Explanation:
mass=kilogram,temperature=Klevin,power=watt,density=kilogram per cubic metre
Explanation:
the unit of mass is kg , temperature is kelvin ,power is watt and density is kilogram per cubic meter.
If a jet travels 350 m/s, how far will it travel each second?
Answer:
350
Explanation:
Since it travels 350 meters per second, the jet will travel 350 meters in one second.
En la siguiente expresión matemáticas w=mg el peso w con relación a se relaciona con la masa m en una proporción
a) Directamente proporcional b) Inversamente proporcional c) Es constante
d) Ninguna de las anteriore
Answer:
a) Directamente proporcional
Explanation:
El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.
Matemáticamente, el peso de un objeto viene dado por la fórmula;
[tex] Peso = mg [/tex]
Donde;
m es la masa del objeto.
g es la aceleración debida a la gravedad.
De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.
Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.
A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 ksi. The bar is subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 ksi. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use Modified Goodman criterion.
Answer:
The correct solution is:
(a) 1.66
(b) 1.05
Explanation:
Given:
Bending stress,
[tex]\sigma_b = 25 \ kpsi[/tex]
Torsional stress,
[tex]\tau= 15 \ kpsi[/tex]
Yield stress of steel bar,
[tex]\delta_y = 60 \ kpsi[/tex]
As we know,
⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]
[tex]= \sqrt{(25)^2+3(15)^2}[/tex]
[tex]=36.055 \ kpsi[/tex]
(a)
The factor of safety against static failure will be:
⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]
By putting the values, we get
[tex]=\frac{60}{36.055}[/tex]
[tex]=1.66[/tex]
(b)
According to the Goodman line failure,
[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]
[tex]S_e = 40 \ kpsi[/tex]
[tex]\sigma_m = \sqrt{3} \tau[/tex]
[tex]=\sqrt{3}\times 15[/tex]
[tex]=26 \ kpsi[/tex]
[tex]Sut = 80 \ kpsi[/tex]
⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]
[tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]
[tex]\eta_y = 1.05[/tex]
how does laser works ?
Explanation:
Lasers produce a narrow beam of light in which all of the light waves have very similar wavelengths. The laser's light waves travel together with their peaks all lined up, or in phase. This is why laser beams are very narrow, very bright, and can be focused into a very tiny spot.
A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
Complete question:
A bullet 2 cm long is fired at 420m/s and passes straight through a 10.0 cm thick board, exiting at 280m/s? What is the average acceleration of the bullet through the board?
Answer:
The average acceleration of the bullet through the board is -4.083 x 10⁵ m/s²
Explanation:
Given;
initial velocity of the bullet, u = 420 m/s
final velocity of the bullet, v = 280 m/s
length of the bullet, d₁ = 2 cm
thickness of the board, d₂ = 10 cm
total distance penetrated by the bullet through the board;
d = d₁ + d₂ = 2 cm + 10 cm = 12 cm = 0.12 m
The average acceleration of the bullet through the board is calculated as;
[tex]v^2 = u^2 + 2ad\\\\2ad = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2d} \\\\a = \frac{(280^2) - (420^2)}{2(0.12)} = -4.083 \times 10^{5} \ m/s^2[/tex]
Therefore, the average acceleration of the bullet through the board is -4.083 x 10⁵ m/s²
What is a measure between the difference in start and end positions?
Answer:
Displacement
General Formulas and Concepts:
Kinematics
Displacement vs Total DistanceExplanation:
Displacement is the difference between the start position and end position.
Total Distance is the entire distance traveled between the start and end position.
Topic: AP Physics 1 Algebra-Based
Unit: Kinematics
A roller coaster has a vertical loop with radius 25.7 m. With what minimum speed should the roller-coaster car be moving at the top of
the loop so that the passengers do not lose contact with the seats?
m/s
Answer:
15.88m/s
Explanation:
At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.
[tex]F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7} = 15.88 m/s[/tex]
The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N
[tex]F = 5.93×10^{13}\:\text{N}[/tex]
Explanation:
Given:
[tex]m_1= 2×10^{16}\:\text{kg}[/tex]
[tex]m_2= 4×10^{22}\:\text{kg}[/tex]
[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]
Using Newton's universal law of gravitation, we can write
[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]
[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]
[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]
A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.235 A. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.40 s.
Required:
a. What average emf is induced in the second coil if it has a diameter of 3.5 cm and N = 7?
b. What is the induced emf if the diameter is 7.0 cm and N = 10?
