Answer:
Before the parachute opens: Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.
Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.
I HOPE THIS WILL HELP YOU IF NOT THEN SORRYHAVE A GREAT DAY :)
A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event
Answer:
The rise in temperature is 0.06 K.
Explanation:
mass of bullet, m = 15 g
initial speed, u = 865 m/s
final speed, v = 534 m/s
mass of water, M = 13.5 kg
specific heat of water, c = 4200 J/kg K
The change in kinetic energy
[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]
According to the conservation of energy, the change in kinetic energy is used to heat the water.
K = m c T
where, T is the rise in temperature.
3473 = 13.5 x 4200 x T
T = 0.06 K
Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]
Answer:
Gap left = Change in length on heating
Gap=Initial length×Coefficient of linear expansion×change in temperature
Gap=10×0.000012×15m
⟹Gap=0.0018 m
this is an example u have to put your equation in it
3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
Learn more about refractive index of light here;
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Which of the following accurately describes circuits?
options:
A)
In a parallel circuit, there's only one path for the current to travel.
B)
In a series circuit, the amount of current passing through each part of the circuit may vary.
C)
In a series circuit, the current can flow through only one path from start to finish.
D)
In a parallel circuit, the same amount of current flows through each part of the circuit.
' C ' is the only correct statement.
Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed
Answer:
c.
magnitude of displacement
Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A. If they are separated by a distance 4.3 m, at what location between the wires is the net magnetic field be twice the strength of the magnetic field from wire 1
Answer:
x= 2*I1*d/(I1+I2) meter
A 2.5 kg block slides along a frictionless surface at 1.5 m/s.A second block, sliding at a faster 4.1 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s. What was the mass of the second block?
Answer:
1.5kg
Explanation:
Given data
mass m1= 2.5kg
mass m2=??
velocity of mass one v1= 1.5m/s
velocity of mass two v2= 4.1m/s
common velocity after impact v= 2.5m/s
Let us apply the formula for the conservation of linear momentum for inelastic collision
The expression is given as
m1v1+ m2v2= v(m1+m2)
substitute
2.5*1.5+ m2*4.1= 2.5(2.5+m2)
3.75+4.1m2= 6.25+2.5m2
collect like terms
3.75-6.25= 2.5m2-4.1m2
-2.5= -1.6m2
divide both sides by -1.6
m2= -2.5/-1.6
m2= 1.5 kg
Hence the second mass is 1.5kg
Flapping flight is very energy intensive. A wind tunnel test
on an 89 g starling showed that the bird used 12 W of
metabolic power to fly at 11 m/s. What is its metabolic power for starting flight?
Answer:
The metabolic power for starting flight=134.8W/kg
Explanation:
We are given that
Mass of starling, m=89 g=89/1000=0.089 kg
1 kg=1000 g
Power, P=12 W
Speed, v=11 m/s
We have to find the metabolic power for starting flight.
We know that
Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]
Using the formula
Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]
Metabolic power for starting flight=134.8W/kg
Hence, the metabolic power for starting flight=134.8W/kg
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making
Solve numerical problem. Please give me step - step explanation Help me out plz
Answer:
You should multiply 60 kg*9.8 and answer will come.
Hope this will help you.
Answer:
yes she is right you should multiple 60*9.8
have a great day God bless you
1. A sequence of potential differences v is applied accross a wire (diameter =0.32 mm length = 11 cm and the resulting current I are measured as follows: V 0.1 0.2 0.3 0.4 0.5 I (MA) 72 144 216 288 360 2) a) plot a graph of v against I.
b) determine the wire's resistence , R.
c) State ohm's law and try to relate it . your results.
Answer:
a. Find the graph in the attachment
b. 720 kΩ
c. The ratio V/I gives us our resistance which is 720 kΩ
Explanation:
a) plot a graph of V against I.
To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.
b) Determine the wire's resistance , R.
The resistance of the wire is determined as the gradient of the graph.
R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)
Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA
R = (144 V - 72 V)/(0.2 - 0.1) mA
R = 72 V/0.1 mA
R = 72 V/(0.1 × 10⁻³ A)
R = 720 × 10³ V/A
R = 720 kΩ
c) State ohm's law and try to relate it your results.
Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.
Mathematically, V ∝ I
V = kI
V/I = k = R
Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.
Since V/I is constant, we thus verify Ohm's law.
Question 5 of 10
What must be the same for two resistors that are connected in parallel?
Answer:
in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .
Explanation:
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.
Answer:
the distance from the cornea where a very distant object will be imaged is 23.35 mm
Explanation:
Given the data in the question;
For a spherical refracting surface;
[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium
[tex]d_0[/tex] is the object distance
[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium
[tex]d_i[/tex] is the image distance
R is the radius of curvature
Now, let [tex]d_0[/tex] = ∞, such that;
[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
we make [tex]d_i[/tex] subject of the formula
[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )
[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )
given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00
so we substitute
[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )
[tex]d_i[/tex] = 9.165 / 0.41
[tex]d_i[/tex] = 23.35 mm
Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?
Answer:
The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.
Explanation:
Consider newtonian mechanics here.
Dynamic equation is
The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.
we'll consider newtonian mechanics here.
so the dynamic equations is S = ut + 0.5at^2
we know u=0.8
S=1.2+0.9
t=1.2/1.5
find a.
What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm
Answer:
The submarine's maximum safe depth in sea water is 801.678 m.
