Answer:
c. yes, 7.7 cm from the 5-A wire
Explanation:
Given;
distance between the two wires, r = 20 cm = 0.2 m
first current, I₁ = 5.0 A
second current, I₂ = 8.0 A
The magnetic field due to the two wires occurs in different directions and it can be zero at this region.
Let x be the distance from 5 A wire where the magnetic field is zero.
[tex]B = \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{5}{x} -\frac{8}{0.2-x} ] = 0\\\\\frac{5}{x} -\frac{8}{0.2-x} = 0\\\\\frac{5}{x} = \frac{8}{0.2-x}\\\\5(0.2-x) = 8x\\\\1 -5x = 8x\\\\1 = 5x \ + \ 8x\\\\1 = 13x\\\\x = \frac{1}{13} \\\\x = 0.077 \ m\\\\x = 7.7 \ cm[/tex]
Therefore, the correct option is c. yes, 7.7 cm from the 5-A wire
List down the types of centripetal force?
Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.
Answer:
roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge
Explanation:
Help plsssssssssss I write it 100 time no one answer
Answer:
1.93×10²⁸ s
Explanation:
From the question given above, the following data were obtained:
Number of electron (e) = 2×10²⁴
Current (I) = 10 A
Time (t) =?
Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:
1 e = 96500 C
Therefore,
2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e
2×10²⁴ e = 1.93×10²⁹ C
Thus, 1.93×10²⁹ C of electricity is passing through the point.
Finally, we shall determine the time. This can be obtained as follow:
Current (I) = 10 A
Quantity of electricity = 1.93×10²⁹ C
Time (t) =?
Q = it
1.93×10²⁹ = 10 × t
Divide both side by 10
t = 1.93×10²⁹ / 10
t = 1.93×10²⁸ s
Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point
39. What is the change in momentum for a 5,000 kg ship in
outer space that experiences no net force over a 1 hr
period?
Answer:
Change in momentum is zero.
Explanation:
The following data were obtained from the question:
Mass (m) = 5000 kg
Time (t) = 1 h
Net force (F) = 0
Change in momentum =?
Force = Rate of change of momentum
0 = change in momentum
Change in momentum = 0
We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.
Captain Jack Sparrow has been marooned on an island in the Atlantic by his crew, and decides to builda raft to escape. The wind seems quite steady, and first blows him due east for 11km, and then 6km ina direction 6degrees north of east. Confident that he will eventually find himself in safety, he fallsasleep. When he wakes up, he notices the wind is now blowing him gently 11degrees south of east -but after traveling for 21km, he finds himself back on the island.
Variable Name Min Max Step Sample Value
thetab 5 10 1 6
a 10 20 11 1
b 5 15 1 6
c 20 30 1 21
thetac 10 15 11 1
Required:
How far (in km) did the wind blow him while he was sleeping?
Answer:
d₃ = 37,729 km, θ= 5.1º North of West
Explanation:
This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction
* first displacement is
d₁ₓ = 11 km
* second offset is
cos 6 = d₂ₓ / d₂
sin 6 = d_{2y} / d₂
d₂ₓ = d₂ cos 6
d_{2y} = d₂ sin 6
d₂ₓ = 6 cos 6 = 5.967 km
d_{2y} = 6 sin 6 = 0.6272 km
* third displacement is unknown
* fourth and last displacement
cos (-11) = d₄ₓ / d₄
sin (-11) = d_{4y} / d₄
d₄ₓ = d₄ cos (-11)
d_{4y} = d₄ sin (-11)
d₄ₓ = 21 cos (-11) = 20.61 km
d_{4y} = 21 sin (-11) = -4.007 km
They tell us that at the end of the tour you are back on the island, so the displacement must be zero
X axis
x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ
0 = 11 +5.967 + d₃ₓ + 20.61
d₃ₓ = -11 - 5.967 - 20.61
d₃ₓ = -37.577 km
Y axis
y = d_{1y} + d_{2y} + d_{3y} + d_{4y}
0 = 0 + 0.6272 + d_{3y} -4.007
d_{3y} = 4.007 - 0.6272
d_{3y} = 3.3798 km
This distance can be given in the form of module and angle
Let's use the Pythagorean theorem for the module
d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]
d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]
d₃ = 37,729 km
Let's use trigonometry for the angle
tan θ = d_{3y} / d₃ₓ
θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]
θ = tan-1 (-3.3798 / 37.577)
θ = 5.1º
Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is
θ= 5.1º North of West
A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.
Answer:
The speed of this particle is constantly [tex]c[/tex].
Explanation:
Position vector of this particle at time [tex]t[/tex]:
[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].
Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:
[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].
Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].
Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:
[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].
The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :
[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].
Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 0.55 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)
Answer:
q = 2 10⁻⁸ C
Explanation:
For this exercise we use the translational equilibrium equation
F_e -A =
F_e = W
the electric force is given by Coulomb's law
F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case they indicate that the loads on the tapes are equal
F_e = k q² / r²
we substitute
k q² / r² = m g
q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]
calculate
q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]
q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]
q = 1,999 10⁻⁸ C
q = 2 10⁻⁸ C
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.
This question is incomplete, the missing diagram is uploaded along this Answer below.
Answer:
a) the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight is - 3.935 lb-ft
Explanation:
Given the data in the question;
(a) determine the work done on the cart by the spring
we calculate the work done on the cart by the spring as follows;
[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )
where k is spring constant ( 3 lb/in )
we substitute
[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )
[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )
[tex]W_{spring}[/tex] = 1/2 × 3( 39 )
[tex]W_{spring}[/tex] = 58.5 lb-in
we convert to pound force-foot
[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft
[tex]W_{spring}[/tex] = 4.875 lb-ft
Therefore, the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight
work done by its weight;
[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )
we substitute in of values from the image below;
[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )
[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13
[tex]W_{gravity}[/tex] = -47.1 lb-in
we convert to pound force-foot
[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft
[tex]W_{gravity}[/tex] = - 3.935 lb-ft
Therefore, the work done on the cart by its weight is - 3.935 lb-ft
a) the work done on the cart by the spring is 4.875 lb-ft.
b) the work done on the cart by its weight is - 3.935 lb-ft.
Calculation of the work done:a. The work done on the cart by the spring is
= 1/2 × 3( (-8)² - (5)² )
= 1/2 × 3( 64 - 25 )
= 1/2 × 3( 39 )
= 58.5 lb-in
Now we have to convert to pound force-foot
So,
= 58.5 × 0.0833333 lb-ft
= 4.875 lb-ft
b) Now
work done by its weight;
= -mgsin∅( x₂ - x₁ )
So,
= -14 × sin(15°)( 5 - (-8) )
= -14 × 0.2588 × 13
= -47.1 lb-in
Now we convert to pound force-foot
= -47.1 × 0.0833333 lb-ft
= - 3.935 lb-ft
Learn more about spring here: https://brainly.com/question/16060099
If the diameter of a moose eye is 40 mm, what is the total refractive power of the anterior portion of the eye?
Answer:
-the ratio of the speed of light
in air to the speed of light in the substance.
-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.
-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5
Explanation:
Messages from the Perseverance Rover on Mars reach Earth in 11 min. The speed of light is 3.00 x 108 m/s. Using this information, how far is Mars from Earth?
Answer:
[tex]d=1.98\times 10^{11}\ m[/tex]
Explanation:
Messages from the Perseverance Rover on Mars reach Earth in 11 min i.e. time is 660 s
The speed of light is [tex]3\times 10^8\ m/s[/tex]
We need to find the distance between Mars and Earth. Let the distance be d.
We know that,
Distance = speed × time
So,
[tex]d=660\times 3\times 10^8\\\\d=1.98\times 10^{11}\ m[/tex]
So, Mars is [tex]1.98\times 10^{11}\ m[/tex] from the Earth.
The current flow in the light bulb is 0.5A
a.Calculate the amount of electric charge that flow through the bulb in 2 hour
b.If one election carries a
charge 1.6 x 10-14 c Find the number of election through the bulb in 2 hour?
Answer:
Explanation:
Given that,
The current in the light bulb, I = 0.5 A
(a) We know that,
Electric current = charge/time
or
Q = It
Put t = 2 hours = 7200 s
So,
Q = 0.5 × 7200
Q = 3600 C
(b) Charge on one electron, [tex]Q=1.6\times 10^{-19}\ C[/tex]
Let there are n electrons flow through the bulb in 2 hours.
I = Q/t
Since, Q = ne
So,
I = ne/t
[tex]n=\dfrac{I\times t}{e}\\\\n=\dfrac{0.5\times 7200}{1.6\times 10^{-19}}\\\\n=2.25\times 10^{22}[/tex]
Hence, this is the required solution.
QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!
Answer:
A. v = √2gh
B. No! The final velocity does not depend on the mass of the car.
C. Yes! the final velocity depends on the steepness of the hill
D. 3.28 m/s
Explanation:
A. Determination of the final velocity.
½mv² = mgh
Cancel out m
½v² = gh
Cross multiply
v² = 2gh
Take the square root of both side
v = √2gh
B. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is no mass (m) in the formula.
