Answer:
Valid
Explanation:
to determine if the claim is valid, we compare the efficiency of the device to that of a Carnot engine.
The following data were given
High Temperature = 920R,
Low temperature =490R
work=4.5hp =4.5*2544.5=11450.25Btu/h
low heat Ql heat= 15000Btu/h
High heat Qh=work +Ql=11450.25Btu/h+15000Btu/h=26450.25Btu/h
Next we cal calculate the efficiency of the Carnot engine
[tex]E_Carnot=1-\frac{T_L}{T_H}\\ E_Carnot=1-\frac{490}{920}\\ E_Carnot=0.467[/tex]
Hence the maximum efficiency at the given temperature is 47%
Next we calculate the efficiency of the device
[tex]E_device=\frac{work}{Q_H} \\E_device=\frac{11450.25}{26450.25} \\E_device=0.433\\[/tex]
which is 43%
since the maximum efficiency of 47% is not exceeded, we can conclude that the claim is valid
. A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b) What is the RC time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant?
Answer:
initial current I₀ = 0.0123 A
RC time constant τ = 0.00075 sec
current after one time constant I = 0.00452 A
voltage on the capacitor after one time constant V = 3.89 V
Explanation:
Given that,
Voltage = 6.16 V
Resistance = 500 Ω
Capacitance = 1.5 µF
(a) What is the initial current?
The initial current can be found using
I₀ = Voltage/Resistance
I₀ = 6.16/500
I₀ = 0.0123 A
(b) What is the RC time constant?
The time constant τ provides the information about how long it will take to charge the capacitor.
τ = R*C
τ = 500*1.5x10⁻⁶
τ = 0.00075 sec
(c) What is the current after one time constant?
I = I₀e^(-τ/RC)
I = 0.0123*e^(-1) (0.00075/0.00075 = 1)
I = 0.00452 A
(d) What is the voltage on the capacitor after one time constant?
V = V₀(1 - e^(-τ/RC))
Where V₀ is the initial voltage 6.16 V
V = 6.16(1 - e^(-1))
V = 6.16*0.63212
V = 3.89 V
That means the capacitor will charge up to 3.89 V in one time constant
Answer:
Explanation:
Given an RC circuit to analyze
R=500Ω
C=1.50-μF uncharged
Emf(V)=6.16V
Series connection
a. Initial current, since the capacitor is initially uncharged then, the voltage appears at the resistor
Using ohms law
V=iR
Then, i=V/R
i=6.16/500
i=0.01232 Amps
i=12.32 mA.
b. The time constant is given as
τ=RC
τ=500×1.5×10^-6
τ=0.00075second
τ=0.75 ms
c. What is current after 1 time constant
Current in a series RC circuit is given as
Time after I time constant is
t=1 ×τ
t= τ
i=V/R exp(-t/RC)
Where RC= τ
i=V/Rexp(-t/ τ)
i=6.16/500exp(-1), since t= τ
i=0.004532A
I=4.532mA
d. Voltage after one time constant
Voltage of a series RC circuit(charging) is given as
Again, t= τ
V=Vo(1 - exp(-t/ τ)),
V=6.16(1-exp(-1))
V=6.16(1-0.3679)
V=6.16×0.632
V=3.89Volts
V=3.89V
V=
You have a device that takes temperature measurements and runs off of solar power. How often it is programmed to take a measurement will affect how much power it uses--more frequent measurements, more power. You have one installed at the equator and one installed in the Antarctic. During which season of the year can you set each device to take more frequent measurements
Answer:
In the Equator:
As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.
In Antarctic:
As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.
Explanation:
Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.55.
To find the force necessary to start the crate moving, we use the formula fs(max) = μsN, where fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force. Plugging in the given values, the force necessary to start the crate moving is 169.36 N.
Explanation:The force necessary to start the crate moving is equal to the maximum static friction force. To calculate this, we use the formula:
fs(max) = μsN
where fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force. The normal force is equal to the weight of the crate, which can be calculated by N = mg, where m is the mass of the crate and g is the acceleration due to gravity. So, the force necessary to start the crate moving is:
fs(max) = μsN = μsmg
Plugging in the given values:
fs(max) = (0.55)(32 kg)(9.8 m/s²) = 169.36 N
A very thin 19.0 cm copper bar is aligned horizontally along the east-west direction. If it moves horizontally from south to north at velocity = 11.0 m/s in a vertically upward magnetic field and B = 1.18 T , what potential difference is induced across its ends ? which end (east or west) is at a higher potential ? a) East b) West
Answer:
2.47 V,East
Explanation:
We are given that
l=19 cm=[tex]19\times 10^{-2} m[/tex]
[tex] 1 cm=10^{-2} m[/tex]
[tex]v=11 m/s[/tex]
B=1.18 T
We have to find the potential difference induced across its ends.
[tex]E=Bvl[/tex]
Using the formula
[tex]E=1.18\times 11\times 19\times 10^{-2}[/tex]
[tex]E=2.47 V[/tex]
Hence, the potential difference induces across its ends=2.47 V
The positive charge will move towards east direction and the negative charge will move towards west direction because the direction of force will be east.Therefore, the potential at east end will be high.