Explanation:
P=Po+(rho)*g*h
Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)
Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2
Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa
Initial Pressure= 1atm= 101.3kPa= 101300Pa
Water Density (rho) = 1000kg/m^3
Gravity= 9.8m/s^2
So rearranging for h= (P-Po)/((rho)*g)
h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m
Write the derivation and unit of impulse
Answer:
impulse applied to an object produces an equivalent vector change in its linear momentum, also in the resultant direction. The SI unit of impulse is the newton second (N⋅s), and the dimensionally equivalent unit of momentum is the kilogram meter per second (kg⋅m/s).
A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2).
a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213
Answer:
4√6 rad/s
Explanation:
Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.
So, W = F
mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft
making k the spring constant subject of the formula, we have
k = mg/x
substituting the values of the variables into the equation, we have
k = mg/x
k = 4 lb × 32 ft/s² ÷ 1/3 ft
k = 32 × 4 × 3
k = 384 lbft²/s²
Now, assuming there is no friction and no external force, we have an undamped system.
So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb
So, substituting the values of the variables into the equation, we have
ω = √(k/m)
ω = √(384 lbft²/s² ÷ 4 lb)
ω = √96
ω = √(16 × 6)
ω = √16 × √6
ω = 4√6 rad/s
15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?
Answer: (15 - 0)/1.8 = 8. 33m/s^2
Explanation:
The acceleration of the racehorse is 8.33 m/s²
The given parameters;
initial velocity of the racehorse, u = 0
final velocity of the racehorse, v = 15 m/s
time of motion of the horse, t = 1.8 s
The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]
Thus, the acceleration of the racehorse is 8.33 m/s²
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What is the need for satellite communication elaborate
The high frequency radio waves used for telecommunications links travel by line of sight and so are obstructed by the curve of the Earth. The purpose of communications satellites is to relay the signal around the curve of the Earth allowing communication between widely separated geographical points.
Explanation:
hope it helps!!
Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not
static equilibrium
Which statement best describes Keisha's error?
An object that is not moving is always in static equilibrium.
O An object that is moving must be in dynamic equilibrium.
An object in either state of equilibrium must have no forces acting on it.
An object in either state of equilibrium must have no net force acting on it.
Answer:
An object in either state of equilibrium must have no net force acting on it.
Explanation:
Answer: An object in either state of equilibrium must have no net force acting on it.
Explanation:
what Is accuracy ............
Answer:
Accuracy is how much the consequence of an estimation adjusts to the right worth or a norm' and basically alludes to how close an estimation is to its concurred esteem
《OAmalaOHopeO》
Answer:
In a set of measurements, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.
Explanation:
_Hope it helps you_
Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?
Answer:
T = 1/f = 1/50(s)
ω = 2πf = 100π (rad/s)
(vote 5 sao nhó :3 )
In which situation should a parent be proactive and act to assume responsibility?
Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.
Explanation: i picked this and it is correct, you’re welcome:)
Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for longer wavelengths of light. The work function of the surface is
Explanation:
Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]
[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]
[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]
The work function [tex]\phi[/tex] is then
[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]
[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]
The work function of the surface is equal to 3.98 × 10⁻¹⁹J.
What are frequency and wavelength?The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.
The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.
The expression between wavelength (λ), frequency, and speed of light (c) is:
c = νλ
Given, the wavelength of the light, ν = 500 nm
The frequency of the light can determine from the above-mentioned relationship:
ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz
The work function = h ν = 6 × 10¹⁴ × 6.626 × 10⁻³⁴
φ = 3.98 × 10⁻¹⁹J
Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.
Learn more about wavelength and frequency, here:
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A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz
The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.
Given the following data:
Period = 645 μsNote: μs represents microseconds.
Conversion:
1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds
645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
[tex]Frequency = \frac{1}{Period}[/tex]
Substituting the value into the formula, we have;
[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
Conversion:
1 hertz = 0.001 kilohertz
1550.39 hertz = X kilohertz
Cross-multiplying, we have;
X = [tex]0.001[/tex] × [tex]1550.39[/tex]
X = 155039 kHz
To 3 significant figures;
Frequency = 155 kHz
Therefore, the corresponding frequency of this sinewave, in kHz is 155.
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(a+b)[tex]x^{2}[/tex]
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Answer:
[tex]v=(6ti+6k)\ m/s[/tex]
Explanation:
Given that,
The position of a particle is given by :
[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]
Let us assume we need to find its velocity.
We know that,
[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]
So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].
A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.
Answer:
The distance between mirror and you is 2 ft.
Explanation:
diameter, d = 8 ft
radius of curvature, R = 4 ft
magnification, m = 0.5
focal length, f = R/2 = 4/2 = 2 ft
let the distance of object is u and the distance of image is v.
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]
Use the formula of magnification
[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]
A rock is suspended by a light string. When the rock is in air, the tension in the string is 37.8 N. When the rock is totally immersed in water, the tension is 32.0 N. When the rock is totally immersed in an unknown liquid, the tension is 20.2 N. What is the Density of the unknown liquid?
When the rock is suspended in the air, the net force on it is
∑ F₁ = T₁ - m₁g = 0
where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So
T₁ = m₁g = 37.8 N
When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then
∑ F₂ = T₂ + B₂ - m₁g = 0
where B₂ is the magnitude of the buoyant force. Then
B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N
B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then
5.8 N = m₂g ==> m₂ ≈ 0.592 kg
If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was
1.00 × 10³ kg/m³ = m₂/V ==> V ≈ 0.000592 m³ = 592 cm³
and this is also the volume of the rock.
When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is
∑ F₃ = T₃ + B₃ - m₁g = 0
which means the buoyant force is
B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N
The mass m₃ of the liquid displaced is then
17.6 N = m₃g ==> m₃ ≈ 1.80 kg
Then the density ρ of the unknown liquid is
ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³