Thus, the final velocity does not depend on the mass of the car.
C. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is height (h) in the formula.
Thus, the final velocity depends on the steepness of the hill
D. Determination of the final velocity.
Height (h) = 0.55 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) =?
v = √2gh
v = √(2 × 9.8 × 0.55)
v = √10.78
v = 3.28 m/s
How does Physics help you as a student?
Answer:
The goal of physics is to understand how things work from first principles. ... Courses in physics reveal the mathematical beauty of the universe at scales ranging from subatomic to cosmological. Studying physics strengthens quantitative reasoning and problem solving skills that are valuable in areas beyond physics
Answer:
you get to understand why things happen this way
Explanation:
for example, are you not curious about why when standing in the bus and when the bus stops, you will might feel like you are going to fall ,
why does this happen because....
newton's laws explains it,
inertia causes you to be reluctant to change your initial state of motion due to your mass so you fall because you are still moving at the 'speed of the bus ' , something in like that
hope this helps,
please mark also
A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.
Answer:
Are you sure it was soccer ball? Or meine hearts
Explanation:
A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.
Answer:
b) a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N
Explanation:
a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle
b) Let's start at point A
Let's use that the acceleration is centripetal
a = v² / r
let's calculate
a = 28² / 15.0
a = 52.26 m / s²
as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards
Point B
a ’= 142/15
a ’= 13.06 m / s²
in this case the acceleration is vertical downwards
c) The values of the normal force
point A
let's use Newton's second law
∑ F = m a
N- W = m a
N = mg + ma
N = m (g + a)
N = 450.0 (9.8 + 52.25)
N = 2.79 10⁴ N
d) Point B
-N -W = m (-a)
N = ma -m g
N = m (a-g)
N = 450.0 (14.0 - 9.8)
N = 1.89 10³ N
what is the average velocity of a van that moves from 0 to 60 m east and 20 seconds
Explanation:
I have a lot to say it was very nice to meet my parents are u doing well I dont want too its been so much I love you so I was like u know I am not a man but you are the auditions I have been in a long long long life is a triangle and a chair for me and my parents think about the way I
Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be +(30±5)pC+(30±5)pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain). What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)?
(a) Because air is a good conductor, the positive charge on the bee’s surface flowed through the air from bee to plant.
(b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee.
(c) The plant became electrically polarized as the charged bee approached.
(d) Bees that had visited the plant earlier deposited a positive charge on the stem.
Answer:
a) True
Explanation:
There are several possible explanations for this positive charge
* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,
to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct
* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.
When reviewing the different statements we have
a) True, it agrees with the second explanation of the phenomenon
b) False. The earth is a deposit of negative charge
c) false. If this is the case the charge should be negative
d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.
Answer:
Explanation:
.
help please i will mark brainlist!!!
Answer:
.50 M
Explanation:
5*.50=2.5 + 2*.25=.5 = 3n
6*.50= 3N
Final answer is .50M
Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.
Answer:It will take about 3000 years
Explanation:
Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 X10^-9 C and the other is given a charge of -18 X 10^-9 C. a. Find the electric force exerted on one sphere by the other. b. The sphere are connected by a conducting wire. After equilibrium has occurred, find the electric force between the two spheres.
Answer:
Explanation:
Force between two charged conducting sphere
= k x Q₁ x Q₂ / r² , k is a constant Q₁ and Q₂ are charges and r is distance between them .
= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²
= 21600 x 10⁻⁹
= 2.16 x 10⁻⁵ N .
b )
After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .
Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2
= - 3 x 10⁻⁹ C .
Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²
= 900 x 10⁻⁹ N .
Please help 25 points!
Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?
A. The 1 Hz wave contains the most energy.
B. The crests of all three waves are of equal height.
C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.
D. The 1 Hz wave has the longest wavelength.
Answer:
B
Explanation:
The crest of all three waves are of equal height
State three factors affecting pressure in liquids
Answer:
Density of liquid
Depth of liquid
Acceleration due to gravity
your "A" never changes, while your "Y" changes depending on strength of gravity. is it mass or weight?
Answer:
A - mass. B - Weight
Explanation:
This is because weight varies with the strength of gravity. Mass is just the amount of matter in an object
You are standing on the bottom of a lake with your torso above water. Which statement is correct?
a. You feel a buoyant force only when you momentarily jump up from the bottom of the lake.
b. There is a buoyant force that is proportional to the weight of your body below the water level.
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
d. There is no buoyant force on you since you are supported by the lake bottom.