The induced potential difference in the copper bar is 2.47 V, with the east end being at a higher potential.
To determine the potential difference induced across the ends of a copper bar moving through a magnetic field, we use the formula:
V = B * l * v
where:
B is the magnetic field strength (1.18 T).l is the length of the bar (0.19 m).v is the velocity (11.0 m/s).Substituting the given values:
V = 1.18 T * 0.19 m * 11.0 m/s = 2.47 V
The potential difference across the ends of the bar is 2.47 V.
To determine which end is at a higher potential, we apply the right-hand rule. Pointing the thumb of your right hand in the direction of the velocity (north), and your fingers in the direction of the magnetic field (upwards), the palm points towards the force acting on positive charges (from west to east).
Therefore, the east end is at a higher potential.
The correct answer is: a) East
A U-shaped tube open to the air at both ends contains water. A quantity of oil of unknown density is slowly poured into the right arm of the tube until the vertical height of the oil column is 20cm. The top of the oil is 8cm higher than the top of the water. Find the density of the oil.
Answer:
[tex]\rho_o=600\ kg.m^{-3}[/tex] is the density of the oil
Explanation:
Given:
height of oil column, [tex]h_o=20\ cm[/tex]oil column height that is more than the water column height in the other arm, [tex]\delta h=8\ cm[/tex]Now from the given it is clear that the height of water column is:
[tex]h_w=h_o-\delta h[/tex]
[tex]h_w=20-8[/tex]
[tex]h_w=12\ cm[/tex]
Now according to the pressure balance condition of fluid columns:
Pressure due to water column = Pressure due to oil column
[tex]P_w=P_o[/tex]
[tex]\rho_w.g.h_w=\rho_o.g.h_o[/tex]
[tex]1000\times 9.8\times 0.12=\rho_o\times 9.8\times 0.2[/tex]
[tex]\rho_o=600\ kg.m^{-3}[/tex] is the density of the oil
Answer:
Explanation:
Let the density of oil is d'.
height of water, h = 20 - 8 = 12 cm
height of oil, h' = 20 cm
density of water, d = 1000 kg/m³
Pressure is balanced
h' x d' x g = h x d x g
0.20 x d' x g = 0.12 x 1000 x g
0.2 d' = 120
d' = 600 kg/m³
Which statement below is NOT true about electric field lines?
A) They can start on + charges
B) They are perpendicular to the electric field at every point
C) They get closer together where the field is stronger
D) Field lines cannot cross
E) They can end on - charges
a wave is 8 meters long and has a frequency of 3 Hz. Find speed
Answer:
The speed is 24 [tex]\frac{meter}{s}[/tex]
Explanation:
A wave is a disturbance that propagates through a certain medium or in a vacuum, with transport of energy but without transport of matter.
The wavelength is the minimum distance between two successive points of the wave that are in the same state of vibration. It is expressed in units of length (m).
Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
The speed of propagation is the speed with which the wave propagates in the middle, that is, the magnitude that measures the speed at which the wave disturbance propagates along its displacement. Relate wavelength (λ) and frequency (f) inversely proportionally using the following equation:
v = f * λ.
In this case, λ= 8 meter and f= 3 Hz
Then:
v= 3 Hz* 8 meter
So:
v= 24 [tex]\frac{meter}{s}[/tex]
The speed is 24 [tex]\frac{meter}{s}[/tex]
The speed of the wave is 24 m/s.
To find the speed of a wave, you can use the relationship between speed, frequency, and wavelength.
The formula is:
Speed (v) = Wavelength (λ) × Frequency (f)Given in this problem:
Wavelength (λ) = 8 metersFrequency (f) = 3 HzWe substitute these values into the formula:
Speed (v) = 8 meters × 3 HzThus, the speed of the wave is:
v = 24 m/sIf you shine a laser on two slits with a separation of 0.215 mm, and the diffraction pattern shines on a background 4.90 m away, what is the wavelength of the laser light if the fringes are separated by 1.60 cm
Answer:
[tex]\lambda = 702\ nm[/tex]
Explanation:
Given,
slit width, d = 0.215 mm
slit separation, D = 4.90 m
Wavelength of the laser light = ?
Fringe width.[tex]\beta = 1.60\ cm[/tex]
Using formula
[tex]\beta =\dfrac{\lambda D}{d}[/tex]
[tex]\lambda=\dfrac{\beta d}{D}[/tex]
[tex]\lambda=\dfrac{0.016\times 0.215\times 10^{-3}}{4.90}[/tex]
[tex]\lambda = 7.02\times 10^{-9}\ m[/tex]
[tex]\lambda = 702\ nm[/tex]
Wavelength of the laser light = [tex]\lambda = 702\ nm[/tex]
The vertical displacement of an ocean wave is described by the function, y = A sin(ωt - kx). k is called the wave number (k = 2π/λ) and has a value of k = 18 rad/m. The remaining values are A = 9.5 m and ω = 14.5 rad/s.
a) Using y = A sin(B), input an expression for B where the wave would be traveling in the -x-direction. sig.gif?tid=7M79-31-9F-4E-8624-20536
b) What is the wave's velocity in m/s?
c) What is the wave's amplitude in m?
