The correct statement about the wind is:
Cool air rushing in to fill an area of low pressureWhat is the wind?Wind is the movement of air currents in relation to the Earth's surface, which is caused by pressure differences and air movement.
Characteristics of the windIt is a meteorological phenomenon originated in the movements of rotation and translation of the Earth.When the rising air cools and loses the moisture it was carrying, due to condensation and rain, the result is dry and cool air.Therefore, we can conclude that the wind is the current of air that occurs in the atmosphere due to natural causes, from high pressure areas to low pressure areas.
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Blue light (450 nm) and orange light
(625 nm) pass through a diffraction
grating with d = 2.88 x 10-6 m. What is
the angular separation between them
for m = 1?
Answer:
3.54
Explanation:
some nerd thing I found it on Yahoo answers
Answer:
3.54º
Explanation:
Find the blue θ first
sin⁻¹(540x10⁻⁹/2.88x10⁻⁶)=8.99°
Then find the orange θ
sin⁻¹(625x10⁻⁹/2.88x10⁻⁶)=12.53°
Take the differences and subtract
12.53°－8.99°=3.54°
16. Two electric bulbs marked 100W 220V and 200W 200V have tungsten
filament of same length. Which of the two bulbs will have thicker
filament?
Answer:
The second bulb will have thicker filament
Explanation:
Given;
First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V
Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V
Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m
Resistance of the first bulb:
[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]
Resistance of the second bulb:
[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]
Resistivity of the tungsten filament is given by the following equation;
[tex]\rho = \frac{RA}{L}[/tex]
where;
L is the length of the filament
R is resistance of each filament
A is area of each filament
[tex]A = \pi r^2[/tex]
where;
r is the thickness of each filament
[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]
Therefore, the second bulb will have thicker filament
What is the definition of the half-life of a radioactive isotope?
answer: The time it takes for half the parent nuclei in a sample to become daughter nuclei.
Answer: The half-life is the amount of time it takes for a given isotope to lose half of its radioactivity. If a radioisotope has a half-life of 14 days, half of its atoms will have decayed within 14 days. In 14 more days, half of that remaining half will decay, and so on.
Somebody, please help me with these by April 6 I'm about to fail my class
1. Describe specific heat capacity and its use in calorimetry.
2. Define latent heat and how it is different than specific heat capacity.
3. Describe how a phase diagram changes when changing from a solid to a liquid.
4. Describe how work is done is related to a change in the volume of a fluid.
—Continue on the next page—
5. What is the relationship of change in internal energy, work, and heat? Explain the significance in terms of energy.
6. As succinctly as possible, explain why it is impossible to have an engine that is 100% efficient.
7. Describe the application of Archimedes principle in everyday terms.
Practice
8. 50 grams of a substance increases its temperature by 10 degrees when 100 J of heat is added. What is the specific heat capacity of the substance?
9. Ten kilograms of a substance has a latent heat of 1000 J/kg. How much energy does it take to change the phase of this substance?
—Continue on the next page—
10. The work done by a piston is measured to be 1000 J. If the pressure is a constant 1000 Pa, what is the change in volume of the piston?
11. Is the process described in #10 isovolumetric? Explain.
12. Is it possible for a substance to change the temperature in an isovolumetric and adiabatic process? Explain.
13. A substance is heated with 1000 J and does 700 J of work on the atmosphere. What is the change in the internal energy of the substance?
14. An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency of this engine?
—Continue on the next page—
15. If a ball of radius 0.1 m is suspended in water, density = 997 kg/m^3, what is the volume of water displaced and the buoyant force?
16. A 10 kg ball of volume = 0.005 m3 is set on a lake. Describe what will happen to the ball.
Application
A group of physics students heats a vial of water, which has a balloon at the top that traps the water vapor and air.
17. Describe how much the internal energy is changed if the students are adding heat to the water via a Bunsen burner. No need to calculate anything.
18. If 50 grams of water start to boil, then how much heat must be added to completely boil off the water? The latent heat of vaporization is 2,260,000 J/kg
—Continue on the next page—
19. Assume this process is 10% efficient, how much work is done by the gas expanding into the atmosphere?
20. Assuming the atmospheric pressure to be constant at 101,325 Pa, by what amount does the volume of the balloon change?
