Answer:
Electromagnetic waves that act like particles would be the violet light
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.
The power of an engine is a measure of
A) the total amount of work it can perform.
B) the rate at which it can perform work.
C) its ability to outperform a horse.
D) its volume.
Answer:
d
Explanation:
Answer:
C
Explanation:
Horsepower refers to the power an engine produces. It's calculated through the power needed to move 550 pounds one foot in one second or by the power needs to move 33,000 pounds one foot in one minute. The power is gauged by the rate it takes to do the work.
Ps:Can I be brainliest.
Have a nice day
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]
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]
Which of the following equations illustrates the law of conservation of
matter?
Answer:
A
Explanation:
the balancing is correct in the first one
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 conductor Which level of organization is all the same type of ecosystem found on Earth
that share common characteristics?
A biome, in simple terms, is a set of ecosystems sharing similar characteristics with their abiotic factors adapted to their environments.
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.
A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. The bicycle is placed on a stationary stand and a resistive force of 121 N is applied tangent to the rim of the tire. (a) What force must be applied by a chain passing over a 8.96-cm-diameter sprocket in order to give the wheel an acceleration of 4.40 rad/s2
Answer:
F2 = 834 N
Explanation:
We are given the following for the bicycle;
Diameter; d1 = 63 cm = 0.63 m
Mass; m = 1.75 kg
Resistive force; F1 = 121 N
For the sprocket, we are given;
Diameter; d2 = 8.96 cm = 0.0896 m
Radius; r2 = 0.0896/2 = 0.0448 m
Radial acceleration; α = 4.4 rad/s²
Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²
Where r1 = (d1)/2 = 0.63/2
r1 = 0.315 m
Thus, I = 1.75 × 0.315²
I = 0.1736 Kg.m²
The torque is given by the relation;
I•α = F1•r1 - F2•r2
Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².
Thus;
0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)
>> 0.76384 = 38.115 - (0.0448F2)
>> 0.0448F2 = 38.115 - 0.76384
>> F2 = (38.115 - 0.76384)/0.0448
>> F2 = 833.73 N
Approximately; F2 = 834 N
The force applied by a chain passing over a 8.96-cm-diameter sprocket in order to give the wheel an acceleration of 4.40 rad/s2 is F2 = 834 N
What is force?
Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
We are given the following for the bicycle;
Diameter; d1 = 63 cm = 0.63 m
Mass; m = 1.75 kg
Resistive force; F1 = 121 N
For the sprocket, we are given;
Diameter; d2 = 8.96 cm = 0.0896 m
Radius; r2 = 0.0896/2 = 0.0448 m
Radial acceleration; α = 4.4 rad/s²
Now the moment of inertia of the wheel which is assumed to be a hoop is given by;
[tex]I=mr_1^2[/tex]
Where
r1 = 0.315 m
I = 1.75 × 0.315²
I = 0.1736 Kg.m²
The torque is given by the relation;
[tex]I\alpha=F_1\times r_1-F_2\times r_2[/tex]
Where [tex]F_2[/tex] is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².
0.1736 × 4.4 = (121 × 0.315) - (0.0448[tex]F_2[/tex])
0.76384 = 38.115 - (0.0448[tex]F_2[/tex])
[tex]F_2= \dfrac{38.115-0.7638}{0.0448}[/tex]
[tex]F_2=833.73\ N[/tex]
Approximately; F2 = 834 N
Thus the force applied by a chain passing over a 8.96-cm-diameter sprocket in order to give the wheel an acceleration of 4.40 rad/s2 is F2 = 834 N
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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 .
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
what is angle of deviance in physics
The angle of deviation is defined as the angle which is obtained from the difference between the angle of incidence and the angle of refraction created by the ray of light travelling from one medium to another that has a different refractive index. I hope this helps you
you're reading from the journal of a European explorer from the early 1600s. In one passage, the explorer describes itting on the Atlantic Ocean with little wind. He describes the area as being quite far north of the equator and having nigh pressure. n which type of global wind was this explorer sailing? doldrums horse latitudes jet stream O polar easterlies
Answer: horse latitudes
Explanation:
Answer:
B
Explanation:
Doldrums are located some 5 degrees away from north while horse latitude is located some thirty degrees away from north latitude and by same degrees from the southern hemisphere. Polar eastariles are cold winds blowing across the Polar Regions and jet stream is fast blowing winds.
Horse latitude is silent winds and the region has high pressure and very little precipitation.
Hence, option B is correct
In a glider stunt at an air show, a towing airplane (motorized plane pulling the gliders) takes off from a level runway with two gliders in tow. The gliders are pulled by two tow ropes, the first glider attached to the back of the towing plane and the second glider attached to the back of the first glider. The mass of each glider is 700 kg. The total resisting force (air drag plus friction with the runway) on each glider is equal to 3700 N. The resisting force can be assumed to be constant during the take off. The maximum allowed tension in the tow rope between the transport plane and the first glider is 12000 N. The maximum allowed tension in the second tow rope is the same as the first tow rope.
If the takeoff speed for the gliders and the plane is 40 m/s, what minimum length of runway is needed for take off? Express your answer in meters.
