Answer:
1.16 Hz
Explanation:
frequency, basically, is the number of wave on 1 second
so, in math we write like this
f = n/t
n = number of waves
t = time to do that (in sec)
f = 140/120 = 7/6 Hz
f = 1.16 Hz
A balloon with a radius of 16 cm has an electric charge of 4.25 10 –9 C.
Determine the electric field strength at a distance of 40.0 cm from the balloon’s centre.
Answer:
239 N/C
Explanation:
Electric field strength at distance R from a charge Q is given by the expression
E = k Q / R² where Q is charge , R is distance of charge from the point . k is a constant .
R = 40 cm , Q = 4.25 x 10⁻⁹
Putting the given values
E = 9 x 10⁹ x 4.25 x 10⁻⁹ / ( 40 x 10⁻²)²
= 239 N/C .
A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a position with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na
Answer:
Explanation:
During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy
mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .
mgr = 1/2 x ( 1/2 m r²+ mr²) x ω²
gr = 1/2 x 1/2 r² x ω² + 1/2r² x ω²
g = 1 / 4 x ω² r + 1 / 2 x ω² r
g = 3 x ω² r/ 4
ω² = 4g /3 r
= 4 x 9.8 / 3 x .25
= 52.26
ω = 7.23 rad / s .
g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the dragster at the end of 2.1 s, (b) the final velocity of the dragster at the end of of twice this time, or 4.2 s, (c) the displacement of the dragster at the end of 2.1 s, and (d) the displacement of the dragster at the end of twice this time, or 4.2 s?
The dragster's velocity v at time t with constant acceleration a is
[tex]v=at[/tex]
since it starts at rest.
After 2.1 s, it will attain a velocity of
[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]
or 92.4 m/s.
Doubling the time would double the final velocity,
[tex]v=a(2t)=2at[/tex]
so the velocity would be twice the previous one, 184.8 m/s.
The dragster undergoes a displacement x after time t with acceleration a of
[tex]x=\dfrac12at^2[/tex]
if we take the starting line to be the origin.
After 2.1 s, it will have moved
[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]
or 88 m.
Doubling the time has the effect of quadrupling the displacement, since
[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]
so after 4.2 s it will have moved 352 m.
A solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins
Answer:
Cube temperature = 526.83 K
Explanation:
Volume of the cube and sphere will be the same.
Now, volume of cube = a³
And ,volume of sphere = (4/3)πr³
Thus;
a³ = (4/3)πr³
a³ = 4.1187r³
Taking cube root of both sides gives;
a = 1.6119r
Formula for surface area of sphere is;
As = 4πr²
Also,formula for surface area of cube is; Ac = 6a²
Thus, since a = 1.6119r,
Then, Ac = 6(1.6119r)²
Ac = 15.5893r²
The formula for radiant power is;
Q' = eσT⁴A
Where;
e is emissivity
σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k
T is temperate in kelvin
A is Area
So, for the cube;
(Qc)' = eσ(Tc)⁴(Ac)
For the sphere;
(Qs)' = eσ(Ts)⁴(As)
We are told (Qc)' = (Qs)'
Thus;
eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)
eσ will cancel out to give;
(Tc)⁴(Ac) = (Ts)⁴(As)
Since we want to find the cube's temperature Tc,
(Tc)⁴ = [(Ts)⁴(As)]/Ac
Plugging in relevant figures, we have;
(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²
r² will cancel out to give;
(Tc)⁴ = [556⁴ × 4π]/15.5893
Tc = ∜([556⁴ × 4π]/15.5893)
Tc = 526.83 K
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes the point x=385 with a speed of 45.0 m/s at t=20.0 s. Find the average velocity and the average acceleration between t=3.0 s and 20.0 s.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
EASY HELP
As a space shuttle climbs, _____.
its mass increases
its mass decreases
its weight increases
its weight decreases
Answer: it's weight decreases
Explanation:
A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward.