Answer:
im sorry i would help but thats too much
The distance from the sun to Earth would be
Which phrase best completes the sentence?
any number of light years
more than one light year
exactly one light year
less than one light year
4
Answer:
less than one lightyear=d
Explanation:
I took the test.:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D::):):):):):):):):):):):):):):):):):):):):):)
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between his clothes and Earth is 0.701. He slides so that his speed is zero just as he reaches the base. (a)How much mechanical energy is lost due to friction acting on the runner
Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
A wooden cylinder (in the form of a thin disk) of uniform density and a steel hoop are set side by side, released from rest at the same moment, and roll down an inclined plane towards a wall at the bottom. The cylinder has a larger radius than the hoop, but the hoop weighs more than the cylinder.
Required:
Who reaches the bottom first and why?
Answer:
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
Explanation:
a. Who reaches the bottom first
The kinetic energy of the objects is given by
K = 1/2mv² + 1/2Iω² where m = mass of object, v = velocity of object, I = moment of inertia and ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy, K = 1/2mv² + 1/2(mr²/2)(v/r)²
K = 1/2mv² + 1/4mv²
K = 3mv²/4
For the steel hoop, I' = mr'² where m' = mass of steel hoop and r' = radius of steel hoop and v' = velocity of steel hoop
So, its kinetic energy, K' = 1/2m'v'² + 1/2(m'r'²)(v'/r')²
K' = 1/2m'v'² + 1/2m'v'²
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
3mv²/4 = m'v'²
v²/v'² = 4m/3m'
v²/v'² = 4/3(m/m')
v/v' = √[4/3(m/m')]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
b. Why
Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
K = K₀ + 1/2Iω² where K₀ = translational kinetic energy of wooden cylinder
K - 1/2Iω² = K₀
3/4mv² - 1/2(mr²/2)(v/r)² = K₀
3/4mv² - 1/4mv² = K₀
K₀ = 1/2mv²
For the steel hoop,
K' = K₁ + 1/2I'ω'² where K₁ = translational kinetic energy of steel hoop
K' - 1/2I'ω'² = K₁
m'v'² - 1/2(m'r'²)(v'/r')² = K₁
m'v'² - 1/2m'v'² = K₁
K₁ = 1/2m'v'²
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
What is Kinetic energy?
The energy of the body due to its movement in a particular direction under the influence of a force like a free-falling body due to gravitaional force is called Kinetic energy.
The kinetic energy of the objects is given by
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}Iw^2[/tex]
where
m = mass of object,
v = velocity of object,
I = moment of inertia and
ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy,
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{mr^2}{2})\dfrac{v}{r}^2[/tex]
[tex]K = \dfrac{3mv^2}{4}[/tex]
For the steel hoop,
I' = mr'²
where
m' = mass of steel hoop and
r' = radius of steel hoop and
v' = velocity of steel hoop
So, its kinetic energy,
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}(m'r'^2)\dfrac{v'}{r'}^2[/tex]
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}m'v'^2[/tex]
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
[tex]\dfrac{3mv^2}{4 }= m'v'^2[/tex]
[tex]\dfrac{v^2}{v'^2} =\dfrac{ 4m}{3m'}[/tex]
[tex]\dfrac{v^2}{v'^2} = \dfrac{4}{3}(\dfrac{m}{m'})[/tex]
[tex]\dfrac{v}{v'} = \sqrt{[\dfrac{4}{3}(\dfrac{m}{m'})][/tex]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
(b) Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
[tex]K = K_o + \dfrac{1}{2}Iw^2[/tex]
where
K₀ = translational kinetic energy of wooden cylinder
[tex]K - \dfrac{1}{2}Iw^2 = K_o[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{2}(\dfrac{mr^2}{2})(\dfrac{v}{r})^2 = K_a[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{4}mv^2 = K_o[/tex]
[tex]K_o = \dfrac{1}{2}mv^2[/tex]
For the steel hoop,
[tex]K' = K_1 + \dfrac{1}{2}I'w'^2[/tex]
where
K₁ = translational kinetic energy of steel hoop
[tex]K' - \dfrac{1}{2}I'w'^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}(m'r'^2)(\dfrac{v'}{r'})^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}m'v'^2 = K_1[/tex]
[tex]K_1= \dfrac{1}{2}m'v'^2[/tex]
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
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If you live in Melbourne, Australia, the local magnetic field has a strength of about 4x10-5 T. The magnetic field vector is directed northward, making an angle of 30 deg above the horizontal. An electron in Melbourne is moving parallel to the ground, in the west direction, at a speed of 9x105 m/s. What are the magnitude and direction of the magnetic force on the electron
Answer:
[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field
Explanation:
B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]
[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]
v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
Magnetic force is given by
[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]
The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.
