3.1Chất điểm chuyển động thẳng với phương trình: x = – 1 + 3t2
– 2t
3
(hệ SI, với t ≥ 0). Chất điểm dừng lại để
đổi chiều chuyển động tại vị trí có tọa độ:
Answer:
eqcubuohwehuuc
Explanation:
the the the the the the the the the the the the the the the the dhueirrjhrhjrirjrheh3jeiiruj fnrhfjjjrjrj fjfiirrjejrjejej jrkrjrjrjrjrjdjdjfjrhruriruru rjridjhwjjsjd
There are only two charged particles in a particular region. Particle 1 carries a charge of 3q and is located on the negative x-axis a distance d from the origin. Particle 2 carries a charge of -2q and is located on the positive x-axis a distance d from the origin.
Required:
Where is it possible to have the net field caused by these two charges equal to zero?
Answer:
The net field will be the sum of the fields created by each charge.
where the charge Q in a position r' is given by:
E(r) = k*Q/(r - r')^2
Where k is a constant, and r is the point where we are calculating the electric field.
Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:
E₁(r) = k*3q/(r - r₁)^2
While for the other charge of -2q in the position r₂ = (d, 0, 0)
The electric field is:
E₂(r) = -k*2*q/(r - r₂)^2
Then the net field at the point r is:
E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2
E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)
Then if the we want to find the points r = (x, y, z) such that:
E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)
Then we must have:
0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)
Also remember that the distance between two points:
(x, y, z) and (x', y', z') is given by:
D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)
Then we can rewrite:
r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)
= √( (x + d))^2 + y^2 + z^2)
and
r - r₂ = √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)
= √( (x - d))^2 + y^2 + z^2)
Replacing that in our equation we get:
0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)
0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)
We want to find the values of x, y, z such that the above equation is true.
2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)
2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]
2*(x + d)^2 + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2
2*(x + d)^2 - 3*(x - d)^2 = 3*y^2 + 3*z^2 - 2*y^2 - 2*z^2
2*(x + d)^2 - 3*(x - d)^2 = y^2 + z^2
2*x^2 + 2*2*x*d + 2*d^2 - 3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2
-x^2 + 10*x*d - d^2 = y^2 + z^2
we can rewrite this as:
- ( x^2 - 10*x*d + d^2) = y^2 + z^2
now we can add and subtract 24*d^2 inside the parenthesis to get
- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) = y^2 + z^2
-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2
-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2
The thing inside the parenthesis is a perfect square:
-(x - 5d)^2 + 24d^2 = y^2 + z^2
we can rewrite this as:
24d^2 = y^2 + z^2 + (x - 5d)^2
This equation gives us the points (x, y, z) such that the electric field is zero.
Where we need to replace two of these values to find the other, for example, if y = z = 0
24d^2 = (x - 5d)^2
√(24d^2) = x - 5d
√24*d = x - 5d
√24*d + 5d = x
so in the point (√24*d + 5d, 0, 0) the net field is zero.
A 285-kg load is lifted 22.0 m vertically with an acceleration a=0.160g by a single cable. Determine
(a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the
load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started
from rest.
Answer:
a) T = 2838.6 N, b) W = 1003.2 J, c) W = 6.22 10⁴ J, d) W = 2.79 10³ J
e) v_f = 2.65 m / s
Explanation:
a) To find the tension of the cable let's use Newton's second law
T - W = m a
T = W + ma
T = m (g + a)
let's calculate
T = 285 (-9.8 - 0.160)
T = 2838.6 N
b) net work is stress work minus weight work
W = F d
W = (T-W) d
W = (m a) d
W = (285 0.160) 22
W = 1003.2 J
c) the work done by the cable
W = T d cos 0
W = 2838.6 22.0
W = 6.22 10⁴ J
d) The work done by the weight
the displacement is upwards and the weight points downwards, so the angle is 180º
W = F. d
W = F d cos 180
W = -285 22.0
W = 2.79 10³ J
e) the final speed of the load. Let's use the relationship between work and the change in kinetic energy
W = ΔK
as part of rest K₀ = 0
W = ½ m v_f²
v_f = [tex]\sqrt{ \frac{2W}{m} }[/tex]
v_f = [tex]\sqrt{\frac{2 \ 1003.2}{285} }[/tex]
v_f = 2.65 m / s
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.
Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?
Answer:
a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b) [tex]a=30.7[/tex]
c) [tex]a=35.91[/tex]
Explanation:
From the question we are told that:
Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
Initial Length [tex]L_1=0.600m[/tex]
Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]
Final Length [tex]L_2=0.900m[/tex]
a)
Generally the rotation with the greater speed is
[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b)
Generally the equation for centripetal acceleration at 8.16 is mathematically given by
[tex]a=\omega_1^2*L_1[/tex]
[tex]a=8.16 rev/s*0.6[/tex]
[tex]a=30.7[/tex]
c)
At 6.35 rev/s
[tex]a=6.35 rev/s*0.9[/tex]
[tex]a=35.91[/tex]
If the temperature stays constant, which change would decrease the amount
of thermal energy in an object?
A. Decreasing its density
B. Increasing its velocity
c. Decreasing its mass
D. Increasing its mass
An object produces a sound wave with a wavelength 75.0cm. if the speed of sound is 350.0m/s, the frequency of the sound is
Answer:
wavelength=75.0
speed of sound(v)=350 .0m/s
frequency(f)=?
we know,
v=f*wavelengh
350.0 =f*750
f. =350/75
=4.667
pls mark me brainlest
An object is positively charged if it has more what
Answer:
An Object is positively charged if it has more Positive Electrons in that object
Which one of the following pairs of units may not be added together, even after the appropriate unit conversions have been made?
a. grams and milligrams
b. miles and kilometers
c. kilometers and kilograms
d. centimeters and yards
Answer:
c. kilometers and kilograms
Explanation:
If two units represent different things, like for example grams and joules, where the first one is a unit for mass, and the second one a unit for energy we can not add them together. (
Let's analyze each one of the given options:
a: grams and milligrams
both grams and milligrams are units of mass, so after the appropriate unit conversions, we can add them.
b: miles and kilometers
Both are units for distance, after the appropriate unit conversions, we can add them.
c: kilometers and kilograms
"Kilometer" is a unit for distance, and "kilogram" a unit for mass, we can not add distance and mass, so these two can not be added together.
d: centimeters and yards
Both are units of distance, after the appropriate unit conversions, we can add them.
Then the only one that we can not add together is option c: kilometers and kilograms
Answer: The correct option is C (kilometers and kilograms).
Explanation:
UNITS is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature. The magnitude of a physical quantity is expressed as, Physical quantity = (numerical value) × (unit). There are six basic units of measurement which includes:
--> metre (m) - unit of length.
--> kilograms (kg) - unit of mass.
--> second (s) - unit of time.
--> ampere (A) - unit of electrical current.
--> kelvin (K) - unit of temperature and
--> mole (mol) - unit of the amount of substance.
Mass is the quantity of matter present in a body which is constant and is measured in KILOGRAMMES or GRAMS. While position is the distance and direction of an object from a particular reference point which is known to vary and is measured in METERS OR KILOMETRES.
From the above definition, KILOGRAMMES is used to measure the magnitude of a different quantity from KILOMETERS and therefore should not be added up after conversation.
To accurately describe the wind, the measurement should include
A) a direction, but not a speed
B)a speed, but not a direction
C) both a speed and a direction
D) neither a speed nor a direction
Answer:
C. both a speed and a direction
CR Physical Science B (GP) 20-21 / 1 Motion
1. A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s.
What is the average speed of the object?
