A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it reach the ground? How far from the point of projection would it strike?
Answer:
20.41 s
3534.80 m
Explanation:
In how many seconds will it reach the ground?
We are given the initial velocity of the body, which is 200 m/s at a 30° angle.
We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.
Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.
200 · sin(30) m/sLet's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.
Now we have one missing variable that we are trying to solve for: time t.
Find the constant acceleration equation that contains v₀, v, a, and t.
v = v₀ + atSubstitute known values into the equation.
0 = 200 · sin(30) + (-9.8)t -200 · sin(30) = -9.8t t = 10.20408163Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.
2t = 20.40816327The body will reach the ground in 20.41 seconds.
How far from the point of projection would it strike?
We want to find the displacement in the x-direction for the body.
Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).
Δx = v₀t + 1/2at²Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².
Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)² Δx = 3534.797567The body will strike 3534.80 m from the point of projection.
Caris parked on road slopes upward at angle θ. The magnitude the force of the road on the is mg cosθ. Is the magnitude of the stalie friction force on the less equal to or greater than μ2mgcosθ? Explain.
Answer:
fr = μ[tex]\miu _{k}[/tex] m g cos θ
the correct answer is less than fr = μ[tex]\miu _{k}[/tex] 2m g cos θ
Explanation:
Let's propose the solution of the problem to be able to answer the question, let's fix a reference system with one axis parallel to the plane (x-axis) and the other perpendicular, let's write Newton's second law
Y Axis
N - W cos θ = 0
N = mg cos θ
X axis.
fr - w sin θ = 0
fr = mg sin θ
the friction force is described by the expression
fr = μ N
fr = μ[tex]\miu _{k}[/tex] m g cos θ
When we analyze this expression we see that the friction force is equal to (μ_k m g cos θ)
If everything is well written in your problem, the correct answer is less, even though I think you have an error when writing the number
A 0.263-kg volleyball approaches a player horizontally with a speed of 17.4 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.8 m/s.
Required:
a. What impulse is delivered to the ball by the player?
b. If the players fist is in contact with the ball for 0.600s, find the magnitude of the average force exerted on the players fist.
Answer:
(A) J = -10.57 kg-m/s (B) 17.61 N
Explanation:
Given that,
Mass of a volleyball, m = 0.263 kg
Initial speed of volleyball, u = 17.4 m/s
Final speed of volleyball, v = -22.8 (in opposite direction)
(a) We need to find the impulse delivered to the ball by the player.
Impulse = change in momentum
J = m(v-u)
Put all the values,
J = 0.263(-22.8-17.4) kg-m/s
= -10.57 kg-m/s
(b) The time of contact with the ball, t = 0.6 s
We need to find the magnitude of the average force exerted on the players first.
Impulse, J = Ft
[tex]F=\dfrac{J}{t}\\\\F=\dfrac{10.57 }{0.6}\\\\F=17.61\ N[/tex]
So, the magnitude of the average force exerted on the player is 17.61 N.
3. A train travels 450 m north traveling at a velocity of 50 m/s . How long did it take the train to make this trip?
Answer:
9 seconds
Explanation:
450 divided by 50 equals 9.
Answer:
9s
Explanation:
Given parameters:
Displacement = 450m
Velocity = 50m/s
Unknown:
Time taken for the train to make the trip = ?
Solution:
Since velocity is displacement divided by the time taken,
Time = [tex]\frac{Displacement}{Velocity}[/tex]
Time = [tex]\frac{450}{50}[/tex] = 9s
The half-life for U 238 is 4.5x109 years.
a) If five half-lives have gone by how many years have gone by?
b) If you start with 240 grams of U 238 and end up with 60 grams, how many years have gone by?
c) If you start with 240 grams of U 238 and 1.8 x 10^10 years go by, how much U 238 is left?
d) If you start with 562 g and six half lives go by how many grams are left?