Answer:
a) ε = 14.7 μv
b) ε = 21 μv
Explanation:
Given the data in the question;
Diameter of solenoid; d = 3 cm
radius will be half of diameter, so, r = 3 cm / 2 = 1.5 cm = 1.5 × 10⁻² m
Number of turns; N = 40 turns per cm = 4000 per turns per meter
Current; [tex]I[/tex] = 0.235 A
change in time Δt = 0.40 sec
Now,
We determine the magnetic field inside the solenoid;
B = μ₀ × N × [tex]I[/tex]
we substitute
B = ( 4π × 10⁻⁷ Tm/A ) × 4000 × 0.235
B = 1.1881 × 10⁻³ T
Now, Initial flux through the coil is;
∅₁ = NBA = NBπr²
and the final flux
∅₂ = 0
so, the εmf induced ε = -Δ∅/Δt = -( ∅₂ - ∅₁ ) / Δt
= -( 0 - NBπr² ) / Δt
= NBπr² / Δt
a)
for N = 7
ε = [ 7 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40
ε = 14.7 × 10⁻⁶ v
ε = 14.7 μv
b)
for N = 10
ε = [ 10 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40
ε = 21 × 10⁻⁶ v
ε = 21 μv
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
a) The mass flow rate through the nozzle can be calculated with the following equation:
[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]
Where:
[tex]v_{i}[/tex]: is the initial velocity = 20 m/s
[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²
[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³
[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]
[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]
[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]
Therefore, the exit area of the nozzle is 23.6 cm².
You can find another example of mass flow rate here: https://brainly.com/question/13346498?referrer=searchResults
I hope it helps you!
a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.
We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:
[tex]\dot m_{in} = \dot m_{out}[/tex] (1)
Where:
[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.
[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.
Given that air flows at constant rate, we expand (1) by dimensional analysis:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)
Where:
[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.
[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.
[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.
a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:
[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]
[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]
[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]
The mass flow rate through the nozzle is 0.265 kilograms per second.
b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]
[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]
[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]
[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]
[tex]A_{out} = 23.202\,cm^{2}[/tex]
The exit area of the nozzle is 23.202 square centimeters.
1. There is a famous intersection in Kuala Lumpur, Malaysia, where thousands of vehicles pass each hour. A 750 kg Tesla Model S traveling south crashes into a 1250 kg Ford F-150 traveling east. What are the initial speeds of each vehicle before collision if they stick together after crashing into each other and move at an angle of 320 and a common velocity of 18 m/s.
Solution :
Let the positive [tex]x-axis[/tex] is along the East and the positive [tex]y[/tex] direction is along the north.
Given :
Mass of the Tesla car, [tex]m_1[/tex] = [tex]750 \ kg[/tex]
Mass of the Ford car, [tex]m_2 = 1250 \ kg[/tex]
Now let the initial velocity of Tesla car in the south direction be = [tex]-v_1j[/tex]
The initial momentum of Tesla car, [tex]p_1 = -750 \ v_1[/tex]
Let the initial velocity of Ford car in the east direction be = [tex]v_2 \ i[/tex]
So the initial momentum of the Ford car is [tex]p_2=1250\ v_2 \ i[/tex]
Therefore, the initial velocity of both the cars is [tex]p_i = p_1+p_2[/tex]
[tex]=1250 \ v_2 \ i - 750\ v_1 \ j[/tex]
Now the final velocity of both the cars is [tex]v = 18 \ m/s[/tex]
So the vector form is :
[tex]v = 18\cos 32\ i-18 \sin 32 \ j[/tex]
[tex]= 15.26 \ i - 9.54 \ j[/tex]
Therefore the momentum after the accident is
[tex]p_f=(m_1+m_2) \times v[/tex]
[tex]=(750+1250) \times (15.26 \ i - 9.54 \ j)[/tex]
[tex]= 30520\ i -19080\ j[/tex]
According to the law of conservation of momentum, we know
[tex]p_i = p_f[/tex]
[tex]1250 \ v_2 \ i - 750\ v_1 \ j[/tex] [tex]= 30520\ i -19080\ j[/tex]
[tex]1250 \ v_2 = 30520[/tex]
[tex]v_2=24.4 \ m/s[/tex]
From, [tex]750\ v_1 = 19080[/tex]
We get, [tex]v_1=25.4 \ m/s[/tex]
Therefore the speed of Tesla car before collision = 25.4 m/s
The speed of ford car before collision = 24.4 m/s
an object of volume has 20ml has a mass of 2.5kg what will be its density
Answer:
0.125
Explanation:
Density =mass/volume
Density =2.5/20
Density =0.125
Answer:
D=m/v
=2.5kg/(20/1000000)m^3
2.5kg÷0.000002m^3
1250000kgm^-3
Explanation:
20 is divided by 1000000 because m^3 is its si unit
What is the principle of potentiometer?
