Answer:
✔️From Newton's second law of motion we know that
✔️acceleration is inversely proportional to the mass of the body
✔️Here in this question increasing the mass of body will decrease its acceleration.
Explanation:
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Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side of firecracker A. You see two flashes of light, from the two explosions, at exactly the same instant of time.
Define event 1 to be "firecracker 1 explodes" and event 2 to be "firecracker 2 explodes." According to your lab partner, based on measurements he or she makes, does event 1 occur before, after, or at the same time as event 2? Explain.
Answer:
See the explanation
Explanation:
Given:
Distance of Firecrackers A and B = 600 m
Event 1 = firecracker 1 explodes
Event 2 = firecracker 2 explodes
Distance of lab partner from cracker A = 300 m
You observe the explosions at the same time
to find:
does event 1 occur before, after, or at the same time as event 2?
Solution:
Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart
So the distance of fire cracker B from the lab partner is:
600 m + 300 m = 900 m
It takes longer for the light from the more distant firecracker to reach so
Let T1 represents the time taken for light from firecracker A to reach lab partner
T1 = 300/c
It is 300 because lab partner is 300 m on other side of firecracker A
Let T2 represents the time taken for light from firecracker B to reach lab partner
T2 = 900/c
It is 900 because lab partner is 900 m on other side of firecracker B
T2 = T1
900 = 300
900 = 3(300)
T2 = 3(T1)
Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.
Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes
So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double slits 0.500 mm apart illuminated by 500-nm light.
Answer:
y = 1.75 cm
Explanation:
In the double-slit experiment the equation for destructive interference is
d sin tea = (m + ½)
λ
let's use trigonometry to find the angle
tan θ = y / L
as all the experiment does not occur at small angles
tan θ = sin θ / cos θ = sin θ = y / L
we substitute
y = (m + 1/2 ) λ L / d
we calculate
y = (3 + ½) 500 10⁻⁹ 5.00 / 0.5 10⁻³
y = 1.75 10⁻² m
y = 1.75 cm
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?
Answer:
La resistencia que debe ser conectada en paralelo con una de 20 Ω para hacer una resistencia combinada de 15 Ω tiene un valor de 60 Ω
Explanation:
Las resistencias son aquellos dispositivos en los circuitos eléctricos que suelen emplearse para oponerse al paso de la corriente eléctrica.
Se denomina resistencia resultante o equivalente al valor de la resistencia que se obtiene al considerar un conjunto de ellas.
Cuando tenes resistencias en paralelo la corriente se divide y circula por varios caminos.
La resistencia equivalente de un circuito de resistencias en paralelo es igual al recíproco de la suma de los inversos de las resistencias individuales:
[tex]R=\frac{1}{\frac{1}{R_{1} } +\frac{1}{R_{2} } +...+\frac{1}{R_{N} }}[/tex]
Esto también puede ser expresado como:
[tex]\frac{1}{R} =\frac{1}{R_{1} } +\frac{1}{R_{2} } +...+\frac{1}{R_{N} }[/tex]
Entonces, en este caso sabes:
R= 15 ΩR1= 20 ΩR2=?Reemplazando:
[tex]\frac{1}{15} =\frac{1}{20}+\frac{1}{R2}[/tex]
y resolviendo:
[tex]\frac{1}{R2} =\frac{1}{15} -\frac{1}{20}[/tex]
[tex]\frac{1}{R2} =\frac{1}{60}[/tex]
se obtiene:
R2=60 Ω
La resistencia que debe ser conectada en paralelo con una de 20 Ω para hacer una resistencia combinada de 15 Ω tiene un valor de 60 Ω
Represent a vector of 100 N in North-East direction
Answer:
please find the attachment to this question.
Explanation:
In this question, we represent the 100N in the North-East direction, but first, we define the vector representation:
It is generally represented through arrows, whose length and direction reflect the magnitude and direction of the arrow points. In this, both size and direction are necessary because the magnitude of a vector would be a number that can be compared to one vector.
