A quarterback, Patrick, throws a football down the field in a long arching trajectory to wide receiver, Tyreek. The football and Tyreek are traveling in the same direction, started at the same spot, and the football was thrown at the same instant that Tyreek began running. Furthermore, both Tyreek and the football have horizontal components of speed of 22.6 mph. Under these circumstances, no matter what angle the football is thrown at, it will land on Tyreek (whether he catches it or not). True or false
Answer:
the statement is False [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
Explanation:
Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.
Let's start by finding the time it takes for the ball to reach the ground
y = [tex]v_{oy}[/tex] t - ½ g t²
when it reaches the ground its height is y = 0
0 = vo sin θ - ½ g t²
0 = t (vo sin θ - ½ g t)
The results are
t = 0 exit point
t = 2 v₀ sin θ/g arrival point
at this point the ball traveled
[tex]x_{ball}[/tex]= v₀ₓ t
x_{ball} = v₀ cos θ 2v₀ sin θ / g
x_{ball}= 2 v₀² cos θ sin θ/ g
Now let's find that distantica traveled the receiver in time
[tex]x_{rec}[/tex] = v₀ t
x_{rec} = v₀ (2 v₀ sin θ / g)
x_{rec} = 2 v₀² sin θ / g
without dividing this into two distances
[tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
therefore the distances are not equal to the ball as long as behind the receiver
Therefore the statement is False
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.
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
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
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;
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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.
You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of electrons from each sphere.
1) What is the charge on each sphere if their gravitational attraction is exactly equal to their electrical repulsion?
2) How many electrons did you remove from each sphere?
Answer:
q = 2.066* 10⁻¹³ C.
n = 1,291,250 electrons.
Explanation:
1)
If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:[tex]F_{g} = F_{c} (1)[/tex]
where Fg is the gravitational attraction, that can be written as follows according Newton's Universal Law of Gravitation:[tex]F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)[/tex]
Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming we can treat both spheres as point charges), as follows:[tex]F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)[/tex]
since m₁ = m₂ = 0.0024 kg, and r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:[tex]G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)[/tex]
Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:[tex]Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)[/tex]
Since both charges are the same, the charge on each sphere is just the square root of (5):Q = 2.066* 10⁻¹³ C.2)
Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:[tex]n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)[/tex]After an initial race George determines that his car loses 35 percent of its acceleration due to air resistance travelling at 38 m/s on flat ground. Assuming that his car travels with a constant acceleration, calculate the maximum speed (Vm), in meters per second, his car can reach on flat ground?\
Answer:
64.2 m/s
Explanation:
We are given that
Speed ,v=38 m/s
We have to find the maximum speed when his car reach on flat ground.
Using dimensional analysis
[tex]F_{res}\propto v^2[/tex]
If 35% acceleration reduced by F(res) at 38 m/s
Then, 100% acceleration can be reduced by F(res) at v' m/s
[tex]\frac{F_1}{F_2}=\frac{v^2}{v'^2}[/tex]
[tex]v'^2=\frac{F_2}{F_1}v^2[/tex]
[tex]v'=v\sqrt{\frac{F_2}{F_1}}[/tex]
Substitute the values
[tex]v'=38\times \sqrt{\frac{100}{35}}[/tex]
[tex]v'=64.2 m/s[/tex]
Hence, the maximum speed when his car can reach on flat ground=64.2 m/s
The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
What is the relation between resistance and speed?
The air resistance is directly proportional to the square of the velocity of an object.
R ∝ v²
The speed of the car was reduced by 35 % at 38 m/s.
So, the speed of the car was reduced by 100% at v' m/s.
The relationship can be given by,
[tex]\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}[/tex]
Put the values in the formula and calculate for [tex]v'[/tex],
[tex]v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}[/tex]
[tex]v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s[/tex]
Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
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Need help in the question that is down
Answer:
4th ans
Explanation:
i will give the brainliest answer to whoever answers this, explain how Sir William Gilbert used models in his investigation of magnetism
Explanation:
Lodestone am iron were the only known magnetic materials in Gilbert's day, and his task was to investigate magnetism. Gilbert was so sure that the earth was a giant lodestone and used the earth as a primary reference, defining the north(magnetic) pole of a needle, or a a nail floating on a piece of cork, to be that which turns towards the Earth's north geographic pole. he wanted to prove this with a model Terella, using short pieces of iron.
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
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
An electron moving along the x axis has a position given by x 16te$t m, where t is in seconds. How far is the electron from the origin when it momentarily stops
This question is incomplete, the complete question is;
An electron moving along the x axis has a position given by x = 16 t[tex]e^{-t}[/tex] m, where t is in seconds. How far is the electron from the origin when it momentarily stops
Answer:
the electron is 5.88 m far from the origin when it momentarily stops
Explanation:
Given that;
the position of the electron is x = 16 t[tex]e^{-t}[/tex] m
now, if the electron stopped after a time t, then its velocity is zero
so
V = dx/dt = 0
d/dt( 16 t[tex]e^{-t}[/tex] m) = 0
16( -t[tex]e^{-t}[/tex] + [tex]e^{-t}[/tex] ) = 0
16(-t + 1) [tex]e^{-t}[/tex] = 0
16(1 - t) [tex]e^{-t}[/tex] = 0
1 - t = 0
t = 1 sec
so
x = 16 × 1 × [tex]e^{-1}[/tex] m
x = 16 × 1 × 0.36787 m
x = 5.88 m
Therefore, the electron is 5.88 m far from the origin when it momentarily stops
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
Please help me I’ll mark brainless .