The expression for B for a wave traveling in the -x direction is B = ωt - kx + π. The wave velocity is 0.805 m/s and the wave amplitude is 9.5 m.
Explanation:To determine the expression for B where the wave would be traveling in the -x direction, we need to consider that the general equation for the wave function is y = A sin(B). In this case, the wave is traveling in the -x direction, which means the phase of the wave is shifted by π (180 degrees). So the expression for B would be B = ωt - kx + π.
The wave's velocity can be calculated using the formula v = ω/k. Substituting the given values, the wave's velocity is v = 14.5 rad/s / 18 rad/m = 0.805 m/s.
The wave's amplitude is given directly as A = 9.5 m.
A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. The same cart again traveling at 0.3 m/s collides with a different stationary object. This time the cart is at rest after the collision. In which collision is the impulse on the cart greater?A. The impulses are the same.B. The second collision.C. The first collision.D. Cannot be determined without knowing the mass of the cart.E. Cannot be determined without knowing the rebound speed of the first collision.
Answer: Impulse is greater in the first case. So, option C is the correct option.
Explanation:
Case 1: Cart is travelling at 0.3 m/s and collide with an stationary object and after collision, cart rebound in opposite direction and another object remains in static condition.
Applying the conservation of linear momentum:
[tex]m_{1} \times u_{1} + m_{2} \times u_{2} = m_{1} \times v_{1} + m_{2} \times v_{2}[/tex]
[tex]m_{1} \times 0.3 + m_{2} \times 0 = m_{1} \times v_{1} + m_{2} \times 0[/tex]
Hence velocity of cart will rebound with the same velocity i.e. 0.3 m/s
Impulse is defined as the change in momentum
Impulse on the cart = [tex]m_{1} \times v_{1} - m_{1} \times u_{1}[/tex] = [tex]m_{1} \times ((-3) - (3)) = m_{1} \times (-6)[/tex] Kg m/s.
Case 2: Initially cart is travelling at 0.3 m/s and after collision it comes to rest.
So, change in momentum or Impulse = [tex]m_{1} \times (0 - 3)[/tex] = [tex]-3 \times m_{1}[/tex] Kg m/s.
Impulse is greater in the first case. So, option C is the correct option.
Answer:
The second impulse is greater then the first impulse.
(B) is correct option.
Explanation:
Given that,
In first case,
Initial speed of cart = 0.3 m/s
Final speed of cart = -0.3 m/s
in second case,
Initial speed of cart = 0.3 m/s
Final speed of cart = 0 m/s
In first case,
We need to calculate the impulse
Using formula of impulse
[tex]I=\Delta p[/tex]
[tex]I=\Delta (mv)[/tex]
[tex]I=mv-mu[/tex]
Put the value into the formula
[tex]I=m(-0.3-0.3)[/tex]
[tex]I= -0.6m\ kg m/s[/tex]
In second case,
We need to calculate the new impulse
Using formula of impulse
[tex]I'=\Delta (mv')[/tex]
[tex]I'=mv'-mu'[/tex]
Put the value into the formula
[tex]I'=m(0-0.3)[/tex]
[tex]I'= -0.3m\ kg m/s[/tex]
So, I'> I
Hence, The second impulse is greater then the first impulse.
Assuming the passive sign convention and an operating frequency of 314 rad/s, calculate the phasor voltage V which appears across each of the following when driven by the phasor current I = 1020° mA: (a) a 2 a resistor; (b) a 1 F capacitor; (c) a 1 Hinductor; (d) a 2 resistor in series with a 1 F capacitor; (e) a 2. resistor in series with a 1 H inductor. (f) Calculate the instantaneous value of each voltage determined in part (e) at t = 0.
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment
The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2 × 10-12 e-940/T, where T is the Kelvin temperature. Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C. Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3.
Answer:
Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C
Knowing
Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x[tex]10^{-11} e^{-255/T}[/tex] and 2x[tex]10^{-12} e^{-940/T}[/tex]
T = -50 °C = 223 K
The reaction rate will be given by [Cl] [O3] 3x[tex]10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3][/tex]
Than, the reaction rate of OH with O3 is
Rate = [OH] [O3] 2x[tex]10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3][/tex]
Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]
Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH
Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3
Knowing
Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x[tex]10^{-11} e^{-255/T}[/tex] and 2x[tex]10^{-12} e^{-940/T}[/tex]
T = -80 °C = 193 K
The reaction rate will be given by [Cl] [O3] 3x[tex]10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3][/tex]
Than, the reaction rate of OH with O3 is
Rate = [OH] [O3] 2x[tex]10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3][/tex]
Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]
Than, considering the concentration of Cl increases by a factor of 100 to about 4 × [tex]10^{5}[/tex] molecules [tex]cm^{-3}[/tex], the result will be that the reaction with OH will be 535 + (100 to about 4 × [tex]10^{5}[/tex] molecules [tex]cm^{-3}[/tex]) times faster than the reaction with Cl
Explanation:
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, 550 J/kg·K, 48 W/m·K), which is initially at a uniform temperature of 170°C and is to be heated to a minimum temperature of 550°C. Heating is effected in a gas-fired furnace, where products of combustion at [infinity] 800°C maintain a convection coefficient of 250 W/m2·K on both surfaces of the plate. How long should the plate be left in the furnace?