Please find the attachment:
In a region of space where the magnetic field of the earth has a magnitude of 80 μT and is directed 30° below the horizontal, a 50-cm length of wire oriented horizontally along an east-west direction is moved horizontally to the south with a speed of 20 m/s. What is the magnitude of the induced potential difference between the ends of this wire?
Answer:
V_ind = 4 × 10^(-4) V
Explanation:
We are given;
Magnetic field; B = 80 μT = 8 × 10^(-5) T
Angle;θ = 30°
Lenght;L = 50 cm = 0.5 m
Speed; v = 20 m/s
Now, formula for the induced potential difference is known as;
V_ind = NBLVsin θ
Where;
V_Ind = Induced potential difference/voltage
N = Number of turns
B = Magnetic field
V = velocity
L = length
Number of turns in this case is 1 since it's just a wire between both ends.
Thus, plugging in the relevant values, we have;
V_ind = 1 × 8 × 10^(-5) × 20 × 0.5 × sin 30
V_ind = 4 × 10^(-4) V
A particle with charge -5 C initially moves at v = (1.00 i^ + 7.00 j^ ) m/s. If it encounters a magnetic field B =80 Tkˆ, find the magnetic force vector on the particle.
Answer:
The magnetic force is [tex]\= F = 400\r j + 2800\r i[/tex]
Explanation:
From the question we are told that
The charge is [tex]q = -5C[/tex]
The velocity is [tex]v = (1.00\ \r i + 7.00 \ \r j )\ m/s[/tex]
The magnetic field is [tex]B = 80 \r k \ \ T[/tex]
Generally the magnetic force is mathematically represented as
[tex]\= F = q \= v \ \ X \ \ \= B[/tex]
=> [tex]\= F = -5 (1.0 \r i + 7.0 \r j ) \ \ X \ \ 80 \r k[/tex]
=> [tex]\= F = -5.0 \r i + 35\r j \ \ \ X \ \ 80\r k[/tex]
=> [tex]\= F = 400\r j + 2800\r i[/tex] N/B - Applied cross - product of unit vector
Two gratings A and B have slit separations dA and dB, respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of A is exactly replaced by the second-order maximum of B.
a.) I already found that the ratio of db/da is 2
b.) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers
Answer:
mA=2,mB=4
mA=2,mB=4 and
mA=3,mB=6
Explanation:
First of all we need to write the equation of the networks
sin θ = mA λ / dA
sin θ = mB λ / dB
Equating we have
mA λ/ dA = mB λ / dB
We are given the ratio as
dB / dA = 2
So
mA 2 = mB
Finally overlapping orders
We have
mA=2,mB=4
mA=2,mB=4
and mA=3,mB=6
The force of gravity will make it easier to stop your car if you are going uphill, and more difficult to stop your car if you are going downhill.
Which statement best describes the liquid state of matter?
ОА.
It has definite shape but indefinite volume.
OB.
It has definite shape and definite volume.
Ос.
It has indefinite shape and indefinite volume.
OD.
It has indefinite shape but definite volume.
Answer:
OB.It has definite shape and definite volume
An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volume doubles.
Required:
What is the gas’s final pressure, in atmospheres, if the gas is diatomic?
Answer:
The pressure is [tex]P_2 = 4.25 \ a.t.m[/tex]
Explanation:
From the question we are told that
The initial pressure is [tex]P_1 = 11.2\ a.t.m[/tex]
The temperature is [tex]T_1 = 299 \ K[/tex]
Let the first volume be [tex]V_1[/tex] Then the final volume will be [tex]2 V_1[/tex]
Generally for a diatomic gas
[tex]P_1 V_1 ^r = P_2 V_2 ^r[/tex]
Here r is the radius of the molecules which is mathematically represented as
[tex]r = \frac{C_p}{C_v}[/tex]
Where [tex]C_p \ and\ C_v[/tex] are the molar specific heat of a gas at constant pressure and the molar specific heat of a gas at constant volume with values
[tex]C_p=7 \ and\ C_v=5[/tex]
=> [tex]r = \frac{7}{5}[/tex]
=> [tex]11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )[/tex]
=> [tex]P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2[/tex]
=> [tex]P_2 = 4.25 \ a.t.m[/tex]
The final pressure of the gas will be 4.244 atm.