Answer:
a. 2.25 × 10^10 yrs
b. 9 × 10^9 yrs
c. 59.5g
d. 8.78g
Explanation:
(a) Original sample(N) = 238g
Half-life = 4.5 × 10^9 yrs
5 half-life 5T½ = 5 × 4.5 × 10^9 yrs
= 2.25 × 10^10 yrs
(b) If N = 240g
N/2 = 120g
N/4 = 60g
Meaning 2T½
= 2 × 4.5 × 10^9 yrs
= 9 × 10^9 yrs
(c) 1.8 × 10^10 ÷ 4.5 × 10^9 = 4
at 4T½ we have N/16 = 238/4 = 59.5
=> the sample left = 59.5g
(d) from the question N = 562g
at 6T½
the amount left will be N/64
= 562÷64 = 8.78g
NB: For questions please you can read more on radioactivity
Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardrums of the first student is 0.58 W/m^2, while at the eardrums of the second student the sound intensity is 1.18 times greater.
Required:
a. What is the ratio of the diameter of the first student's eardrum to that of the second student?
b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?
Answer:
a. d₁/d₂ = 1.09 b. 0.054 mW
Explanation:
a. What is the ratio of the diameter of the first student's eardrum to that of the second student?
We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²
So, I ∝ I/d²
I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.
Given that I₂ = 1.18I₁
I₂/I₁ = 1.18
Since I₁/I₂ = d₂²/d₁²
√(I₁/I₂) = d₂/d₁
d₁/d₂ = √(I₂/I₁)
d₁/d₂ = √1.18
d₁/d₂ = 1.09
So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09
b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?
We know intensity, I = P/A where P = acoustic power and A = area = πd²/4
Now, P = IA
= I₂A₂
= I₂πd₂²/4
= 1.18I₁πd₂²/4
Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m
So, P = 1.18I₁πd₂²/4
= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4
= 0.691244π × 10⁻⁴ W/4 =
2.172 × 10⁻⁴ W/4
= 0.543 × 10⁻⁴ W
= 0.0543 × 10⁻³ W
= 0.0543 mW
≅ 0.054 mW
(a) The ratio of the diameter of the first student's eardrum to that of the second student is 1.09.
(b) The acoustic power, in microwatts, striking each of his (and the other student's) eardrums is of 0.054 mW.
Given data:
The sound intensity of first student is, [tex]I_{1}= 0.58 \;\rm W/m^{2}[/tex].
And sound intensity of second student is, [tex]I_{2} = 1.18 \times I_{1}[/tex].
The diameter of second eardrum is, [tex]d_{2} = 1.01 \;\rm cm=1.01 \times 10^{-2} \;\rm m[/tex]
(a)
The power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²
So,
I ∝ I/d²
I₁/I₂ = d₂²/d₁²
Here
I₁ is the intensity at eardrum of first student.
d₁ is the diameter of first student's eardrum.
I₂ is the intensity at eardrum of second student.
d₂ is the diameter of second student's eardrum.
Since, I₂ = 1.18I₁
I₂/I₁ = 1.18
Also, I₁/I₂ = d₂²/d₁²
=√(I₁/I₂) = d₂/d₁
=d₁/d₂ = √(I₂/I₁)
=d₁/d₂ = √1.18
d₁/d₂ = 1.09
Thus, we can conclude that the ratio of the diameter of the first student's eardrum to that of the second student is 1.09.
(b)
We know that the expression for the intensity of sound is,
I = P/A
P = IA
Here,
P is the acoustic power and A is the area. (A = πd²/4)
P = I₂A₂
P= I₂πd₂²/4
P= 1.18I₁πd₂²/4
Since,
I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m
Substituting the values as,
P = (1.18I₁πd₂²) /4
P = 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4
P = (0.691244π × 10⁻⁴ W) /4
P = (2.172 × 10⁻⁴ W) /4
P = 0.543 × 10⁻⁴ W
P ≅ 0.054 mW
Thus, we can conclude that the acoustic power, in microwatts, striking each of his (and the other student's) eardrums is of 0.054 mW.
Learn more about the acoustic power here:
https://brainly.com/question/14932158
The sound from a trumpet travels at 351 m/s in the frequency of the note is 294 Hz, what is the wavelength of the sound wave?
Answer:
1.19m
Explanation:
Given parameters:
Speed of the trumpet sound = 351m/s
Frequency of the note = 294Hz
Unknown:
Wavelength of the sound = ?
Solution:
To solve this problem, we use the wave - velocity equation.
V = F x ∧
V is the velocity of the body
F is the frequency
∧ is the wavelength
So;
351 = 294 x ∧
∧ = [tex]\frac{351}{294}[/tex] = 1.19m
In an attempt to deliver a parcel on time, the dispatch rider had to ride 10km 15 degrees SE, he then rode 11 km 30 degrees NE and then takes a shortcut at 22km W .find the rider's displacement
The total displacement of the dispatch rider is calculated as 43km.