Answer:
The answer to the question is;
The plate be left in the furnace for 905.69 seconds.
Explanation:
To solve the question, we have to check the Bi number as follows
Bi = [tex]\frac{hL}{k} = \frac{250\frac{W}{m^{2} K} *0.05 m}{48\frac{W}{mK} } = 0.2604[/tex]
As the Bi number is > 0.1 we have to account for the variation of temperature with location in the mass.
We perform nonlumped analysis
The relation for heat transfer given by
Y = [tex]\frac{T_f-T_{inf}}{T_i- T_{inf}}[/tex]
=[tex]\frac{550-800}{170- 800}[/tex] = 0.3968 = C₁ exp (ζ₁² F₀)
where
C₁ and ζ₁ are coefficients of a series solution
We therefore look for the values of C₁ and ζ₁ from Bi tables to be
ζ₁ = 0.4801 +(0.26-0.25) (0.5218-0.4801)/(0.3-0.25) ≈ 0.4884 and
C₁ = 0.4801 +(0.26-0.25) (1.0450 - 1.0382)/(0.3-0.25) ≈ 1.03956 and
This gives the relation
0.3968 = 1.03956 exp (ζ₁² F₀)
or ζ₁² [tex](\frac{\alpha t}{L^2})[/tex]
where
α = Thermal diffusivity of solid = k/(ρ·c[tex]_p[/tex]) = [tex]\frac{48}{7830*550}[/tex] = 1.1146×10⁻⁵
c[tex]_p[/tex] = Specific heat capacity of solid at constant pressure = 550 J/kg·K
ρ = Density of the solid = 7830 kg/m³
=㏑[tex](\frac{0.3968 }{1.03956 })[/tex] = -0.9631 from where we have
t = [tex]\frac{0.9631 *0.05^{2} }{0.4884^2*1.11*10^{-5}}[/tex] = 905 seconds.
A thin-walled cylindrical steel water storage tank 30 ft in diameter and 62 ft long is oriented with its longitudinal axis vertical. The tank is topped with a hemispherical steel dome. The wall thickness of the tank and dome is 0.68 in. If the tank is pressurized to 55 psig and contains water 55 ft above its base, and considering the weight of the tank, determine the maximum state of stress in the tank and the corresponding principal stresses (normal and shear). The weight density of water is 62.4 lbf/ft3.
Answer:
ρ
=
55.0
π
⋅
15.0
2
⋅
62.0
Explanation:
An isolated charged conducting sphere of radius 11.0 cm creates an electric field of 4.90 ✕ 104 N/C at a distance 20.0 cm from its center. (a) What is its surface charge density?
Answer:
Surface charge density = 1.43 × 10⁶μC/m²
Explanation:
surface charge density = Q /A_____(1)
where charge Q is the uniformly distributed on surface area A and d surface charge density σ
The electric field due to uniformly charge sphere of charge Q a distance r from the center of the sphere
[tex]E = k\frac{Q}{r^2}[/tex]______(2)
where k is 8.988 × 10⁹N.m²C²
The surface area of the sphere is 4πR² ______(3)
The Capacitance is 4πε₀R
where ε₀ = 8.8542 × 10⁻¹²C/Nm² is a constant
Given that,
R = 11cm = 0.11m
E = 4.90 ✕ 10⁴ N/C
r = 20.0cm = 0.20m
substitute for Q in eqn(2) and for A in eqn(3)
surface charge density = [tex]\frac{Er^2 }{k(4\pi R^2)} \\\frac{(4.9 * 10^4)(0.20)^2}{4\pi (8.988 * 10^9)(0.11)^2 }[/tex]
= 1.43 * 10⁶C/m²
Surface charge density = 1.43 μC/m²
An isolated, charged conducting sphere of radius 11.0 cm creates an electric field of 4.90×10⁴ N/C at a distance 20.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance?
Answer:(a) 1.47 x 10⁻⁶ C/m² or 1.47 μC/m²
(b) 12.07 x 10⁻¹² F or 12.07 pF
Explanation:The surface charge density, σ, of a surface (sphere, in this case) of area A which has a charge Q uniformly distributed on it is given by;
σ = [tex]\frac{Q}{A}[/tex] -----------------(i)
Also, the electric field, E, due to the charge Q, at a distance r, from the center of the sphere to another point on the sphere is given as;
E = k x [tex]\frac{Q}{r^2}[/tex] --------------(ii)
Where;
k = Coulomb's constant = 8.99 x 10⁹Nm²/C²
(a) i. First calculate the charge on the sphere as follows;
From the question;
r = 20.0cm = 0.20m
E = 4.90 x 10⁴ N/C
Substitute these values into equation (ii) as follows;
4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.20^2}[/tex]
4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.04}[/tex]
4.90 x 10⁴ = 224.75 x 10⁹ x Q
Q = [tex]\frac{4.90*10^4}{224.75*10^9}[/tex]
Q = 0.022 x 10⁻⁵
Q = 0.22 x 10⁻⁶ C
(a) ii. Also calculate the area A, of the sphere as follows;
A = 4π R²
Where;
R = radius of the sphere = 11.0cm = 0.11m
Substitute this value into equation above;
A = 4π (0.11)² [Take π = 3.142]
A = 4(3.142)(0.0121)
A = 0.15m²
(a) iii. Now calculate the surface charge density, of the sphere as follows;
Substitute the values of A and Q into equation (i) as follows;
σ = [tex]\frac{0.22 * 10^{-6}}{0.15}[/tex]
σ = 1.47 x 10⁻⁶C/m²
Therefore the surface charge density is 1.47 x 10⁻⁶C/m²
==============================================================
(b) The capacitance C, of an isolated charged sphere with radius R, is given by;
C = Aε₀ / R ----------------(iii)
Where;
R = 11.0cm = 0.11m
A = area of the sphere = 0.15m² [as calculated above]
ε₀ = permittivity of free space = 8.85 x 10⁻¹² C/Nm² [a known constant]
Substitute these values into equation (iii) as follows;
C = 0.15 x 8.85 x 10⁻¹² / 0.11
C = 12.07 x 10⁻¹²F
Therefore, the capacitance of the charged sphere is 12.07 x 10⁻¹²F
Two blocks, joined by a string, have masses of 6.0 kg and 9.0 kg. They rest on a frictionless horizontal surface. A 2nd string, attached only to the 9 kg block, has horizontal force = 30 N applied to it. Both blocks accelerate. Find the tension in the string between the blocks.