(A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s
Answer:
a) 98 m/s
b) 19.6 s
c) 1449.8 m
d) 29.6 s
e) 168.6 m/s
f) 46.8 s
Explanation:
Given that
Acceleration of the rocket, a = 5 m/s²
Altitude of the rocket, s = 960 m
a)
Using the equation of motion
v² = u² + 2as, considering that the initial velocity, u is 0. Then
v² = 2as
v = √2as
v = √(2 * 5 * 960)
v = √9600
v = 98 m/s
b)
Using the equation of motion
S = ut + ½at², considering that initial velocity, u = 0. So that
S = ½at²
t² = 2s/a
t² = (2 * 960) / 5
t² = 1920 / 5
t² = 384
t = √384 = 19.6 s
c)
Using the equation of motion
v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that
0 = 98² + 2(-9.8) * s
9600 = 19.6s
s = 9600/19.6
s = 489.8 m
The maximum altitude now is
960 m + 489.8 m = 1449.8 m
d)
Using the equation of motion
v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that
0 = 98 +(-9.8 * t)
98 = 9.8t
t = 98/9.8
t = 10 s
Total time then is, 10 + 19.6 = 29.6 s
e) using the equation of motion
v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,
v² = 0 + 2 * 9.8 * 1449.8
v² = 28416.08
v = √28416.08
v = 168.6 m/s
f) using the equation of motion
S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s
1449.8 = 0 + ½ * 9.8 * t²
2899.6 = 9.8t²
t² = 2899.6/9.8
t² = 295.88
t = √295.88
t = 17.2 s
total time in air then is, 17.2 + 29.6 = 46.8 s
A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.90 m above the water.
A. Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
b) Use energy conservation to find his speed just he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.90 m/s .
c) Use energy conservation to find his speed just he hits the water if he manages to jump downward at 2.90 m/s .
Answer:
Explanation:
The Law of Energy Conservation states that K1 + U1 = K2 + U2
m= 72.0 kg
h= 3.90 m
a)
K1 + U1 = K2 + U2
0 + mgh = 1/2mvf^2 + 0
mass cancels out so gh=1/2vf^2
(9.8 m/s^2)(3.9 m)=(.5)(vf^2)
vf= 8.74 m/s
b)
1/2mv^2 + mgh = 1/2mv^2 + 0
mass cancels again
(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)
vf= 9.21 m/s
c)
This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s
A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 C. Now a valve is opened, and half of the mass of the gas is allowed to escape. if the final pressure in the tank is 2.2 atm. The final temperature in the tank is: Hint: make sure you convert the units of temperature and pressure to the proper units
Answer:
Final Temperature = 71 °C
Explanation:
In this case, the ideal gas equation is written as;
PV = mRT
Where;
P is pressure
V is volume
m is mass
R is gas constant
T is temperature
We will take the volume to be constant.
So, in the initial state, we have;
P1•V = m1•R•T1 - - - eq(1)
In the final state, we have;
P2•V = m2•R•T2 - - - - eq(2)
Combining eq (1) and eq(2),we have;
P1•m2•R•T2 = P2•m1•R•T1
Dividing both sides by R gives;
P1•m2•T2 = P2•m1•T1
Making T2 the subject gives;
T2 = (P2•m1•T1)/(P1•m2)
Now, we are given;
m1 = 2kg
m2 = ½*2 = 1kg
P1 = 4 atm
P2 = 2.2 atm
T1 = 40°C = 273 + 40 K = 313K
Plugging in this values into the T2 equation, we have;
T2 = (2.2 × 2 × 313)/(4 × 1)
T2 = 344 K
Converting to °C, we have;
T2 = 344 - 273 = 71 °C
1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts
experience on the Moon or on the space station? Explain
Answer:
by using it's buoyant or floating effect by Archimedes.
the buoyant force act on the astronauts body and make he/ she feels like in low gravity.
the buoyant force equation is
F = Density of liquid x earth gravitational field x volume of astronauts body and suit.
the Weight of astronauts in the pools will be less than in the land or air.
Weight in water = weight in air/land - buoyant force
so the astronauts will feel like in the outer space with low gravity.
Crystalline germanium (Z=32, rho=5.323 g/cm3) has a band gap of 0.66 eV. Assume the Fermi energy is half way between the valence and conduction bands. Estimate the ratio of electrons in the conduction band to those in the valence band at T = 300 K. (See eq. 10-11) Assume the width of the valence band is ΔΕV ~ 10 eV.