Data;
10km 15 degrees SE11km 30 degrees NE22km WTotal DisplacementTo calculate the total displacement of the dispatch rider, we can simply add up the total distance covered by the rider.
This becomes;
[tex]10+11+22 = 43km[/tex]
The total displacement of the dispatch rider is calculated as 43km.
Learn more on displacement here;
https://brainly.com/question/13416288
https://brainly.com/question/1397750
how do forces act when they are not in contact with each other?
Answer:they seperate from eachother but they don't touch eachother at all
Explanation:
Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be added to the gas in the form of heat by applying a flame to the outside of the container. Conversely, energy can also be removed from the gas in the form of heat by immersing the container in ice water. Energy can be added to the system in the form of work by pushing the piston in, thereby compressing the gas. Conversely, if the gas pushes the piston out, thereby pushing some atmosphere aside, the internal energy of the gas is reduced by the amount of work done.
Complete Question
Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be added to the gas in the form of heat by applying a flame to the outside of the container. Conversely, energy can also be removed from the gas in the form of heat by immersing the container in ice water. Energy can be added to the system in the form of work by pushing the piston in, thereby compressing the gas. Conversely, if the gas pushes the piston out, thereby pushing some atmosphere aside, the internal energy of the gas is reduced by the amount of work done.
[tex]pV=nRT[/tex]
so the absolute temperature T is directly proportional to the product of the absolute pressure p and the volume V,Here n denotes the amount of gas moles,which is a constant because the gas is confined and R is the universal constant
What is the [tex]\triangle U[/tex] as the system of ideal gas goes from point A to point B on the graph recall u is proportional to T
Answer:
[tex]\triangle T=0[/tex]
[tex]\triangle V=0[/tex]
The gas A and B have same internal energy
Explanation:
From the question we are told that
[tex]Pa=u atm\\Va=1m^3\\Pb=1 atm\\Vb=4m^3[/tex]
Generally the equation of temperature is mathematically given as
[tex]Ta=\frac{Pv}{nR}[/tex]
[tex]Ta=\frac{u*1}{nR}[/tex]
And
[tex]Tb=\frac{PbVb}{nR}[/tex]
[tex]Tb=\frac{u*1}{nR}[/tex]
Generally the change in temperature [tex]\triangle T[/tex] is mathematically given as
[tex]\triangle T=Tb-Ta=Tb=\frac{u*1}{nR}-\frac{u*1}{nR}[/tex]
[tex]\triangle T=0[/tex]
Generally the change in internal energy [tex]\triangle V[/tex]
[tex]\triangle V=nC_v \triangle T\\[/tex]
[tex]\triangle V=0[/tex]
Therefore with
[tex]\triangle T=0[/tex]
[tex]\triangle v=0[/tex]
The gas A and B have same internal energy
A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coasting. Describe each of the following events as an elastic, an inelastic, or a perfectly inelastic collision between the skater and the Frisbee.
Required:
a. She catches the Frisbee and holds it.
b. She tries to catch the Frisbee, but it bounces off her hands and falls to the ground in front of her.
c. She catches the Frisbee and immediately throws it back with the same speed (relative to the ground) to her friend.
Answer:
a) perfectly inelastic, b) collision is inelastic, c) elastic
Explanation:
In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved
Initial instant. Before the skater touches the frisbee
p₀ = M v₁ + m v₂
where M and m are the masses of the skater and frisbee, respectively
for the final moment they give us several possibilities, in all case the moment is conserved
p₀ = [tex]p_{f}[/tex]
case a)
Final instant. grabs the frisbee and holds it
p_{f} = (M + m) v '
p₀ = p_{f}
We can see that this shock is perfectly inelastic, it holds the fressbee
case b)
final instant.
This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.
case c)
final instant. Grab the fressbee and resend it
[tex]p_{f} = M v_{1f} + m v_{2f}[/tex]
this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the mean value of your results to three significant digits. ________
Answer: The mean value = 9.85m/s².
Explanation:
Mean = [tex]\dfrac{\text{Sum of n observations}}{n}[/tex]
The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.