Answer:
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
Explanation:
Buoyancy can be defined as a force which is created by the water displaced by an object.
Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.
Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.
The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.
A bicyclist rides 5.0 km due east, while the resistive forcefrom the air has a magnitude of 3.0 N and points due west. Therider then turns around and rides 5.0 km due west, back to herstarting point. The resistive force from the air on the return triphas a magnitude of 3.0 N and points due east.
a) Find the work done by the resistive force during the roundtrip.
Based on answer in part A.
b) Is the resistive force a conservative force? explain.
Answer:
a) Find the work done by the resistive force during the roundtrip.
W=-30kJ
b) Is the resistive force a conservative force? explain.
The resistive force is not a conservative force since the work done during the round trip is not zero
Explanation:
The worf done on object y a constant force F is given by:
W= (F cos ∅)S
Where S is the displacement and ∅ is the angle between the force and the displacement.
The displacement of the bicycle during each part of the trip is s=5000m and teh magnitude of teh resistance force is F=3.0N
∅1=180° he angle between the displacement and the force
W1=W2
W1 = (3.0 cos180) 5000m
W1=-15.O kJ
W=W1+W2
W=-30kJ
The resistive force is not a conservative force since the work done during the round trip is not zero
(a) The work done by the resistive force is 15,000 J
(b) The work done the resistive force is non-conservative since the resultant resistive force in not zero.
Work doneWork is said to be when an applied force displaces an object from its initial position.
Work done by resistive forceThe work done by the resistive force is calculated as follows;
W = FΔr
W = 3 x (5,000 - 0)
W = 15,000 J
Thus, the work done the resistive force is non-conservative since the resultant resistive force in not zero.
Learn more about conservative force here: https://brainly.com/question/15357875
1. A block with mass 20 kg is
sliding up a plane (Ukinetic=0.3,
inclined at 10°) at a speed of
2 m/s to the right (positive
X-direction). How far does it
go up along the plane before
it comes to rest momentarily?
Answer: 0.435 m
Explanation:
Given
mass m=20 kg
initial speed u=2 m/s
coefficient of kinetic friction [tex]\mu_k=0.3[/tex]
deceleration which opposes the motion is given by
[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]
[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]
using [tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]
If you have a 0.125 kg lead piece at
20.0°C, how much heat must you
add to melt it? (Remember, you
must warm it to its melting point
first.)
Material
Lead
Melt Pt (°C)
327
L (1/kg)
2.32.104
Boil Pt (°C) Lv (1/kg)
1750 8.59.105
c (1/(kg*c)
128
(Unit = J)
Answer:
7,812 J
Explanation:
Using the relation:
Q = mcΔθ
Q = quantity of heat
C = specific heat capacity of lead
Δθ = temperature change (T2 - T1)
M = mass of substance
Q = mass * specific heat * Δθ
Q = 0.125kg * 128 * (327 – 20)
Q = 0.125 * 128 * 307
Q = 4912 J
For melting:
Q = mass * Hf
0.125 * (2.32 * 10^4)
= 2,900 J
Total = 4,912 J + 2,900 J = 7,812 J
Standing at a crosswalk, you hear a frequency of 530 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 424 Hz. Determine the ambulance's speed from these observations.
Answer:
_s = 37.77 m / s
Explanation:
This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is
f ’= f₀ [tex]\frac{v}{v - v_s}[/tex]
where d ’= 530 Make
when the ambulance passes away from the observer the relationship is
f ’’ = f₀ [tex]\frac{v}{v + v_s}[/tex]
where d ’’ = 424 beam
let's write the two expressions
f ’ (v-v_s) = fo v
f ’’ (v + v_s) = fo v
let's solve the system, subtract the two equations
v (f ’- f’ ’) - v_s (f’ + f ’’) = 0
v_s = v [tex]\frac{ f' - f''}{ f' + f''}[/tex]
the speed of sound is v = 340 m / s
let's calculate
v_s = 340 [tex](\frac{ 530 -424}{530+424} )[/tex]
v_s = 340 [tex](\frac{106}{954}[/tex])
v_s = 37.77 m / s
Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.
Answer:
Kinetic energy increases and potential energy decrease when velocity of an object increase.
A student is conducting an experiment to compare the resistivity of two unknown materials by using two wires, each made of one of the materials and each connected in a circuit. The student measures the potential difference across and current in the wires. What must be the same to be able to compare the resistivities using just the potential difference and current measurements?
Answer:
is there a. b. c or d?
Explanation:
A 50kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3m/s. What was the force acting on the mass?
Answer:
75N
Explanation:
a = v/t = 3/2
F = ma = 50(3/2) = 75