O 4 m/s
O
2 m/s
O 8 m/s
10 m/s
Answer:
Average speed = total distance traveled / total time traveled
V = (2 + 80) / (2 + 40) = 82 / 42 = 2 m / s
A team of people who traveled to the North Pole by dogsled lived on butter because they needed to consume 6 000 dietitian's Calories each day. Because the ice there is lumpy and irregular, they had to help the dogs by pushing and lifting the load. Assume they had a 16-hour working day and that each person could lift a 500-N load. How many times would a person have to lift this weight 1.00 m upwards in a constant gravitational field, where (g = 9.80m/s2) where to do the work equivalent to 6 000 Calories?
Answer:
The right solution is "50200 days".
Explanation:
Given:
Calories intake,
= 6000 kcal,
or,
= [tex]2.52\times 10^7 \ J[/tex]
Force,
= 500 N
As we know,
⇒ [tex]Work \ done = Force\times distance[/tex]
Or,
⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]
By putting the values, we get
[tex]=\frac{2.52\times 10^7}{500}[/tex]
[tex]=0.502\times 10^5[/tex]
[tex]=50200 \ m[/tex]
hence,
The number of days will be:
= [tex]\frac{50200}{1}[/tex]
= [tex]50200 \ days[/tex]
Which of the following is acceleration toward the center of a circular motion? O A. Centripetal acceleration O B. Uniform circular motion O C. Centrifugal force D. Centripetal force
PLEASE HELP ASAP!!
We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
Question: A car of mass 500kg travelling at 12m/s enters a stretch of road where there's a constant resistive force of 8000N. The car comes to a stop due to this resistive force. Calculate the distance travelled by the car before stopping.
Answer:
ans: 2.25 meter
explanation
use following equations
F = ma
V = U + aT
S = UT + 1/2 aT^2
Speed of light in water
Answer:
225,000 kilometers per second
Explanation:
Have a nice day
What must be the same for any two resistors that are connected in a series
Baseball runner with a mass of 70kg, moving at 2.7m/s and collides head-on into a shortstop with a mass of 85kg and a velocity of 1.6m/s. What will be the resultant velocity of the system when they make contact with each other
Answer:
The speed of the combined mass after the collision is 2.1 m/s.
Explanation:
mass of runner, m = 70 kg
speed of runner, u = 2.7 m/s
mass of shortstop, m' = 85 kg
speed of shortstop, u' = 1.6 m/s
Let the velocity of combined system is v.
Use conservation of momentum
Momentum before collision = momentum after collision
m u + m' u' = (m + m') v
70 x 2.7 + 85 x 1.6 = (70 + 85) v
189 + 136 = 155 v
v = 2.1 m/s
A car accelerates from 0-30m/s in 5 s. It has a mass of 1200kg. What force does the engine produce? *
A) 36000N
B) 7200N
C) 60000N
D) 2000N
include explanation please
Answer:
a = ?
u = 0
v = 30
by using v = u + at equation we can find " a "
30 = 0 + 5a
6 m/s = a
by using f = ma equation we can find force produce by engine ,
f = ?
a = 6
m = 1200
f = 1200 × 6
f = 7200 N
so the answer is "B"
Imagine that you are standing on a spherical asteroid deep in space far from other objects. You pick up a small rock and throw it straight up from the surface of the asteroid. The asteroid has a radius of 9 m and the rock you threw has a mass of 0.113 kg. You notice that if you throw the rock with a velocity less than 45.7 m/s it eventually comes crashing back into the asteroid.
Required:
Calculate the mass of the asteroid.
Answer:
M = 1.409 10¹⁴ kg
Explanation:
In this exercise we have that the prioress with a minimum speed can escape from the asteroid, therefore we can use the conservation of energy relation.
Starting point. When you drop the stone
Em₀ = K + U
Em₀ = ½ m v² - G m M / r
where M and r are the mass and radius of the asteroid
Final point. When the stone is too far from the asteroid
Em_f = U = - G m M / R_f
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v² - G m M / r = - G m M / R_f
½ v² = G M (1 / r - 1 /R_f)
indicate that for the speed of v = 45.7 m /s, the stone does not return to the asteroid so R_f = ∞
½ v² = G M (1 /r)
M = [tex]\frac{v^2 r}{2G}[/tex]
let's calculate
M = [tex]\frac{45.7^2 \ 9}{ 2 \ 6.67 \ 10^{-11}}[/tex]
M = 1.409 10¹⁴ kg
Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
What is the y component of a vector that is 673 m at -38o?