A student using a stopwatch finds that the time for 10 complete orbits of a ball on the end of a string is 25 seconds. The period of the orbiting ball is
Answer:
T = 2.5 s
Explanation:
Given that,
Number of complete orbits = 10
Time, t = 25 seconds
We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.
So,
[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]
So, the period of the orbiting ball is equal to 2.5 seconds.
A 3.5 kg object gains 76 J of potential energy as it is lifted vertically. Find the new height of the object?
Answer:
1.72 m
Explanation:
Potential energy = mgh, where m is mass, g is acceleration due to gravity (9.8), and h is height
76 = (3.5)(9.8)h
76=44.1h
h=1.72335600907 ≈1.72 m
Answer:
:r
Explanation:r
Tobnbv346468this Ishmael
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin wedge. One edge of the wedge is vertical, and the tip makes an angle of 33. The block that hangs vertically weighs 2.85 kg, and the block on the incline weighs 2.94 kg. If the two blocks do not move, what is magnitude of the force of friction on the second second block
Answer:
13.6 N
Explanation:
Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.
Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.
We assume the motion is downwards in the direction of the hanging block, M.
We now write equations of motion for each block.
So
Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.
Now, since both blocks do not move, a = 0.
So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0
Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)
From (3), T = Mg
Substituting T into (4), we have
T - mgcos57 - F = 0
Mg - mgcos57 - F = 0
So, Mg - mgcos57 = F
F = Mg - mgcos57
F = (M - mcos57)g
Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.
We find F, thus
F = (2.94 kg - 2.85 kgcos57)9.8 m/s²
F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²
F = (2.94 kg - 1.552 kg)9.8 m/s²
F = (1.388 kg)9.8 m/s²
F = 13.6024 kgm/s²
F ≅ 13.6 N
A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water
Answer:
B. it increases
Explanation:
As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).
Answer:
B is the correct answer.
Explanation:
One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. The force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00 x 103 m/s, given the collision lasts 6.00 x 10-8 s is Fill input: x 106 N.
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ [tex]F=m(\frac{\Delta v}{\Delta t} )[/tex]
On substituting the given values, we get
⇒ [tex]=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )[/tex]
⇒ [tex]=6666.67 \ N[/tex]
Look at the diagram showing the different wavelengths in sunlight.
A diagram showing the human eye and visible light. Visible light is broken down by color with wavelength in nanometers. Red is 700, orange is 600, yellow is 580, green is 550, blue is 475, indigo is 450, violet is 400.
Which has a wavelength of 350 nanometers?
red light
violet light
infrared light
ultraviolet light
Answer:
ultraviolet light
plz mark me as brainliest.
Answer:
Ultra violet
Explanation:
An 80.0-kg skydiver jumps out of a balloon at an altitude of 1,000 m and opens his parachute at an altitude of 200 m. A. Assuming the total friction (resistive) force on the skydiver is constant at 50.0 N with the parachute closed and constant at 3,600 N with the parachute open, find the speed of the skydiver when he lands on the ground. B. At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s
Answer:
[tex]24.9\ \text{m/s}[/tex]
[tex]206.7\ \text{m}[/tex]
Explanation:
m = Mass of skydiver = 80 kg
[tex]x_1[/tex] = Height for which the parachute is closed = 1000-200 = 800 m
[tex]x_2[/tex] = Height for which the parachute is open = 200 m
[tex]f_1[/tex] = Resistive force when parachute is closed = 50 N
[tex]f_2[/tex] = Resistive force when parachute is open = 3600 N
v = Velocity of skydiver on the ground
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
h = Height from which the skydiver jumps = 1000 m
The energy balance of the system will be
[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times 800-3600\times 200=\dfrac{1}{2}\times 80\times v^2\\\Rightarrow v=\sqrt{\dfrac{2(80\times 9.81\times 1000-50\times 800-3600\times 200)}{80}}\\\Rightarrow v=24.9\ \text{m/s}[/tex]
The velocity fo the skydiver when he lands will be [tex]24.9\ \text{m/s}[/tex]
x = Height where the person opens the parachute
v = 5 m/s
[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times (1000-x)-3600\times x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow 80\times 9.81\times 1000-50000+50x-3600x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow x=\dfrac{80\times 9.81\times 1000-50000-\dfrac{1}{2}\times 80\times 5^2}{3550}\\\Rightarrow x=206.7\ \text{m}[/tex]