Given information:
The initial pressure of the gas is [tex]P_1=11.2\rm\;atm[/tex].
The initial temperature of the gas is [tex]T_1=299\rm\; K[/tex].
Let the initial volume of the gas be V. So, the final volume will be double or 2V.
The given diatomic gas is expanded adiabatically. So, the equation of the adiabatic process will be,
[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}[/tex]
where [tex]\gamma[/tex] is the ratio of specific heats of the gas which is equal to 1.4 for a diatomic gas.
So, the final pressure [tex]P_2[/tex] can be calculated as,
[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}\\11.2\times V^{1.4}=P_2\times (2V)^{1.4}\\P_2=4.244\rm\;atm[/tex]
Therefore, the final pressure of the gas will be 4.244 atm.
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Why is area a scalar quantity?
Answer:
Area is a scalar quantity because there is no need of direction to define and also follow the algebraic summation. When we talk about vector there exists a frame of reference with a certain origin. It actually depends on the fact of the physical area, but if that factor changes to a non-directional object such as a rug spread on the floor, you can consider the area or a region as a scalar. (*Scalars are quantities that are fully described by a magnitude (or numerical value) alone.)
Hope this helps!
An object floats in water with 58 of its volume submerged. The ratio of the density of the object to that of water is
Complete Question
An object floats in water with 5/8 of its volume submerged. The ratio of the density of the object to that of water is:
(a) 8/5
(b) 1/2
(c) 3/8
(d) 5/8
(e) 2/1
Answer:
The correct option is d
Explanation:
From the question we are told that
The ratio of the volume of the object submerged to the total volume of the object is [tex]\frac{V_w}{V_o} = \frac{5}{8}[/tex]
Generally the buoyancy force acting on the object is equal to the weight of the water displaced and this is mathematically represented as
[tex]F_b = W[/tex]
Now the mass of the water displaced is mathematically represented as
[tex]m_w = \rho_w * V_w[/tex]
While the mass of the object is mathematically represented as
[tex]m_o = \rho_o * V_o[/tex]
So
[tex]F_b = W \ \equiv \ \rho * V_o * g = \rho * V_w * g[/tex]
=> [tex]\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}[/tex]
From the question that it volume of the water displace (equivalent to the volume of the object in water ) to the volume of the total object is
[tex]\frac{V_w}{V_o} = \frac{5}{8}[/tex]
So
[tex]\frac{\rho_o}{\rho_w} = \frac{5}{8}[/tex]
A source of emf is connected by wires to a resistor, and electrons flow in the circuit. The wire diameter is the same throughout the circuit. Compared to the potential energy of an electron before entering the source of emf, the potential energy of an electron after leaving the source of emf is
Answer
The potential energy is less
Explanation:
From the question we are told that
The source of the emf is by wires to a resistor.
Now the potential energy of electron before leaving the source emf will be greater than the potential energy of an electron after leaving the source of emf because the resistor connected to the source emf will reduced the potential energy as it will convert some of the energy to heat
The potential energy of an electron after leaving the source of emf is lesser.
What is electro motive force?Electro motive force is the voltage or potential difference of an electrical energy device such as battery.
The source of the emf is by wires to a resistor.Now, the potential energy of an electron before leaving the source emf will be greater than the potential energy of an electron.
After leaving the source of emf because the resistor connected to the source of emf will reduce the potential energy as it will convert some of the energy to heat.
Thus the potential energy of an electron after leaving the source of emf is lesser.
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A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875 Hz. a) what is the max speed of that point in SI units? b) what is the max acceleration of the point in SI units?
Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
Answer:
a
[tex]v _{max } = 6.82 \ m/s[/tex]
b
[tex]a_{max} = 37489.5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.24 \ mm = 1.24 * 10^{-3} \ m[/tex]
The frequency is [tex]f = 875 \ Hz[/tex]
Generally the maximum speed is mathematically represented as
[tex]v _{max } = A * 2 * \pi * f[/tex]
=> [tex]v _{max } = 1.24*10^{-3} * 2 * 3.142 * 875[/tex]
=> [tex]v _{max } = 6.82 \ m/s[/tex]
Generally the maximum acceleration is mathematically represented as
[tex]a_{max} = A * (2 * \pi * f)[/tex]
=> [tex]a_{max} = 1.24*10^{-3} * (2 * 3.142 * 875 )^2[/tex]
=> [tex]a_{max} = 37489.5 \ m/s^2[/tex]
a torque of 6Nm is required to accelerate a wheel from rest to 7.5rev/s In a time of 5.0s.find the moment of inertia of the wheel
Your question has been heard loud and clear
Torque = 6Nm
Angular acceleration= 7.5 revolutions / second
Moment of inertia = ?
Torque = Moment of inertia * Angular acceleration
Therefore , moment of intertia = Torque / Angular acceleration
Moment of inertia = 6 / 7.5 = 0.8 Kgm^2
Moment of inertia of wheel = 0.8 Kgm^2
Thankyou
The allowed energies of a simple atom are 0.0 eV, 4.0 eV, and 6.0 eV. Part A What wavelength(s) appear(s) in the atom's emission spectrum
Answer:
3.1 × 10^- 7 m and 2.1 × 10^-7 m
Explanation:
First we must convert each value of energy to Joules by multiplying its value by 1.6 ×10^-19. After that, we can now obtain the wavelength from E= hc/λ
Where;
h= planks constant
c= speed of light
λ= wavelength of light
For 6.0ev;
E= 6.0 × 1.6 ×10^-19
E= 9.6 × 10^-19 J
From
E= hc/λ
λ= hc/E
λ= 6.6 × 10^-34 × 3 × 10^8/9.6 × 10^-19
λ= 2.1 × 10^-7 m
For 4.0 eV
4.0 × 1.6 × 10^-19 = 6.4 × 10^-19 J
E= hc/λ
λ= hc/E
λ= 6.6 × 10^-34 × 3 × 10^8/6.4 × 10^-19
λ= 3.1 × 10^- 7 m
(a) The wavelength of the atom's emission spectrum when the energy is 4 eV is [tex]3.1 \times 10^{-7} \ m[/tex]
(b) The wavelength of the atom's emission spectrum when the energy is 6 eV is
[tex]2.1 \times 10^{-7} \ m[/tex]
The wavelength of the atom's emission spectrum is calculated as follows;
[tex]E = hf\\\\E = \frac{hc}{\lambda}[/tex]
where;
λ is the wavelengthh is Planck's constantFor 4 eV;
[tex]\lambda = \frac{hc}{E} \\\\\lambda = \frac{(6.626 \times 10^{-34}) \times 3\times 10^8}{4 \times 1.602 \times 10^{-19}} \\\\\lambda = 3.1 \times 10^{-7} \ m[/tex]
For 6 eV;
[tex]\lambda = \frac{hc}{E} \\\\\lambda = \frac{(6.626 \times 10^{-34}) \times 3\times 10^8}{6 \times 1.602 \times 10^{-19}} \\\\\lambda = 2.1 \times 10^{-7} \ m[/tex]
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Zoning laws establish _______.
Answer:
Zoning ordinances detail whether specific geographic zones are acceptable for residential or commercial purposes. Zoning ordinances may also regulate: - size
- placement
- density
- height of structures
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PLEASE MARK IT THE BRAINLIEST!
Answer:
B
Explanation:
B on edg.
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object
Answer:
a). C = b/2 and C = b/4
b). [tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
c). m = 63.4 kg (approx.)
Explanation:
Ex. 2.4.4
The total force acting on mass m is [tex]$ F = F_{spring }= -kx $[/tex] , where x is the displacement from the equilibrium position.
The equation of motion is [tex]$ m {\overset{..}x} + kx = 0 $[/tex]
or [tex]$ {\overset{..}x}+ \frac{k}{m}x=0 $[/tex] or [tex]$ {\overset{..}x} + w_0^2 x = 0 $[/tex] , where [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex]
The solution is [tex]$ x = A \cos (w_0t + \phi) $[/tex] , where A and Ф are constants.