Number of measurements =9
Sum of measurements = 88.69
Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]
Hence, the mean value = 9.85m/s².
A racecar reaches 24 m/s in 6 seconds at the start of a race. What is the acceleration of the car?
Answer:
4m/s^2
Explanation:
Convert 20 C to F
-40 C to F
40C to F
Answer:
20 C to F
Ans: 68F
-40 C to F
Ans:-40F
40C to F
Ans:104F
Two satellites are approaching the Earth from opposite directions. According to an observer on the Earth, satellite A is moving at a speed of 0.648c and satellite B at a speed of 0.795c. What is the speed of satellite A as observed from satellite B
Answer:
Explanation:
Satellite A and satellite B are approaching the earth from opposite directions , that means they are approaching each other . The velocity of satellite A and B are .648c and .795c respectively . Their velocities are comparable to velocity of light so they will follow relativistic laws .
Their relative velocity will be given by the following relation .
[tex]V_r=\frac{u+v}{1+\frac{uv}{c^2} } }[/tex]
where u and v are velocities of vehicles coming from opposite direction and c is velocity of light .
[tex]V_r=\frac{.795c+.648c}{1+\frac{.648c\times .795c}{c^2} } }[/tex]
[tex]V_r=\frac{1.443c}{1+.515 } }[/tex]
= .952c
Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of 0.624c and determines its lifetime to be 159 ns. (a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory
Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
A 76.O kg person is being pufed away from a burning building as shown below
LLLL
IRLA
If the tension in rope 1.11. As 193.5 N. calculate, without using the component method,
the tension in rope 2.
15a
Answer:
Final Answer
T1=736 NT1=736 N
T2=194
what Species can change over time to adapt to their environment
Answer:
Explanation:
Most environments have many niches. If a niche is "empty" (no organisms are occupying it), new species are likely to evolve to occupy it. This happens by the process of natural selection. By natural selection, the nature of the species gradually changes to become adapted to the niche.
Two quantities that have the same dimension but have a different physical concepts.
Please help me
Answer:
In general, if the dimensions are same, the quantities do represent the same physical content. Like work and energy have the same dimensions and represent inter-convertible quantity. ... These two quantities represent the same quantity - same meaning and content.
Explanation: i hope this helped.
Hlo
what is a force........??
Explanation:
In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.
Formula
Newton's Second Law
F = m * a
F = force
m = mass of an object
a = acceleration
A proton is given an acceleration of 1.5x109 m/s² when it is placed in an electric field.
What is the strength of the electric field?
Answer:
The strength of the electric field is [tex]E=15.66\: N/C[/tex]
Explanation:
Here the electric force is equal to Newton's second law.
[tex]F_{e}=ma[/tex]
Let's recall that electric force is the electric field times the charge, so we have:
[tex]qE=ma[/tex]
[tex]E=\frac{ma}{q}[/tex] (1)
Where:
m is the proton mass
q is the proton charge
a is the acceleration
Using the equation (1) we have:
[tex]E=\frac{1.67*10^{-27}1.5x10^{9}}{1.6*10^{-19}}[/tex]
Therefore, the strength of the electric field is [tex]E=15.66\: N/C[/tex]
I hope it helps you!
What is the great egg drop experiment about?
Answer:
The egg drop experiment is about building a structure around an egg with different materials so that when it is dropped from a high place it doesn't break.
Explanation:
I have done this experiment before so I know. Have a cool awesome great splendid Supercalifragilisticexpialidocious day ok bye.
A billiard ball collides with another ball with a force of 30 N after rolling for 3 seconds. What is the
impulse during the collision?
Answer:
The impulse is 90Ns.
Explanation:
Given that the formula for impulse is F×t, where F represents force and t is time (in seconds) :
Impulse = F × t
Impulse = 30 × 3 = 90Ns
Scenario
You have recently joined the team at A&L, an engineering firm with a broad portfolio. A&L has recently been hired to help plan a supply drop following a natural disaster. Due to conditions on the ground, the supply drop will be done from the air. Your supervisor has asked you to use that information, as well as your knowledge of kinematics, to create a supply drop plan detailing how far the payload should be from the drop site when it is delivered. Additionally, due to adverse conditions in the area, your supervisor has asked you to prepare for two contingencies.