The height at which the parachute is to be opened is [tex]206.7\ \text{m}[/tex]
A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces. At the east end of the beam, a 200\ N200 N forces pushes downward. At the west end of the beam, a 200\ N200 N force pushed upward. What is the angular acceleration of the beam
Answer: [tex]240\ rad/s^2[/tex]
Explanation:
Given
Length of beam [tex]l=2\ m[/tex]
mass of beam [tex]m=5\ kg[/tex]
Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude
[tex]\tau =F\times l=200\times 2=400\ N.m[/tex]
Also, the beam starts rotating about its center
So, the moment of inertia of the beam is
[tex]I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2[/tex]
Torque is the product of moment of inertia and angular acceleration
[tex]\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2[/tex]
N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches the bottom side of the board at a point 85 cm from the end under the rock. The board is 4.5 m long, has a mass of about 22 kg, and projects essentially horizontally out over a river. Is it safe for an adult with a mass of 62 kg to stand at the unsupported end of the board
Answer:
it is safe to stand at the end of the table
Explanation:
For this exercise we use the rotational equilibrium condition
Στ = 0
W x₁ - w x₂ - w_table x₃ = 0
M x₁ - m x₂ - m_table x₃ = 0
where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m
the mass of the man is 62 kg and the distance
x₂ = 4.5 - 0.85
x₂ = 3.65 m
the mass of the table (m_table = 22 kg) is at its geometric center
x_{cm} = L/2 = 2.25 m
x₃ = 2.25 -0.85
x₃ = 1.4 m
let's look for the maximum mass of man
m_{maximum} = [tex]\frac{ M x_1 -m_{table} x_3}{ x_2}[/tex]
let's calculate
m_{maximum} = [tex]\frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}[/tex](380 0.85 - 22 1.4) / 3.65
m_{maximum} = 80 kg
we can see that the maximum mass that the board supports without turning is greater than the mass of man
m_{maximum}> m
consequently it is safe to stand at the end of the table
Which option identifies the specific knowledge that the team in the following scenario must possess?
A team of engineers is designing a space probe that will go to Saturn and collect atmospheric samples. The temperature and atmosphere on Saturn are much different from the conditions on Earth.
(A) The team must have a vast knowledge of thermodynamics.
(B) The team must have a vast knowledge of propulsion.
(C) The team must have a vast knowledge of fluid power systems.
(D) The team must have a vast knowledge of acoustics.
Answer:
The team must have a vast knowledge of thermodynamics
Explanation:
Just took the test!!!
Answer:
C. Thermodynamics
Explanation:
A water balloon weighing 4.5 N rests on a table. The balloon has an area of 2.6 x 10-3
m² in contact with the table. What pressure does the balloon exert on the table?
Answer:
the pressure the balloon exerts on the table is 1,730.77 N/m²
Explanation:
Given;
weight of the water balloon, F = 4.5 N
area of the balloon, A = 2.6 x 10⁻³ m²
The pressure the balloon exerts on the table is calculated as follows;
[tex]P = \frac{F}{A}[/tex]
substitute the given values and solve for pressure, P;
[tex]P = \frac{4.5}{2.6 \times 10^{-3}} \\\\P = 1,730.77 \ N/m^2[/tex]
Therefore, the pressure the balloon exerts on the table is 1,730.77 N/m²
A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and
rebounds with a velocity of 22 m/s. Is the change in momentum of the cart
Explanation:
mass, m = 5kg
initial velocity, u = 16m/s
final velocuty, v = -22m/s
change in momentum, ∆p = ?
∆p = m (v-u)
5(-22-16)
5(38)
∆p = 190kgm/s
check the calculations!
Explain why it is not advisable to be in a garage when the car engine is being
heated.
Answer:
You can breathe in too much carbon monoxide, which will eliminate the flow of oxygen to your bloodstream and can kill you.
Explanation:
If 10 Coulombs flow through a circuit every 2 seconds, what is the current?
A. Not enough info
B. 5 A
C. 10 A
D. 1 A
Answer:
not enought info
Explanation:
tbh I just know it's not 5 10 or 1
Answer:
B. 5 A
Explanation:
10/2= 5
Educere
answer asap!!! i suck at acceleration
Answer: 2.67
Explanation: it said he went from 0 to 8 in 3 seconds so if we divide eight By three we get 2.67 rounded to the nearest hundredth so you accelerated that 2.67 m/s
Which of the following best describes our
atmosphere?
A. envelope of gases that surround Earth
B. a specific range of altitude where plant life flourishes
C. The air, water, and land that form our planet
D. the water vapor in the air surrounding our planet
A wire carries a current of 4.2 A at what distance from the wire does the magnetic field have a magnitude of 1.3×10^ -5 t
Answer:
the distance is 6.46 cm.