Answer:
Explanation:
30 N force is pulling total mass of 15 kg , so acceleration in the system of masses
= 30 / 15
= 2 m / s²
Let us now consider forces acting on 9 kg . 30 N is pulling it in forward direction . Tension T in the string attached to it is pulling it in reverse direction
so net force on it
30 - T
Applying Newton's law of motion on it
30 - T = mass x acceleration
30 - T = 9 x 2
30 - 18 = T
T = 12 N
Using Newton’s second law, the applied force is used to find the acceleration of the whole system. We then calculate the force (tension) required to move the 6 kg block using this acceleration which is 12N.
Explanation:In Physics, specifically Newtons’ second law of motion, the tension in the string between the two blocks can be calculated by using the equation F=ma, where F is the force, m is the mass, and a is the acceleration. The force applied on the 9 kg block can be considered to cause an acceleration in the entire system (both blocks) given the fact that they are connected by a string. First, find the acceleration of the whole system by using the formula a = F/total mass = 30N/(9kg+6kg) = 2 m/s². Then, to find the tension in the string between the blocks, we calculate the force required to move the 6 kg block with that acceleration: T = ma = 6kg * 2 m/s² = 12 N. Therefore, the tension in the string between the blocks is 12N.
Learn more about Tension in a string here:https://brainly.com/question/34111688
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A small ranger vehicle has a soft, ragtop roof. When the car is at rest, the roof is flat. When the car is cruising at highway speeds with its windows rolled up, does the roof bow upwards, remain flat, or bow downwards
Answer:
roof bow upwards
Explanation:
The top of the roof of the small ranger vehicle will bow upwards. This is as a result of gas pressure on the soft ragtop roof.
As air begins to fill the vehicle, pressure resonates in all direction proportionally. The pressure of the air will be greater than that which the roof can withstand and this forces the roof sky up. It is a common scene when we see roof of ragtop vehicles bowing upwards into the sky.A 5.93 5.93 kg ball is attached to the top of a vertical pole with a 2.03 2.03 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.25 4.25 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take g = 9.81 g=9.81 m/s2.
Final answer:
To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.Therefore angle is approximately 45.6°.
Explanation:
To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.
At the top of the circle, the tension in the string is equal to the weight of the ball, which is given by T = mg. We can use this equation to find the angle the string makes with the pole.
tan(theta) = T/ (m*v²/ R), where m is the mass of the ball, v is the speed of the ball, and R is the length of the string. Plugging in the values given, we get tan(theta) = (5.93 kg * 4.25 m/s²) / (5.93 kg * 9.81 m/s²). Solving for theta, we find that the angle is approximately 45.6°.
ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed point. Find the magnitude of the tension when themass is at the top if its speed at the top is 8.5 m/s.(
Answer:
[tex]6.046N[/tex]
Explanation:
The net force exerted on the mass is the sum of tension force and the external force of gravity.
[tex]F_n_e_t=F_g+F_t[/tex]
[tex]F_t[/tex] is the tension force.[tex]F_g=9.8N/kg[/tex] is the force of gravity.
[tex]F_n_e_t=ma_c=mv^2/r\\[/tex]
where [tex]r[/tex] is the rope's radius from the fixed point.
From the net force equation above:
[tex]F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N[/tex]
Hence the tension force is 6.046N
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have wavelengths 58-cm in the tube, what is the frequency (in Hertz) of the tuning fork
Answer:
594.8 Hz
Explanation:
Parameters given:
Speed of sound, v = 345 m/s
Wavelength = 58 cm = 0.58 m
Speed of a wave is given as:
Speed = wavelength * frequency
Therefore:
Frequency = Speed/Wavelength
Frequency = 345/0.58
Frequency = 594.8 Hz
The frequency of the tuning fork is calculated using the formula Vw = fa, where Vw is the speed of sound, f is frequency, and a is wavelength. Substituting their given values, we get a frequency of approximately 595 Hz.