Directions
The plane carrying the supplies will be cruising at a constant velocity of 250 miles per hour relative to the ground and at a height of 2,650 meters above the target site. Using this information, create a supply drop plan including all required information and calculations outlined below. As you are completing your supply drop plan, remember that correct SI units are a required component of your calculations and descriptions.
Construct a diagram that describes the horizontal and vertical motion of the payload. Your diagram should visually represent the initial velocity and height of the payload as it approaches the drop site. To create the diagrams, you may use drawing tools, or you may photograph or screenshot a drawing of your own. This diagram should include the following elements:
The initial velocity of the payload
The initial height of the payload
Horizontal distance to the drop site
Using your understanding of kinematic equations and the given variables in the scenario, calculate the horizontal and vertical motion of the payload to ensure it arrives at the drop site. In your calculations, account for both the horizontal and vertical motion of the payload. Your calculations should address the following:
Initial velocity of the payload when launched
The velocity of the payload when it hits the ground
Following your calculations, describe the equations used in calculating the vertical and horizontal motion of an object. In your descriptions, address the following:
What is the relationship between vertical and horizontal motion in kinematics equations?
How did you analyze the vertical motion of the payload in your solution?
How did you analyze the horizontal motion of the payload in your solution?
What other kinematics principles did you consider in analyzing the motion of the payload?
Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{ \frac{2 y_o}{g} }[/tex]
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
what evidence supports the information consolidation theory?
PLEASE HELP ME ITS DUE NEXT PERIOD
Explanation:
Memory loss in retrograde amnesia has long been held to be larger for recent periods than for remote periods, a pattern usually referred to as the Ribot gradient. One explanation for this gradient is consolidation of long-term memories. Several computational models of such a process have shown how consolidation can explain characteristics of amnesia, but they have not elucidated how consolidation must be envisaged. Here findings are reviewed that shed light on how consolidation may be implemented in the brain. Moreover, consolidation is contrasted with alternative theories of the Ribot gradient. Consolidation theory, multiple trace theory, and semantization can all handle some findings well but not others.
Can someone help me with this?
Answer:
C would make the most sense.
Explanation:
ok so basically the reason why i chose C is because i've always been told that whatever i type, send, or receive on my phone is stored into the phone's memory and can be brought back up with keystrokes. so yeah, i'd go with C but it's DEFINATELY not A or D.
How do you work out the spring constant of a motorcycle spring when the force is 240N and the spring is 2cm?
Answer:
1200N/m
Explanation:
Given parameters:
Force on the motorcycle spring = 240N
Extension = 2cm or 0.02m
Unknown:
Spring constant = ?
Solution:
To a spring the force applied is given as :
F = K e
F is the applied force
K is the spring constant
e is the extension
240 = k x 0.02
k = 1200N/m
A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with uniform linear charge density λ2 is located on the y-axis with their centers crossing at the origin. In what direction is the electric field at point z = a on the positive z-axis if λ1 and λ2 are positive?
A-) The positive z-direction
B-) All directions are possible parallel to the x-y plane
C-) The negative z-direction
D-) Halfway between the x-direction and the y-direction
C-) Cannot be determined
Answer:
A.The positive z-direction
Explanation:
We are given that
Linear charge density of long line which is located on the x-axis=[tex]\lambda_1[/tex]
Linear charge density of another long line which is located on the y-axis=[tex]\lambda_2[/tex]
We have to find the direction of electric field at z=a on the positive z-axis if [tex]\lambda_1[/tex] and [tex]\lambda_2[/tex] are positive.
The direction of electric field at z=a on the positive z-axis is positive z-direction .
Because [tex]\lambda_1[/tex] and [tex]\lambda_2[/tex] are positive and the electric field is applied away from the positive charge.
Hence, option A is true.
A.The positive z-direction
A car is traveling at 40 m/s. The brakes are applied an after 3 seconds the car is traveling 13 m/s. What is the acceleration of the car?
Answer:
- 9m/s²
Explanation:
Given parameters:
Initial velocity = 40m/s
Time taken = 3s
Final velocity = 13m/s
Unknown:
Acceleration of the car = ?
Solution:
Acceleration is the rate of change of velocity with time taken.
Acceleration = [tex]\frac{Final velocity - Initial velocity }{Time taken}[/tex]
Acceleration = [tex]\frac{13 - 40 }{3}[/tex] = - 9m/s²
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