A is amplitude of motion
[tex]$ w_0$[/tex] is the angular frequency of motion
Ф is the phase angle.
Now, [tex]$ w_0 = 2 \pi f_0 = \sqrt{k/m} $[/tex]
or [tex]$ m = \frac{k}{4\pi f_0^2} $[/tex]
Given [tex]$ f_0 = 0.8 Hz , k = 4 N/m $[/tex]
a). [tex]$ m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$[/tex]
b). [tex]$ w_0^2 = k/m $[/tex]
or [tex]$ m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2 $[/tex]
Ex. 2.4.5
a). Total force acting on the mass m is [tex]$F = F_{spring}+f $[/tex]
[tex]$ = -kx-bv $[/tex]
The equation of motion is [tex]$ m {\overset{..}x}= -kx-b{\overset{.}x} $[/tex]
or [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex] , angular frequency of the undamped oscillation.
γ = b/2m is called the damping coefficient (γ=C)
[tex]$ k = m w_0^2 = 4 \pi^2 m f_0^2 $[/tex]
for 1 kg weight (= 9.8 N), [tex]$ f_0$[/tex] = 1.1 Hz
k = 4 x (3.14)² x (9.8) x 1.1² = 4.6 x 10² N/m
For 2 kg weight (= 19.6 N), [tex]$ f_0$[/tex] = 0.8 Hz
k = 4 x 9.8596 x 2 x 9.8 x 0.8² = 5 x [tex]$ 10^7$[/tex] N/m
[tex]$ \gamma = \frac{b}{2m_1} = \frac{b}{2m_2} $[/tex]
or [tex]$ \gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2} $[/tex]
γ = b/2 (for 1 kg) and γ = b/4 (for 2 kg)
C = b/2 and C = b/4
b). [tex]$ w_0^2 = \frac{k}{m} \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2} $[/tex]
For two particle problem,
[tex]$ w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}} $[/tex]
[tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
where, μ is the reduced mass.
This time period is same for both the particles.
c). [tex]$ m =\frac{k}{4 \pi^2 f_0^2}$[/tex]
[tex]$ = \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2} = 63.4\ kg $[/tex] ( approx.)
The howler monkey is the loudest land animal and can be heard up to a distance of 2.5 km. Assume the acoustic output of a howler to be uniform in all directions. At 2.5 km away from the monkey, what would be the intensity of the sound
Answer:
10⁻¹² W / m²
Explanation:
The feeble sound that a man can hear is of the vale which measures 0 on decibel scale . The intensity of sound in terms of J / m² .s is 10⁻¹² W / m² .
So the intensity of sound of monkey at 2.5 km must be 10⁻¹² W / m² .
Parallel rays of monochromatic light with wavelength 582 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 4.40×10^−4 W/m^2.
Required:
What is the intensity at a point on the screen that is 0.710 mm from the center of the central maximum?
Answer:
What is the intensity is 1.3349 × 10⁻⁷ w/m²
Explanation:
Given that;
λ = 582 nm = 582 × 10⁻⁹
R = 75.0 cm = 0.75 m
d = 0.640 mm = 0.000640 m
a = 0.434 mm = 0.000434 m
I₀ = 4.40×10⁻⁴ W/m²
y = 0.710 mm = 0.00071 m
Now to get our tanФ we say
tanФ = y/R = 0.00071 / 0.75 = 0.0009466
Ф is so small
∴ tanФ ≈ sinФ
So
∅ = 2πdsinФ / λ
we substitute
∅ = ( 2π × 0.000640 × 0.0009466 ) / 582 × 10⁻⁹
= 6.54 rad
Now
β = 2πasinФ / λ
we substitute
β = ( 2π × 0.000434 × 0.0009466 ) / 582 × 10⁻⁹
β = 4.435 rad
I = I₀ cos²(∅/2) [(sin(β/2))/(β/2)]²
we substitute
I = 4.40×10⁻⁴ cos(3.27)² [ (sin(2.2175)) / (2.2175) ]²
= 4.40×10⁻⁴ × 0.9967 × 0.0003044
= 1.3349 × 10⁻⁷ w/m²
A vertical scale on a spring balance reads from 0 to 220 N. The scale has a length of 15.0 cm from the 0 to 220 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.70 Hz. Ignoring the mass of the spring, what is the mass m of the fish?