Answer:
minimum length of runway is needed for take off 243.16 m
Explanation:
Given the data in the question;
mass of glider = 700 kg
Resisting force = 3700 N one one glider
Total resisting force on both glider = 2 × 3700 N = 7400 N
maximum allowed tension = 12000 N
from the image below, as we consider both gliders as a system
Equation force in x-direction
2ma = T -f
a = T-f / 2m
we substitute
a = (12000 - 7400 ) / (2 × 700 )
a = 4600/1400
a = 3.29 m/s²
Now, let Vf be the final speed and Ui = 0 ( as starts from rest )
Vf² = Ui² + 2as
solve for s
Vf² = 0 + 2as
2as = Vf²
s = Vf² / 2a
given that take of speed for the gliders and the plane is 40 m/s
we substitute
s = (40)² / 2×3.29
s = 1600 / 6.58
s = 243.16 m
Therefore, minimum length of runway is needed for take off 243.16 m
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.
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
A 0.6 kg basketball is 3.0 high calculate its potential Energy PE=mgh
Choose true or false for each statement regarding a parallel plate capacitor.. The voltage of a disconnected charged capacitor increases when the plate area is decreased. The electric field is dependent on the charge density on the plates. The voltage of a connected charged capacitor remains the same when the plate area is decreased.
Answer:
Explanation:
The voltage of a disconnected charged capacitor increases when the plate area is decreased.
When plate area decreases , capacitance C decreases , but charge Q remains constant .
Q = C V where C is capacitance and V is voltage .
when C decreases , V increases for keeping Q constant .
So the statement is true.
The electric field is dependent on the charge density on the plates.
This statement is true .
The voltage of a connected charged capacitor remains the same when the plate area is decreased .
For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .
So the statement is true .
Is epidermis made of living cells or dead cells and why?
Answer:
Dead
Explanation:
there are no blood vessels in the epidermis so the cells get their nutrients by diffusion from the connective tissue below
Which 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.
A 50 kg object traveling at 100 m/s collides (perfectly elastic) with a 50 kg object initially at rest.
What is the sum of the momentum of both objects prior to the collision?
a. 100 kgm/s b. 500 kgm/s C. 5,000 kgm/s d. 100,000 kgm/s
Answer:
a
Explanation:
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.
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
Aunt Jane weights 45 Newtons. What is her mass?
Answer:
10.116 Pounds/45 newtons = 10.1164024 pounds/force
Explanation:
Divide the newtons by the rate of acceleration, which will give you the mass of the object. The mass will be in kilograms, because a single newton represents the amount of force needed to move one kilogram one meter. For our example, we will divide 10 N by 2 m/s/s, which give us a mass of 5 kg
help asap According to your data, what trend exists between the independent variable and the dependent variable? Make sure to use terms like “positive”, “negative”, or “neutral” to describe the trend. Add your reasoning. 100pts
Answer:
Explanation:
Please post if there is a graph showing the data.
Without a graph, in general the trend can be described as:
"positive" when the dependent variable increases with an increase in independent variable
"negative" when the dependent variable decreases with an increase in independent variable
"neutral" when the dependent variable remains the same with an increase or decrease in independent variable
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
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 boy is playing with a ball of mass 72g attached to a string. He is moving it at constant speed in a horizontal circle of radius 0.9m. a. State the net work done on the ball in one complete revolution. (1) b. The tension in the string is 60N. Determine the angular velocity of the ball. (4) c. Calculate the time it takes for the ball to complete 20 revolutions. (2)
Answer:
Explanation:
radius of circle r = 0.9 m.
(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .
( b )
Tension in string T = m ω²r
Putting the values
60 = .072 x ω² x 0.9
ω² = 926
ω = 30.4 rad /s
angle made in 20 revolutions θ = 20 x 2π = 126.6 rad
time taken = θ / ω
= 126.6 / 30.4
= 4.16 s .
The ballistic pendulum is a device used to measure the speed of a fast - moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet embeds in the block, and the entire system swings up to a height h. A Walther PPK, the gun used by James Bond, has an average muzzle velocity of 950 m/s. In a ballistic pendulum, how high would we expect the block to travel when shot by a Walther PPK, given the mass of a 0.32 ACP is 5 grams, a the mass of the block is 2kg
Answer:
h = [tex](\frac{m}{m+M} )^2 \ \frac{v_o^2}{2g}[/tex]
Explanation:
To solve this problem, let's work in parts, let's start with the conservation of the moment, for this we define a system formed by the block and the bullet, in such a way that the forces during the collision have been internal and the moment is conserved.
initial instant. Before the crash
p₀ = m v₀
final instant. Right after the crash, but before the pendulum started to climb
m_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v = [tex]\frac{m}{m+M} \ v_o[/tex]
Now we work the pendulum system with embedded block, we use the concept of conservation of energy
starting point. Lower
Em₀ = K = ½ (m + M) v²
final point. higher, when it stops
Em_f = U = (m + M) g h
as there is no friction, energy is conserved
Em₀ = Em_f
½ (m + M) v² = (m + M) g h
h = [tex]\frac{v^2}{2g}[/tex]
we substitute the speed value of the block plus bullet set
h = [tex]( \frac{m}{ m+M} \ v_o )^2 \ \frac{1}{2g}[/tex]
h = [tex](\frac{m}{m+M} )^2 \ \frac{v_o^2}{2g}[/tex]