Answer:
60
Explanation:
Work Done= Force×Displacement in the direction of the force
W.D. = 10×6
W.D. = 10×0.6
W.D. = 6m
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°
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.
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:
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!
Someone help me like please thank you
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.
To know more about Kinetic energy follow
https://brainly.com/question/25959744
A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 6.82 s. What is the
Answer:
The answer is "[tex]11.55780\ m[/tex]"
Explanation:
Using formula:
[tex]= 2 \pi f= \frac{2\pi}{T} =\sqrt{\frac{g}{L}}[/tex]
L = length of pendulum.
[tex]= T =2 \pi \sqrt{\frac{L}{g}}[/tex]
Calculate the value for L:
[tex]L= g (\frac{T}{2 \pi})^2 \\\\[/tex]
[tex]= (9.80 \ \frac{m}{s^2}) (\frac{6.82 \ s}{2 \pi})^2\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{4 \times \pi^2})\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124\ s^2}{4 \times 9.8596 })\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{ 39.4384 })\\\\= \frac{455.82152}{39.4384} \ m\\\\=11.55780\ m[/tex]
The height of the tower = 11.55780 m
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 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
In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 527 nm. An interference pattern is observed on the screen. The distance between the screen and the double-slit is 1.54 m and the distance between the two slits is 0.102 mm. A light wave propogates from each slit to the screen. What is the path length difference between the distance traveled by the waves for the fifth-order maximum (bright fringe) on the screen
Answer:
Λ = 5.14 10⁻⁴ m
Explanation:
This is a double slit experiment, which for the case of constructive interference
d sin θ = m λ
let's use trigonometry
tan θ = y / L
as the angles are very small
tan θ = [tex]\frac{sin \theta}{cos \theta}[/tex] = sin θ
sin θ = y / L
we substitute
d y / L = m λ
y = m λ L / d
we calculate for the interference of order m = 5
y = 5 527 10⁻⁹ 1.54/0.102 10⁻³
y = 3.978 10⁻² m
Now we can find the difference in length between the two rays, that of the central maximum and this
let's use the Pythagorean theorem
L’= [tex]\sqrt{L^2 +y^2}[/tex]
L ’= [tex]\sqrt{1.54^2 +(3.978 \ 10^{-2})^2 }[/tex]
L ’= 1.54051 m
optical path difference
Λ = L’- L
Λ = 1.54051 - 1.54
Λ = 5.14 10⁻⁴ m
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:
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:
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
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
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 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.
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
g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest at the top of an incline plane. The height of top of the incline plane is 1.2 m. Find translational speed of each cylinder upon reaching the bottom and determine which cylinder has the greatest translational speed upon reaching the bottom. Moment of inertia of hollow cylinder about its axis passing through the center is mr2 and for solid cylinder mr2/2
Answer:
a. i. 3.43 m/s ii. 2.8 m/s
b. The thin-walled cylinder
Explanation:
a. Find translational speed of each cylinder upon reaching the bottom
The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy
So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.
i. translational speed of thin-walled cylinder upon reaching the bottom
So, For the thin-walled cylinder, I = mr², we find its translational velocity, v
So, mgh = 1/2mv² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²)(v/r)²
mgh = 1/2mv² + 1/2mv²
mgh = mv²
v² = gh
v = √gh
v = √(9.8 m/s² × 1.2 m)
v = √(11.76 m²/s²)
v = 3.43 m/s
ii. translational speed of solid cylinder upon reaching the bottom
So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'
So, mgh = 1/2mv'² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²
mgh = 1/2mv'² + mv'²
mgh = 3mv'²/2
v'² = 2gh/3
v' = √(2gh/3)
v' = √(2 × 9.8 m/s² × 1.2 m/3)
v' = √(23.52 m²/s²/3)
v' = √(7.84 m²/s²)
v' = 2.8 m/s
b. Determine which cylinder has the greatest translational speed upon reaching the bottom.
Since v = 3.43 m/s > v'= 2.8 m/s,
the thin-walled cylinder has the greatest translational speed upon reaching the bottom.
A 2.5 Coulomb charge is placed in an electric field where it experiences an electrical force of 50.N. What is the value of the electrical field at the place where the charge is located?
Remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits to receive full credit. (5 points)
Answer: E= 20 N/C
Explanation: Charge q = 2.5 C , force F = 50 N. F = qE and
E= F/q= 50N / 2.5 C = 20 N/C
NEED TO SUBMIT THIS IN 10 MINS, PLS HELP!!!!
Answer:
Your answer is B
because it's on sneel's law.
that is sin of incident ray / sin of refracted ray is refractive index
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::):):):):):):):):):):):):):):):):):):):):):)
Acep
we
White light is incident normally on a thin soap film (n = 1.33) suspended in air:
(a) What are the two minimum thickness that will destructively
reflect yellow light
of wavelength 590 nm?
I hope this will help you. Here is the link http://web.physics.ucsb.edu/~phys6c/summer2006/hw4solutions
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
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
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
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]
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
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.
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 carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Learn more about angular velocity here: https://brainly.com/question/6860269
Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.