Mass of fish is 5.09Kg
Explanation:
First to find the spring constant K using
k = F/s
= 220/0.15 = 1466.7 N/m
So using the formula
T = 2π√(m/k)
f = 1/T = 1/2πx √(k/m)
f² x 4π²= k/m
So
m = k/(f² x π²)
m = 1466.7/(2.7² x 4π²)
m = 5.09 kg
A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concerning this process is correct? (There may be more than one correct choice.)
A. The entropy gained by the iron is equal to the entropy lost by the ocean.
B. The ocean gains less entropy than the iron loses.
C. The change in the entropy of the iron-ocean system is zero.
D. The entropy lost by the iron is equal to the entropy gained by the ocean.
E. The ocean gains more entropy than the iron loses.
Answer:
E. The ocean gains more entropy than the iron loses.
Explanation:
When there is a spontaneous process , entropy of the system increases . Here hot iron is losing entropy and ocean is gaining entropy . Net effect will be gain of entropy . That means entropy gained by ocean is more than entropy lost by iron .
Hence option E is correct .
An airplane is traveling at 400 mi/h. It touches down at an airport 2000 miles away. How long was the airplane airborne?
Answer:
5 hours
Explanation:
Given that
Speed of the airplane, v = 400 mile/hr
Distance of the airport, s = 2000 miles
This is quite a straightforward question that deals with one of the basic formulas in physics.
Speed.
speed is said to the the ratio of distance covered with respect to the time taken. This can be mathematically expressed as
Speed, v = distance covered, d / time taken, t
v = d / t
In the question above, we're looking for the time taken. So, so make t, subject of formula.
t = d / v, now we proceed to substituting the earlier given values into this equation.
t = 2000 / 400
t = 5 hrs,
therefore we can conclude that the airplane was airborne for 5 hours
Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure to label each axis and note the units used in the measurements. Be sure to draw a smooth curve through the points. Do not just connect the dots. Upload your data and graph in the essay box below and answer the following questions. Did the car travel at a constant speed? What was the average speed of the car? What are some practical applications for determining the motion of an object?
Answer:
yes it was a constant speed and the car traveled 10 meters in 20 seconds.
Explanation:
Answer:
It's a constant speed and the car traveled 10 meters in 20 seconds.
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Explanation:
The non reflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 636-nm light. If the lens glass has an index of refraction of 1.56, what is the minimum thickness of the coating that will accomplish this task
Answer:
The minimum thickness is [tex]t = 1.0192 *10^{-7} \ m[/tex]
Explanation:
From the question we are told that
The refractive index is [tex]n = 1.29[/tex]
The wavelength of the light is [tex]\lambda = 636 \ nm = 636 *10^{-9} \ m[/tex]
The refractive index of the glass lens is [tex]n_g = 1.56[/tex]
Generally the condition for destructive interference of a films is
[tex]2t = [ m + \frac{1}{2} ] \frac{\lambda}{n}[/tex]
for minimum m = 0
[tex]2t = [ 0 + \frac{1}{2} ] \frac{\lambda}{n}[/tex]
=> [tex]2t = [ 0 + \frac{1}{2} ] \frac{636 *10^{-9}}{1.56}[/tex]
=> [tex]t = 1.0192 *10^{-7} \ m[/tex]
The 2-kg collar is attached to a spring that has an un-stretched length of 3.0 m. If the collar is drawn to point B and releases from rest, what is the speed when it arrives at point A. Note that k = 3.0 N/m and neglect friction.