Explanation:
Given
current in the wire, I = 4.2 A
magnitude of the magnetic field, B = 1.3 x 10⁻⁵ T
The distance from the wire is determined by using Biot-Savart Law;
[tex]B = \frac{\mu_o I}{2\pi r} \\\\r = \frac{\mu_o I}{2\pi B}[/tex]
Where;
r is the distance from the wire where the magnetic field is experienced
[tex]r = \frac{\mu_o I}{2\pi B}\\\\r = \frac{4\pi \times 10^{-7} \times 4.2 }{2\pi \times 1.3 \times 10^{-5}}\\\\r = 0.0646 \ m\\\\r = 6.46 \ cm[/tex]
Therefore, the distance is 6.46 cm.
A wheel rotates about a fixed axis with an initial angular velocity of 24 rad/s. During a 4 s interval the angular velocity decreases to 14 rad/s. Assume that the angular acceleration is constant during the 4 s interval. How many radians does the wheel turn through during the 4 s interval
Answer:
[tex]\theta=76\ rad[/tex]
Explanation:
Hoven that,
Initial angular velocity of the wheel = 24 rad/s
Final angular velocity = 14 m/s
Time, t = 4 s
We need to find how many radians does the wheel turn through during the 4 s interval. Let the displacement is [tex]\theta[/tex]. Using second equation of rotational kinematics to find it such that,
[tex]\theta=\omega_i t+\dfrac{1}{2}\alpha t^2[/tex]
Where
[tex]\alpha[/tex] is angular acceleration
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{14-24}{4}\\\\\alpha =-2.5\ rad/s^2[/tex]
So,
[tex]\theta=24\times 4+\dfrac{1}{2}\times (-2.5)\times 4^2\\\\\theta=76\ rad[/tex]
So, it will turn 76 radian during the 4 s interval.
Explain why your image never disappears and never flips over as you bring the convex mirror
close to your eye.
Explanation:
When you get closer to the mirror than the focal point a virtual image is formed behind the mirror and this image is not inverted. That's why the image flips as you get closer. ... With a virtual image the light rays never come to a focus so there is no place you can put a piece of paper to see the image.
If a current of 1.10 A flows through a 7.00 Ω resistor of length 3.00 m, what is the electric field strength inside the resistor?
Answer:
the electric field strength inside the resistor is 2.57 V/m
Explanation:
Given;
current flowing through the wire, I = 1.10 A
resistance of the wire, R = 7.00 Ω
length of the wire, L = 3.00 m
The emf created inside the resistor is calculated as;
V = IR
V = 1.10 x 7
V = 7.7 V
The electric field strength inside the resistor is calculated as;
E = V/L
E = 7.7 / 3
E = 2.57 V/m
Therefore, the electric field strength inside the resistor is 2.57 V/m
Larry is making a model of the Solar System. What objects will Larry need to put in his model of the Solar System? Name three types of objects. Describe where Larry should place Earth within the Solar System. es ) your answer below:
Answer:
1) It seems that he would need the central gravitational force
(the sun)
2) Also the planets would need to be included (orbits around the sun)
Mercury, Venus, Earth, Mars, Jupiter, Saturn, etc.
3. Then, many of the planets have significant objects (moons) rotating about them.
Those would seem to be objects to be included in a model of the solar system.
1) He would need the central gravitational force (the sun)
2) The planets would need to be included: Mercury, Venus, Earth, Mars, Jupiter, Saturn, etc.
3) Many of the planets have specific moons rotating about them.
Larry should put the Earth between the planets Venus, and Mars.
A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.6 A runs through the wire, in the x direction. Along 0.17 m of the length of the wire there is a magnetic field of 0.52 tesla in the y direction, due to a large magnet nearby. At other locations in the circuit, the magnetic field due to external sources is negligible. What is the magnitude of the magnetic force on the wire
Answer:
the magnitude of the magnetic force on the wire is 0.2298 N
Explanation:
Given the data in the question;
we know that, the magnitude of magnetic force is given as;
|F[tex]_{mg}^>[/tex] | = I([tex]B^>[/tex] × [tex]L^>[/tex] )
given that
I = 2.6 A
[tex]B^>[/tex] = 0.17
[tex]L^>[/tex] = 0.52
so we substitute
|F[tex]_{mg}^>[/tex] | = 2.6( 0.17i" × 0.52j" )
|F[tex]_{mg}^>[/tex] | = 0.2298 N
Therefore, the magnitude of the magnetic force on the wire is 0.2298 N