Answer:
= 8.2*10⁻¹²
Explanation:
Probability of finding an electron to occupy a state of energy, can be expressed by using Boltzmann distribution function
[tex]f(E) = exp(-\frac{E-E_f}{K_BT} )[/tex]
From the given data, fermi energy lies half way between valence and conduction bands, that is half of band gap energy
[tex]E_f = \frac{E_g}{2}[/tex]
Therefore,
[tex]f(E) = exp(-\frac{E-\frac{E_g}{2} }{K_BT} )[/tex]
Using boltzman distribution function to calculate the ratio of number of electrons in the conduction bands of those electrons in the valence bond is
[tex]\frac{n_{con}}{n_{val}} =\frac{exp(-\frac{[E_c-E_g/2]}{K_BT} )}{exp(-\frac{[E_v-E_fg/2}{K_BT} )}[/tex]
[tex]= exp(\frac{-(E_c-E_v}{K_BT} )\\\\=exp(\frac{-(0.66eV)}{(8.617\times10^-^5eV/K)(300K)} )\\\\=8.166\times10^-^1^2\approx8.2\times10^{-12}[/tex]
The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?
Answer:
61.6° west of South
Explanation:
The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.
Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side and the hypotenuse is 8
cos(θ) = [adjacent/hypotenuse]
Cos θ = 3.8/8
Cos θ = 0.475
θ = cos^-1 (0.475)
θ = 61.6°
Therefore the angle is 61.6° west of South.
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy Stored = 36000 J = 36 KJ
Explanation:
The power of a battery is given by the formula:
P = IV
where,
P = Power delivered by the battery
I = Current Supplied to the battery
V = Potential Difference between terminals of battery = 12 volt
Now, we multiply both sides by the time period (t):
Pt = VIt
where,
Pt = (Power)(Time) = Energy Stored = E = ?
It = Battery Current Rating = 50 A.min
Converting this to A.sec;
It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
A swimmer heading directly through a 200m wide river reaches the opposite shore in 6 min 40s. She is washed downstream 480 m. How fast can you swim in calm water?
Answer :v=480m400s=1.2ms
2002+4802=H2
The hypotenuse H=520m
A quicker way to get the length of the hypotenuse is to recognize that this is a simple 5–12–13 triangle where the sides are multiples of 5, 12, and 13:
5(40) = 200m, 12(40)= 480m, 13(40)= 520m
We know that the swimmer travelled 520 m in 400 seconds, so her average speed was:
VR=520m400sec= 1.3ms
hope i got it right!! xx
Explanation:
What is the relationship between electric force and distance between charged objects and the amount of charge?
Explanation:
The relationship between electric force and distance between charged objects is given by the formula as follows :
[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]
k is electrostatic constant and d is distance between charges
The electric force between charges is inversely proportional to the square of distance between them.
A note on a piano vibrates 262 times per second . What is the period of the wave ?
In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.
Calculate the equivalent resistance of the circuit. [Show all work, including the equation and substitution with units.]
Answer:
Explanation:
The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]
What is equivalent resistance?The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.
What is equivalent resistance in series?Resistors are in series whenever the current flows through the resistors sequentially. It is given by
[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]
What is equivalent resistance in parallel?Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.
The equivalent resistance is the total resistance measured in a parallel. It is given by
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]
Given:
Resistor, [tex]R_{1} = 100 ohm[/tex]
Resistor, [tex]R_{2} = 200 ohm[/tex]
Voltage, [tex]V = 10 Volt[/tex]
Since, resistors are connected in parallel, the equivalent resistor is given by,
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]
[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]
[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]
Hence, the equivalent resistor is [tex]66.66 ohm[/tex].
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a) Write the names of the materials used in the ohm law according to the Figure 1?
b) If the voltage of a circuit is 12 V and the resistance is 40 , What is the generated power?
Answer:
a. i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Power generated is 3.6 W.
Explanation:
Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.
i.e V = IR
i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Given that; V = 12 V and R = 40 Ohm's.