Complete Question
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Answer:
[tex]v = 3.4 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the collar is [tex]m = 2 \ kg[/tex]
The original length is [tex]L = 3.0 \ m[/tex]
The spring constant is [tex]k = 3.0 \ N/m[/tex]
Generally the extension of the spring is mathematically evaluated as
[tex]e = 4 -3 = 1 \ m[/tex]
Now with Pythagoras theorem we can obtain the length from A to B as
[tex]AB = \sqrt{5 ^2 + 4^2}[/tex]
[tex]AB = 6.4 \ m[/tex]
The extension of the spring at B is
[tex]e_b = 6.4 - 3 = 3.4 \ m[/tex]
According to the law of energy conservation
The energy stored in the spring at point A + the kinetic energy of the spring = The energy stored on the spring at B
So
[tex]\frac{1}{2} * k * e + \frac{1}{2} * m* v^2 = \frac{1}{2} * k * e_b[/tex]
substituting values
[tex]\frac{1}{2} * 3 * 1^2 + \frac{1}{2} * 2* v^2 = \frac{1}{2} * 3 * 3.4^2[/tex]
=> [tex]v = 3.4 \ m/s[/tex]
Sodium has a work function of 2.46 eV.
(a) Find the cutoff wavelength and cutoff frequency for the photoelectric effect.
(b) What is the stopping potential if the incident light has a wavelength of 181 nm?
Answer:
Explanation:
given, work function of Φ = 2.46 eV.
converting the eV to joule, we have
2.45 * 1.6*10^-19 J
The cutoff wavelength is the wavelength where the incoming light does not have enough energy to free an electron, i.e. all of
the photon’s energy will be channeled into trying overcoming the work function barrier.
It is mathematically given as
Φ = hf
f = Φ/h
f = (2.46 * 1.6*10^-19) / 6.63*10^-34
f = 3.936*10^-19 / 6.63*10^-34
f = 5.94*10^14 Hz as our cut off frequency
λf = c,
λ = c/f
λ = 3*10^8 / 5.94*10^14
λ = 5.05*10^-7
λ = 505 nm as our cut off wavelength
K(max) = hf - Φ
K(max) = hc/λ - Φ
K(max) = [(6.63*10^-34 * 3*10^8) / 181*10^-9] - 3.936*10^-19
K(max) = (1.989*10^-25/181*10^-9) - 3.936*10^-19
K(max) = 1.1*10^-18 - 3.936*10^-19
K(max) = 7.064*10^-19 J or 4.415 eV
V(s) = K(max) / e
V(s) = 4.612 V
a father and his son want to play on a seesaw. where on the seesaw should each of them sit to balance the torque
A father and his son want to play on a seesaw. Due to his larger size than that of the son, the father should outweigh the boy on the opposing sides.
What is seesaw?A seesaw is a long, narrow board with a single pivot point, which is often situated in the middle of both ends. As one end rises, the other falls.
What is outweigh?Rugby continues to have far more health advantages than hazards.
A father and his son want to play on a seesaw. Due to his larger size than that of the son, the father should outweigh the boy on the opposing sides.
To know more about seesaw and outweigh
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20 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What is the ratio of the diameter of a single copper wire to d, if its length is also l and it has the same resistance?
Please show work and type out answer.
Answer:
D = 4.47d
Explanation:
given that
1/R(eq) = 20/R
R(eq) = R/20
also, we know that the formula for resistance is given by the relation
R = pl/A, where A is πd²/4
If we substitute the value of A, we have
R = pl/(πd²/4)
R = 4pl/πd²
Now, we substitute this in the earlier derived equation
R(eq) = (4pl/πd²) / 20
R(eq) = pl/5πd²
To find the resistance of a single wire made of the same material, the resistance is
R(D) = 4pl / πD²
R(eq) = R(D), and thus
pl/5πd² = 4pl/πD²
1/5d² = 4/D²
D² = 20d²
D = √20d²
D = 4.47 d
Consider two different isotopes of the same neutral element. Which statements about these isotopes are true?
a. Both isotopes contain the same number of protons.
b. Both isotopes contain the same number of nucleons.
c. isotopes contain the same number of neutrons.
d. Both isotopes contain the same number of orbital electrons.
d. The sum of the protons and neutrons is the same for both isotopes.
Answer:
a. d.
Explanation:
isotopes have a diff number of neutrons