Answer:
The principle of a potentiometer is that the potential dropped across a segment of a wire of uniform cross-section carrying a constant current is directly proportional to its length. The potentiometer is a simple device used to measure the electrical potentials (or compare the e.m.f of a cell).
Explanation:
I hope it will help you
26. A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
Answer:
Explanation:
The formula for determining the Emf induced in a loop is:
[tex]\varepsilon = \dfrac{d \phi}{dt}[/tex]
[tex]\varepsilon = \dfrac{d (B*A)}{dt}[/tex]
[tex]\varepsilon = A \times \dfrac{dB}{dt}[/tex]
[tex]\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}[/tex]
where;
square area A = ( l²)
l² = 6.0 cm = 6.0 × 10⁻²
∴
[tex]\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S[/tex]
[tex]\varepsilon =18 \times 10^{6} \ V[/tex]
Recall that:
The resistivity of copper = [tex]1.68 \times 10^{-8}[/tex] ohm m
We can as well say that the length of the copper wire = perimeter of the square loop;
The perimeter of the square loop = 4L
Thus, the length of the copper wire = 4 (6.0 × 10⁻² )m
= 24× 10⁻² m
Finally, the current in the loop is determined from the formula:
V = IR
where,
V = voltage
I = current and R = resistance of the wire
Making "I" the subject:
I = V/R
where;
[tex]R = \dfrac{\rho \times l}{A}[/tex]
[tex]R = \dfrac{\rho \times l}{\pi * r^2}[/tex]
[tex]R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}[/tex]
[tex]R = 0.001283 \ ohms[/tex]
∴
[tex]I = \dfrac{18*10^{-6}}{0.001283}[/tex]
I = 14.029 mA
how interfacial angles are determined using contact goinometer
A spherical, concave shaving mirror has a radius of curvature of 0.983 m. What is the magnification of a person's face when it is 0.155 m from the vertex of the mirror (answer sign and magnitude)
Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex] .................2
put here value and we get
[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]
v = 0.155 mso
Magnification will be here as
m = [tex]- \frac{v}{u}[/tex]
m = [tex]\frac{0.155}{0.155}[/tex]
m = 1Answer:
The magnification is 1.5.
Explanation:
radius of curvature, R = - 0.983 m
distance of object, u = - 0.155 m
Let the distance of image is v.
focal length, f = R/2 = - 0.492 m
Use the mirror equation
[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]
The magnification is given by
m = - v/u
m = 0.226/0.155
m = 1.5
A meter stick has only two forces acting on it, of equal sizes at the ends that are in opposite directions. (I have magically turned gravity Off.) Is the stick in equilibrium? Explain your answer.
Answer:
Explanation:
Equilibrium is a state in which the algebraic sum of all forces acting on an object is zero. Thus the object has no force acting on it. The types are: stable, unstable and neutral equilibrium. While a torque is a turning force which are equal but acts in an opposite direction. When applied to on object, it constitute a turning effect. Example is the force applied on a tap, handle wheel of a car etc.
In the given question, the condition stated shows that the stick would experience a torque, thus not in equilibrium. Since the forces at its ends are in opposite directions, then it continues to rotate about its axis.
A converging lens is used to focus light from a small bulb onto a book. The lens has a focal length of 10.0 cm and is located 40.0 cm from the book. Determine the distance from the lens to the light bulb.
Answer:
[tex]u=13.3cm[/tex]
Explanation:
From the question we are told that:
Focal Length [tex]F=10.0cm[/tex]
Distance [tex]d=40cm[/tex]
Generally the equation for Focal length is mathematically given by
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{10}=\frac{1}{u}+\frac{1}{40}[/tex]
[tex]\frac{1}{u}=\frac{3}{40}[/tex]
[tex]u=13.3cm[/tex]
Focal length is the distance from the center of the lens to principle foci. The distance of the from the lens to the light bulb is 13.3 cm.