P = IV
From Ohm's law, I = [tex]\frac{V}{R}[/tex]
So that;
P = [tex]\frac{V^{2} }{R}[/tex]
= [tex]\frac{12^{2} }{40}[/tex]
= [tex]\frac{144}{40}[/tex]
= 3.6 W
The power is 3.6 W.
You are helping your friend prepare for the next skateboard exhibition by determining if the planned program will work. Your friend will take a running start and then jump onto a heavy-duty 13-lb stationary skateboard. The skateboard will glide in a straight line along a short, level section of track, then up a curved concrete wall. The goal is to reach a height of at least 10 feet above the starting point before coming back down the slope. Your friend's maximum running speed to safely jump on the skateboard is 24 feet/second. Your friend weighs 155-lbs. What is the height hf that your friend will reach according to his plan?
Answer:
8.3 feet
Explanation:
The kinetic energy of the system on the ground is ...
KE = Σ(1/2)(mv^2) = (1/2)(155)(24^2) +(1/2)(13)(0^2) = 44640 lb·ft²/s²
The potential energy at the highest point is the same:
PE = mgh
44640 = (155 +13)(32)h
h = 44640/5376 = 8.30 . . . . feet
_____
We haven't worried too much about the conversion between pounds mass and pounds force. Whatever factor may be involved will divide out when computing the maximum height. We have used g=32 ft/s².
__
To achieve a 10 ft height, the running speed would need to be 26.34 ft/s, about 10% higher.
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, counting the ground level as the first,
A. What is the energy E of the emitted photon in electron volts?、
B. What is the wavelength in nanometers of the emitted photon?
C. What is the radius of the hydrogen atom in nanometers in its initial 5th energy level?
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
[tex]E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})[/tex] (1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
[tex]E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV[/tex]
B. The energy of the emitted photon is given by the following formula:
[tex]E=h\frac{c}{\lambda}[/tex] (2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
[tex]2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J[/tex]
Next, you use the equation (2) and solve for λ:
[tex]\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm[/tex]
C. The radius of the orbit is given by:
[tex]r_n=n^2a_o[/tex] (3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
[tex]r_5=5^2(2.380)=59.5[/tex]
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
A) The energy E of the emitted photon in electron volts is; E = 2.856 eV
B) The wavelength in nanometers of the emitted photon is; λ = 434.4nm
C) The radius of the hydrogen atom in nanometers in its initial 5th energy level is; rₙ = 1.323 nm
A) Formula for the energy E of the emitted photons is;
E = -13.6([tex]\frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}[/tex])
We are given;
n₂ = 5
n₁ = 2
Thus;
E = -13.6([tex]\frac{1}{5^2} - \frac{1}{2^2}[/tex])
E = 2.856 eV
B) The formula for the wavelength is;
λ = hc/E
where;
h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s
c is speed of light = 3 × 10⁸ m/s
E is energy of photon
λ is wavelength of the photon
Earlier we saw that E = 2.856 eV. Converting to Joules gives;
E = 4.5758 × 10⁻¹⁹ J
Thus;
λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4.5758 × 10⁻¹⁹)
λ = 4.344 × 10⁻⁷ m
Converting to nm gives;
λ = 434.4nm
C) Formula for the radius of the hydrogen atom is;
rₙ = n²a₀
where;
a₀ is bohr's radius = 5.292 × 10⁻¹¹ m
n = 5
Thus;
rₙ = 5² × 5.292 × 10⁻¹¹
rₙ = 1.323 × 10⁻⁹
rₙ = 1.323 nm
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Chapter 24, Problem 20 GO A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.9 m from his television set. A reporter at the press conference is located 4.1 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Therefore, the distance between politician and TV set is 2536kmExplanation:
Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.
The reporter hears the sound is
4.1 / 343 = 0.01195 s later
The viewer hears the sound from the TV is
2.9 / 343 = 0.00845s
the difference is 0.00845 sec
the question is how far the TV signal can travel in that time.
the distance between politician and TV set is
= 0.00845 * 3*10^8 m
= 2536 km
d = 2536km
Therefore, the distance between politician and TV set is 2536kmcalculate the volume of marble if its diameter is 10mm
Answer:
The volume of the marble is [tex]523.33\ mm^2[/tex].