Explanation:
The frequency of sound can be calculated with the formula Vw = fa, where Vw is the speed of sound, f is frequency, and a is the wavelength.
From your question, we know that the speed of sound (Vw) in air is 345 m/s and the wavelength (a) is 58 cm (or 0.58 m when converted to meters to match the speed of sound's units).
Frequency (f) can be calculated by rearranging to f = Vw / a. Substituting the values: f = 345 m/s / 0.58 m = 594.83 Hz. So, the tuning fork's frequency is approximately 595 Hz.
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Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.
Answer:
av=0.333m/s, U=3.3466J
b.
[tex]v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s[/tex]
Explanation:
a. let [tex]m_A[/tex] be the mass of block A, and[tex]m_B=10.0kg[/tex] be the mass of block B. The initial velocity of A,[tex]\rightarrow v_A_1=2.0m/s[/tex]
-The initial momentum =Final momentum since there's no external net forces.
[tex]pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]
Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):
[tex]v_A_1-v_B_1=v_{B2}-v_{A2}[/tex]
-Applying the conservation of momentum. The blocks have the same velocity after collision:
[tex]v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s[/tex]
#Total Mechanical energy before and after the elastic collision is equal:
[tex]K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J[/tex]
Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s
b. Taking the end collision:
From a above, [tex]m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0[/tex]
We plug these values in the equation:
[tex]m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]
[tex]2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s[/tex]
(a) The maximum energy stored in the spring bumpers during the collision is 3.00 J, and the velocity of both the blocks is 0.50 m/s.
(b) After they move apart, block A has a velocity of -1.00 m/s and block B has a velocity of 1.00 m/s.
You can follow these simple steps to find the required solution -
(a) Maximum Energy Stored in the Spring Bumpers
To find the maximum energy stored in the spring bumpers, we will use the conservation of momentum and energy principles.
Initially, block A (mass 2.00 kg) is moving at 2.00 m/s, and block B (mass 6.00 kg) is at rest. The total initial momentum ([tex]p_{initial[/tex]) is:
[tex]p_{\text{initial}} = m_A \cdot v_A + m_B \cdot v_B = 2.00 \, \text{kg} \cdot 2.00 \, \text{m/s} + 6.00 \, \text{kg} \cdot 0 \, \text{m/s} = 4.00 \, \text{kg} \cdot \text{m/s} \\[/tex]At the point of maximum compression, both blocks momentarily move with the same velocity ([tex]v_{common[/tex]). Using the conservation of momentum:
[tex]p_{\text{final}} = (m_A + m_B) \cdot v_{\text{common}} = 4.00 \, \text{kg} \cdot \text{m/s} \\[/tex]Solving for [tex]v_{common[/tex]:
[tex]v_{\text{common}} = \frac{4.00 \, \text{kg} \cdot \text{m/s}}{2.00 \, \text{kg} + 6.00 \, \text{kg}} = 0.50 \, \text{m/s} \\[/tex]Next, determine the initial kinetic energy ([tex]KE_{initial[/tex]):
[tex]KE_{\text{initial}} = 0.5 \cdot m_A \cdot (v_A)^2 + 0.5 \cdot m_B \cdot (v_B)^2 = 0.5 \cdot 2.00 \, \text{kg} \cdot (2.00 \, \text{m/s})^2 + 0.5 \cdot 6.00 \, \text{kg} \cdot (0 \, \text{m/s})^2 = 4.00 \, \text{J} \\[/tex]The kinetic energy at the point of maximum compression ([tex]KE_{final[/tex]) is:
[tex]KE_{\text{final}} = 0.5 \cdot (m_A + m_B) \cdot (v_{\text{common}})^2 = 0.5 \cdot 8.00 \, \text{kg} \cdot (0.50 \, \text{m/s})^2 = 1.00 \, \text{J} \\[/tex]The maximum energy stored in the spring bumpers ([tex]E_{spring[/tex]) is the difference between [tex]KE_{initial[/tex] and [tex]KE_{final[/tex]:
[tex]E_{\text{spring}} = KE_{\text{initial}} - KE_{\text{final}} = 4.00 \, \text{J} - 1.00 \, \text{J} = 3.00 \, \text{J} \\[/tex](b) Velocity of Each Block After Collision
After they have moved apart, assuming an elastic collision where both kinetic energy and momentum are conserved, the final velocities ([tex]v_A_{final[/tex] and [tex]v_B_{final[/tex]) can be found using the respective equations:
For block A:
[tex]v_{A_{\text{final}}} = \frac{(m_A - m_B)}{(m_A + m_B)} \cdot v_{A_{\text{initial}}} + \frac{(2 \cdot m_B)}{(m_A + m_B)} \cdot v_{B_{\text{initial}}} \\[/tex][tex]v_{A_{\text{final}}} = \frac{(2.00 \, \text{kg} - 6.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 2.00 \, \text{m/s} + \frac{(2 \cdot 6.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 0 \, \text{m/s} = -1.00 \, \text{m/s} \\[/tex]For block B:
[tex]v_{B_{\text{final}}} = \frac{(2 \cdot m_A)}{(m_A + m_B)} \cdot v_{A_{\text{initial}}} + \frac{(m_B - m_A)}{(m_A + m_B)} \cdot v_{B_{\text{initial}}} \\[/tex][tex]v_{B_{\text{final}}} = \frac{(2 \cdot 2.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 2.00 \, \text{m/s} + \frac{(6.00 \, \text{kg} - 2.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 0 \, \text{m/s} = 1.00 \, \text{m/s} \\[/tex]Therefore, the final velocities are:
[tex]v_{A_{\text{final}}} = -1.00 \, \text{m/s} \quad \text{and} \quad v_{B_{\text{final}}} = 1.00 \, \text{m/s}[/tex]The width of a particular microwave oven is exactly right to support a standing-wave mode. Measurements of the temperature across the oven show that there are cold spots at each edge of the oven and at three spots in between. The wavelength of the microwaves is 12 cm . How wide is the oven?