Answer:
D_y = 414.38m
Explanation:
D_y = D*sin(x)
D_y = 673m*sin(38°)
D_y = 414.38m
With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds?
a) What will be its speed when it hits the ground?
b) How high does the ball rise?
Answer:
◆ See the attachment photo.
◆ Don't forget to thanks
◆ Mark as brainlist.
Convert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
Answer:
1) 0.3675
2) 0.367
3) 72.6186
4) 0.5
5) 0.000671
Answer:
1) 367.5 mg = 0.3675 g
2) 367 mL = 0.367 L
3) 28.59 in = 72.61 cm
4) 8 0z = 0.5 lb
5) 0.671 mm = 0.0000671 m
Two small silver spheres, each of mass m=6.2 g, are separated by distance d=1.2 m. As a result of transfer of some fraction of electrons from one sphere to the other, there is an attractive force F=900 KN between the spheres. Calculate the fraction of electrons transferred from one of the spheres: __________
To evaluate the total number of electrons in a silver sphere, you will need to invoke Avogadro's number, the molar mass of silver equal to 107.87 g/mol and the fact that silver has 47 electrons per atom.
Answer:
4.60 × 10⁻⁸
Explanation:
From the given information;
Assuming that q charges are transferred, then:
[tex]F = \dfrac{kq^2}{d^2}[/tex]
where;
k = 9 ×10⁹
[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]
[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]
q = 0.012 C
No of the electrons transferred is:
[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]
[tex]= 7.5 \times 10^{16} \ C[/tex]
Initial number of electrons = N × 47 × no of moles
here;
[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]
no of moles = 0.0575 mol
∴
Initial number of electrons = [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]
= 1.63 × 10²⁴
The fraction of electrons transferred [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]
= 4.60 × 10⁻⁸
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
Kirchhoff's junction rule is a statement of: Group of answer choices the law of conservation of momentum the law of conservation of charge the law of conservation of energy. the law of conservation of angular momentum. Newton's second law
Answer:
the law of conservation of energy.
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.
The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.
Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.
One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.
Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.
This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.
The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.
This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
Answer:
Explanation:
Here's what we know and in which dimension:
y dimension:
[tex]v_0=30[/tex] m/s
v = 0 (I'll get to that injust a second)
a = -9.8 m/s/s
The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.
x dimension:
Δx = 70 m
v = ??
Velocity is our unknown.
Solving for the time in the y dimension:
[tex]v=v_0+at[/tex] and filling in:
0 = 30 + (-9.8)t and
-30 = -9.8t so
t = 3.1 seconds
We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.
In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.
Δx = vt and
70 = v(6.2) so
v = 11.3 m/s
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
Its Acceleration during the upward Journey ?
A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW from a geothermal source and rejects energy to the atmosphere. Find the power to the air and how much air should be flowed to the cooling tower (kg/s) if its temperature cannot be increased more than 10oC.
Answer:
- the power to the air is 850 MW
- mass flow rate of the air is 84577.11 kg/s
Explanation:
Given the data in the question;
Net power generated; [tex]W_{net[/tex] = 150 MW
Heat input; [tex]Q_k[/tex] = 1000 MW
Power to air = ?
For closed cycles
Power to air Q₀ = Heat input; [tex]Q_k[/tex] - Net power generated; [tex]W_{net[/tex]
we substitute
Power to air Q₀ = 1000 - 150
Q₀ = 850 MW
Therefore, the power to the air is 850 MW
given that ΔT = 10 °C
mass flow rate of air required will be;
⇒ Q₀ / CpΔT
we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K
we substitute
⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]
⇒ ( 850 × 10³ ) / 10.05
⇒ 84577.11 kg/s
Therefore, mass flow rate of the air is 84577.11 kg/s