Explanation:
Marble is spherical in shape. The diameter of marble is 10 mm. It radius will be 5 mm.
The volume of spherical shaped object is given by :
[tex]V=\dfrac{4}{3}\pi r^3[/tex]
Plugging all the values, we get :
[tex]V=\dfrac{4}{3}\times 3.14\times (5)^3\\\\V=523.33\ mm^2[/tex]
So, the volume of the marble is [tex]523.33\ mm^2[/tex].
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 70 kg and the truck exerts a constant force of 20 N. How fast will you be going after 15 seconds, in m/s and MPH?
Explanation:
It is given that,
Total mass is 70 kg
The truck exerts a constant force of 20 N.
Then the net force is given by :
F = ma
a is acceleration of rider
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]
Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :
[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]
Since, 1 m/s = 2.23 mph
4.28 m/s = 9.57 mph
So, the speed of the rider is 4.28 m/s or 9.57 mph.
The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magnitude of the magnetic field at twice the distance from the conductor
Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday's Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.
Answer:
The average emf induce is [tex]V = 2.625 * 10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The radius of the coil is [tex]r = 0.30 \ m[/tex]
The number of turns is [tex]N = 420 \ turns[/tex]
The frequency of the transition radio wave is [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]
The magnetic field is [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]
The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]
The period of the transmitted radio wave is [tex]T = \frac{1}{f}[/tex]
So
[tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]
The potential difference can be mathematically represented as
[tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]
[tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]
Where [tex]B_{min} = 0T[/tex]
substituting values
[tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]
[tex]V = 2.625 * 10^{-5} \ V[/tex]
Countries create quotas and tariffs to increase the volume of trade with their neighbors.
Oooooo, that statement is not true. Countries create quotas and tariffs to LIMIT the volume of trade with other countries, including their neighbors.
Answer:
False
Explanation:
I took the text :)
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.
- 2x
- 1/4
- 1/2
- 4x
Answer:
The correct option is;
- 4x
Explanation:
From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source
The inverse square law can be presented as follows;
[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]
As the distance, r, increases, the surface it covers also increases by the power of 2
Therefore, where the distance increases from r to 2·r, we have;
When, I, remain constant
[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]
The surface increases to 4·S by the inverse square law
Therefore, the correct option is 4 × x.
Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new pipe which has a diameter half that of the first section of pipe. Determine the speed and pressure of the water in the new, reduced in size pipe.
Answer:
The velocity is [tex]v_2 = 6.8 \ m/s[/tex]
The pressure is [tex]P_2 = 204978 Pa[/tex]
Explanation:
From the question we are told that
The speed at which water is travelling through is [tex]v = 1.7 \ m/s[/tex]
The pressure is [tex]P_1 = 205 k Pa = 205 *10^{3} \ Pa[/tex]
The diameter of the new pipe is [tex]d = \frac{D}{2}[/tex]
Where D is the diameter of first pipe
According to the principal of continuity we have that
[tex]A_1 v_1 = A_2 v_2[/tex]
Now [tex]A_1[/tex] is the area of the first pipe which is mathematically represented as
[tex]A_1 = \pi \frac{D^2}{4}[/tex]
and [tex]A_2[/tex] is the area of the second pipe which is mathematically represented as
[tex]A_2 = \pi \frac{d^2}{4}[/tex]
Recall [tex]d = \frac{D}{2}[/tex]
[tex]A_2 = \pi \frac{[ D^2]}{4 *4}[/tex]
[tex]A_2 = \frac{A_1}{4}[/tex]
So [tex]A_1 v_1 = \frac{A_1}{4} v_2[/tex]
substituting value
[tex]1.7 = \frac{1}{4} * v_2[/tex]
[tex]v_2 = 4 * 1.7[/tex]
[tex]v_2 = 6.8 \ m/s[/tex]
According to Bernoulli's equation we have that
[tex]P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}[/tex]
substituting values
[tex]205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}[/tex]
[tex]P_2 = 204978 Pa[/tex]
1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
Explanation:
It is given that,
Mass of a car is 900 kg
Radius of curve is 600 m
Speed of the car in the curve is 25 m/s
We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]
So, the centripetal force exerted on a car is 937.5 N.
Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.