What name is given to "an object or group of objects that we wish to consider for a physics problem?
Answer:
System
Explanation:
A system is an object or group of objects that we wish to consider for a physics problem. Examples of systems are universe, house and its surroundings etc.
There are three different types of systems which are:
Isolated system: In this type of system there is no energy or matter exchange with the surroundings. Closed system: In this type of system, only energy is exchanged with the surroundings, matter is not exchanged. Open system: in this system both matter and energy are exchanged with the surroundings..A student in gym class swings from a rope and they are moving 5 m/s at the bottom of their swing. What is the height they reach above the floor before swinging back down?
1/2v^2=gh g = 9.8 m/s^2
A 2.55 m
B. 1.28 m
c. 5m
D. 12.5 m
Answer:
A
Explanation:
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
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.
There is always one way to calculate the magnitude of the net vector of any direction vectors.
True
False
Answer and I will give you brainiliest
Answer:
False
Explanation:
Hope it help mark as BrainlistAnswer:
False
Explanation:
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.
What does Newtons 1st Law Desribe?
A. the tendency of moving objects to stay in motion
B. How force balances out of zero net force
C. The change in an objects motion over time
D. The attractive force between two objects due to their masses
Answer:
A. the tendency of moving objects to stay in motion
Two cars, a Porsche Boxster convertible and a Toyota Scion xB, are traveling at constant speeds in the same direction. Suppose, instead, that the Boxster is initially 170 m behind the Scion. The speed of the Boxster is 24.4 m/s and the speed of the Scion is 18.6 m/s. How much time does it take for the Boxster to catch the Scion
Answer:
It will take 29.31 seconds for the Boxster to catch the Scion
Explanation:
Given the data in the question;
lets say Toyota Scion xB is car A and Porsche Boxster convertible is B and Toyota Scion xB is car A
the distance travelled by car A is
x = [tex]V_{A}[/tex] × t
where [tex]V_{A}[/tex] is the speed of the car and t is time
the distance travelled by car B before reaching car A will be;
x + x₀ = [tex]V_{B}[/tex] × t
Now lets replace x by [tex]V_{A}[/tex] × t
so
([tex]V_{A}[/tex] × t) + x₀ = [tex]V_{B}[/tex] × t
x₀ = ([tex]V_{B}[/tex] × t) - ([tex]V_{A}[/tex] × t)
x₀ = t ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])
t = x₀ / ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])
so we substitute
t = 170 m / (24.4 - 18.6)
t = 170 / 5.8
t = 29.31 s
Therefore; it will take 29.31 s for the Boxster to catch the Scion
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 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:
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Two skaters, each of mass 45 kg, approach each other along parallel paths separated by 3.2 m. They have equal and opposite velocities of 1.1 m/s. The first skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes.. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. What is their angular speed?
Answer:
Explanation:
The skaters in a circle of radius; R = L/2
R = 3.2/2 = 1.6 m
From Ii*Wi = If*Wf
Note; i is initial while f is final
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
Magnetic dipole X is fixed and magnetic dipole Y is free to move. Dipole Y will initially:
A. Move toward X but not rotate
B. Move away from X but not rotate
C. Move toward X and rotate Move away from X and rotate
D. Rotate but not translate
Answer: The correct option is A.
Move toward X but not rotate
Explanation:
This is because from the question, X is fixed and y is free to move. Since magnetic dipole of X is fixed, that is it can't move and that of y is free to move, therefore y will move toward x because it's forces of attraction is linear and not rotational and besides X and Y are on a linear path, therefore Y will move towards X that is fixed and it will not rotate since it's linear.
Suppose a car is traveling at 18.6 m/s, and the driver sees a traffic light turn red. After 0.500 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 4.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light
Answer:
s = 52.545 m
Explanation:
First, we calculate the distance covered during the 0.5 s when the driver notices the light and applies the brake.
[tex]s_{1} = vt\\[/tex]
where,
s₁ = distance covered between noticing light and applying brake = ?
v = speed = 18.6 m/s
t = time = 0.5 s
Therefore,
[tex]s_{1} = (18.6\ m/s)(0.5\ s)\\s_{1} = 9.3\ m\\[/tex]
Now, we calculate the distance for the car to stop after the application of brakes. For that we use 3rd equation of motion:
[tex]2as_{2} = V_{f}^{2} - V_{i}^{2}\\\\[/tex]
where,
s₂ = distance covered after applying brake = ?
a = deceleration = - 4 m/s²
Vf = final speed = 0 m/s
Vi = initial speed = 18.6 m/s
Therefore,
[tex]2(- 4\ m/s^{2})s_{2} = (0\ m/s)^{2} - (18.6\ m/s)^{2}\\\\s_{2} = \frac{(18.6\ m/s)^{2})}{8\ m/s^{2}}\\\\s_{2} = 43.245\ m[/tex]
So the total distance covered by the car before stopping is:
[tex]s = s_{1} + s_{2}\\s = 9.3\ m + 43.245\ m\\[/tex]
s = 52.545 m
how much work is required to make a 1400 kg car increase its speed from 10 m/s to 20 m/s?
what average force is required if the car travels 15 m during this speed change?
