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v
v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]
we substitute the values
v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]
v = 0.559 v₀ₐ - 26.40 (1)
Now as the two vehicles united we can use the relationship between work and kinetic energy
the total mass is
M = mₐ + m_b
M = 46,875 + 35,156 = 82,031 slug
starting point. Jsto after the crash
K₀ = ½ M v²
final point. When they stop
K_f = 0
The work is
W = - fr x
the negative sign is because the friction forces are always opposite to the displacement
Let's write Newton's second law
Axis y
N-W = 0
N = W
the friction force has the expression
fr = μ N
we substitute
-μ W x = Kf - Ko
-μ W x = 0 - ½ (W / g) v²
v² = 2 μ g x
v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5
v = 28.98 ft / s
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.
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.
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
10. A person is about to kick a soccer ball. Consider the leg and foot to be a single rigid body and that it rotates about a fixed axis through the knee joint center. Immediately prior to impact the leg and foot are in the vertical direction and the distal end of the foot has an acceleration of 10 g in the horizontal direction. The muscle force vector makes an angle of 15 degrees with the vertical and has a moment arm of 5 cm from the knee joint center. Assume the person is 1.7 m tall and has a mass of 75 kg. Find the force in the muscle
Answer:
i don't know but i hope you get it right
Explanation
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.)
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
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.
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.
HELP PLEASE DUE IN 3 MINUTES
Answer:
Tectonic Plate Movement
Explanation:
Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.
Answer:
tectonic plates movement
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]
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
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 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
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
Which one is it? Help ASAP
Answer:
extreme heat, because no physical damage can demagnetize a magnet
Explanation:
Answer:
the 3rd one
Explanation:
A toy projectile is fired vertically from the ground upward with a velocity of +29 meters per second. It arrives at its maximum altitude in 3.0 seconds. How high does the projectile go?
Answer:
[tex]\boxed{\text{\sf \Large 42 m}}[/tex]
Explanation:
Use height formula
[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]
u is initial velocity
θ = 90° (fired vertically upward)
g is acceleration of gravity
[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]
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]
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
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:
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
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]
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:
:)
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
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
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.
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]
You have a hand-crank generator with a 100-turn coil (each turn having an area of 0.0350 m2), which can spin through a uniform magnetic field that has a magnitude of 0.0500 T. You can turn the crank at a maximum rate of 3 turns per second, but the hand crank is connected to the coil through a set of gears that makes the coil spin at a rate 6 times larger than the rate at which you turn the crank. What is the maximum emf you can expect to get out of this generator
Answer:
Explanation:
Maximum emf produced = nωAB. where n is no of turns , A is area , B is magnetic field and ω is angular velocity .
n = 100 , A = .035 m²
B = .05 T
ω = 2π f , f is no of revolution per second by coil
= 2 x 3.14 x 6 x 3
= 113.04 rad /s
Maximum emf produced = 100 x 113.04 x .035 x .05
= 19.78 volt .
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