Answer:
Explanation:
In standing wave pattern we find region of nodes where vibration is minimum or cold spots . The distance between any two consecutive node is half the wave length . There are 5 cold spot or node in between which is equal to 4 half wave length .
width of oven = 4 x half wave length
= 4 x (12 / 2 )
= 24 cm
Consider a magnetic force acting on an electric charge in a uniform magnetic field. Which of the following statements are true? Check all that apply. Check all that apply. An electric charge moving parallel to the magnetic field experiences a magnetic force. The direction of the magnetic force acting on a moving electric charge in the magnetic field is perpendicular to the direction of motion. An electric charge moving perpendicular to the magnetic field experiences a magnetic force. The magnetic force is exerted on a stationary electric charge in the uniform magnetic field. The direction of the magnetic force acting on a moving charge in the magnetic field is perpendicular to the direction of the magnetic field.
Answer: 5) "The direction of the magnetic force acting on a moving charge in the magnetic field is perpendicular to the direction of the magnetic field"
3) "An electric charge moving perpendicular to the magnetic field experiences a magnetic force"
Explanation:
When a charge of magnitude q, moving with a velocity v is placed in a magnetic field of strength B, it experiences a force, the magnitude of this force F is given mathematically as
F =qvB sinx
Where x is the angle between the magnetic field and the velocity of the charge.
From this equation, we can see that the force is zero when magnetic field strength B is parallel to velocity (x=0) or when velocity v is zero.
Also the force F is maximum when the angle between magnetic field strength B and velocity is 90 ( that's they are perpendicular).
By the right hand rule, the force, velocity and strength of magnetic field are perpendicular to each other.
These points have made statement 3 and 5 of the questions to be correct.
In the lab, you did not include friction in your calculations for the acceleration. Explain why it was not necessary. What would your equation for acceleration look like if you were to include friction
Final answer:
Friction was not included in the lab calculations because it opposes motion between surfaces, resulting in a smaller acceleration. The equation for acceleration without friction is a = g sinθ, while the equation for acceleration with friction is a = (g sinθ - μk cosθ) / (1 + μk sinθ), where μk is the coefficient of kinetic friction.
Explanation:
The reason friction was not included in the calculations for acceleration in the lab is because friction always opposes motion between surfaces, resulting in a smaller acceleration when it is present. In the absence of friction, all objects slide down a frictionless incline with the same acceleration, regardless of mass. The equation for acceleration without friction is a = g sinθ, where g is the acceleration due to gravity and θ is the angle of the incline.
If friction were to be included in the calculations, the equation for acceleration would be different. It would depend on the coefficient of friction (μ) and the normal force (N) of the object. The equation for acceleration with friction is a = (g sinθ - μk cosθ) / (1 + μk sinθ), where μk is the coefficient of kinetic friction. This equation takes into account the opposing force of friction and provides a more accurate representation of the object's acceleration.
You are given a sample of several compounds to separate by paper chromatography. You draw a pencil line exactly 1.0 cm from the bottom of the paper, and place a spot of sample on it. You dry the sample, then develop it in a solvent. When the chromatogram is taken out of the solvent, the paper is wet up to 8.8 cm from the bottom of the sheet. The compound you are interested in shows up as a spot 7.5 cm from the bottom of the paper. Calculate the following: How far did the compound move?
Answer:
The compound moves 6.5 cm in total.
Explanation:
Before solving this problem, let's first write down all lengths we know of from the question:
Starting point of sample = 1.0 cm from bottom of paper
Paper wet up to = 8.8 cm from bottom of paper
Ending point of the sample = 7.5 cm from bottom of paper
With these lengths stated, we can easily calculate the length which the compound moved through:
Length compound moved = Ending point - Starting point
Length compound moved = 7.5 - 1.0
Length compound moved = 6.5 cm
Thus, we can see that the compound moved 6.5 cm between the time the paper was put into, and taken out of the solvent.
While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock and the car begins to slide against the road. The car collides with the rabbit at a final speed of vf, after T seconds of braking. What is the coefficient of kinetic friction of the tires against the road?
Answer:
μk = (Vf - Vc)/(T×g)
Explanation:
Given
Vi = initial velocity of the car
Vf = final velocity of the car
T = Time of application of brakes
g = acceleration due to gravity (known constant)
Let the mass of the car be Mc
Assuming only kinetic frictional force acts on the car as the driver applies the brakes,
The n from Newtown's second law of motion.