The distance can be determined by the formula,
[tex]\bold {\dfrac 1{f} = \dfrac 1{u} + \dfrac 1{v} }[/tex]
Where,
f - focal length = 10 cm
u - distance of object = ?
v = distance of image = 40 cm
Put the values in the equation,
[tex]\bold {\dfrac 1{10} = \dfrac 1{u} + \dfrac 1{40} }\\\\\bold {\dfrac 1{u} = \dfrac 3{40}}\\\\\bold {\dfrac 1{u} = 13.3 cm}[/tex]
Therefore, the distance of the from the lens to the light bulb is 13.3 cm.
To know more about the focal length,
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on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
The coefficients of friction between a race cars tyres and the track surface are
the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).
Explanation:
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What is the Ah rating of a battery that can provide 0.8 A for 76 h?
Answer:
6.08
Explanation:
Given that,
Current, I = 0.8 A
Time, t = 76 h
We need to find the Ah rating of a battery. It can be calculated by taking the product of current and time. So,
Ah = (0.8)(76)
= 6.08 Ah
So, the Ah rating of the battery is 6.08.
Please help, I really need this. Thanks
Answer
Delta Q = change in thermal energy = c M * change in temperature
change in temperature = Q / (c * M)
change in temperature = -12 J / (390 J / Kg*deg * .012 kg
change in temp = -12 / (390 * .012) = - 2.56 deg C
what are three effects of gravity
Answer:
effect on motation.effect on direction
The instrument includes a light source, which is passed through a Choose... , which isolates a single wavelength to pass through an aperture to reach the Choose... . Then, the light travels to the Choose... , which measures the intensity of light reaching it.
Answer:
Following are the response to the given question:
Explanation:
It's being used to measure the amount of light absorbed after traveling through a test tube (the amount of solar radiation received). For several quantitative estimations, this technique is widely employed. Spectrometer and Spectrometer were two devices that are used together to light intensity and light intensity.
It creates and diffuses phosphorescent light into the selected frequency, while the Spectrometer measures the strength of attenuation by the sample solution.
Diffraction beams or prisms are being used to convert polychromatic illumination into monochrome light.
Afterward, the sunlight has a certain hue. Once it reaches the specimen cuvette, it begins absorption. It falls on a sensor that transforms its intensity into such an electronic current.
Here are some ways to fill in such gaps:
In order to reach the specimen cuvette, the light from the light source must be routed via an aperture in order to be isolated by either a diffraction pattern. Light travels to the detector, which detects its intensity.
What is mechanical advantage?
O A. Output force divided by input force
O B. Input work divided by output work
O C. Output work divided by input work
O D. Input force divided by output force
Answer:
0A.Output force divided by input force.
2. What is the average speed of an athlete who runs 1500 m in 4 minutes?
Answer:
375 is the answer.
Explanation:
Speed : Distance / Time taken
S: m/ s
s: 1500/4
375 m / s answer
Answer:
375m per minute
Explanation:
if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency 2 rev/s; What is the maximum emf induced in this loop? If its resistance is 0.00336 ohms, how much current is induced in this loop? And what is the maximum power dissipated in the loop due to its rotation in Jupiter's magnetic field?
Answer:
a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W
Explanation:
This exercise is about Faraday's law
fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]
where the magnetic flux is
Ф = B x A
the bold are vectors
A = π r²
we assume that the angle between the magnetic field and the normal to the area is zero
fem = - B π 2r dr/dt = - 2π B r v
linear and angular velocity are related
v = w r
w = 2π f
v = 2π f r
we substitute
fem = - 2π B r (2π f r)
fem = -4π² B f r²
For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T
we reduce the magnitudes to the SI system
f = 2 rev / s (2π rad / 1 rev) = 4π Hz
we calculate
fem = - 4π² 428 10⁻⁶ 4π 0.10²
fem = - 16π³ 428 10⁻⁶ 0.010
fem = - 2.1514 10⁻⁴ V
for the current let's use Ohm's law
V = I R
I = V / R
I = -2.1514 10⁻⁴ / 0.00336
I = - 64.0 10⁻³ A
Electric power is
P = V I
P = 2.1514 10⁻⁴ 64.0 10⁻³
P = 1.38 10⁻⁶ W
a volcano that may erupt again at some time in the distant future is
a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air
Answer:
Explanation:
The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:
initial velocity is 20 m/s
Here's what we know that we are NOT told:
a = -9.8 m/s/s and
final velocity is 0 at an object's max height in parabolic motion.
We will use the equation:
[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:
0 = 20 + (-9.8)t and
-20 = -9.8t so
t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.