Fk = Mc×a
Fk = μk×Mc×g
a = (Vf - Vc)/T
Equating both preceding equation.
μk×Mc×g = Mc × (Vf - Vc)/T
Mc cancels out.
μk = (Vf - Vc)/(T×g)
Final answer:
To find the coefficient of kinetic friction of the tires against the road, you need the initial and final speeds of the car, the time it took to decelerate, and the acceleration due to gravity. Without specific values for these variables, the problem cannot be solved directly, but the formula given provides a method to calculate the coefficient if such values are known.
Explanation:
The question asks for the coefficient of kinetic friction between the tires and the road when a car, initially traveling at speed vi, decelerates to a final speed vf over time T seconds due to the driver slamming on the brakes upon seeing a rabbit.
The formula to calculate the coefficient of kinetic friction (μ_k) can be derived from Newton's second law of motion and the equation of motion that relates initial velocity, final velocity, acceleration, and time. The formula for the coefficient of kinetic friction is μ_k = (vi - vf) / (g × T), where g is the acceleration due to gravity (9.81 m/s2).
To solve this problem, you need the initial and final velocities of the car (vi and vf), the deceleration time (T), and knowledge that the acceleration due to gravity (g) is approximately 9.81 m/s2. However, actual calculations cannot be performed without specific values for vi, vf, and T.
This equation illustrates that the coefficient of kinetic friction is directly related to the deceleration rate of the car on the road surface.
An electric field can exist solely due to a source charge. An electric force requires at least two charges, the source charge to set up the field and the test charge to feel the field.
A. TRUE
B. FALSE
Answer:
TRUE.
Explanation:
The first part of the statement is understood as certain since the definition of the electric field warns that its existence is dependent on the charge of the source. At the same time, we know that for there to be an electric force it is necessary to consider two charges. Either for the generation of the force of repulsion or for the force of attraction.
The statement is true. An electric field can exist due to a single source charge, while an electric force requires a test charge to respond to the field generated by the source charge.
Explanation:The statement is
TRUE
. The phenomenon takes place within the study of
electromagnetism
. An
electric field
indeed can exist solely due to a source charge. The source charge creates an electric field in the space around it. However, to cause an electric force, there must be at least one other charge (test charge) that is capable of sensing the presence of this field. The greater the source charge, the stronger the electric field. The test charge responds to the field by experiencing a force. The direction of the electric force is the same as the direction of the electric field if the test charge is positive, but opposite if the test charge is negative.
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Basidiomycete fungi ballistically eject millions of spores into the air by releasing the surface tension energy of a water droplet condensing on the spore. The spores are ejected with typical speeds of 1.11 m/s, allowing them to clear the "boundary layer" of still air near the ground to be carried away and dispersed by winds, (a) If a given spore is accelerated from rest to 1.11 m/s in 7.40 us, what is the magnitude of the constant acceleration of the spore (in m/s) while being ejected? m/s(b) Find the maximum height of the spore (in cm) if it is ejected vertically. Ignore air resistance and assume that the spore is ejected at ground level. cm(c) Find the maximum horizontal range of the spore (in cm) if it is ejected at an angle to the ground. Ignore air resistance and assume that the spore is ejected at ground level CM
a) [tex]1.5\cdot 10^5 m/s^2[/tex]
b) 6.3 cm
c) 12.6 cm
Explanation:
a)
The acceleration of an object is the rate of change of its velocity; it is given by:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time interval taken for the velocity to change from u to t
In this problem for the spore, we have:
u = 0 (the spore starts from rest)
v = 1.11 m/s (final velocity of the spore)
[tex]t=7.40\mu s = 7.40\cdot 10^{-6}s[/tex] (time interval in which the spore accelerates from zero to 1.11 m/s)
Substituting, we find the acceleration:
[tex]a=\frac{1.11-0}{7.40\cdot 10^{-6}}=1.5\cdot 10^5 m/s^2[/tex]
b)
Since the upward motion of the spore is a free fall motion (it is subjected to the force of gravity only), it is a uniformly accelerated motion (=constant acceleration, equal to the acceleration due to gravity: [tex]g=9.8 m/s^2[/tex]). Therefore, we can apply the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where:
v = 0 is the final velocity of the spore (when it reaches the maximum height, its velocity is zero)
u = 1.11 m/s is the initial velocity (the velocity at which it is ejected)
[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration (negative because it is downward)
s is the vertical displacement of the spore, which corresponds to the maximum height reached by the spore
Solving for s, we find:
[tex]s=\frac{v^2-u^2}{2a}=\frac{0^2-(1.11)^2}{2(-9.8)}=0.063 m = 6.3 cm[/tex]
c)
If the spore is ejected at a certain angle [tex]\theta[/tex] from the ground, then its motion is a projectile motion, which consists of two independent motions:
- A uniform horizontal motion, with constant horizontal velocity
- A uniformly accelerated motion along the vertical direction (free fall motion)
The horizontal range of a projectile, which can be derived from the equations of motion, is given by:
[tex]d=\frac{v^2 sin(2\theta)}{g}[/tex]
where
v is the initial velocity
[tex]\theta[/tex] is the angle or projection
g is the acceleration of gravity
From the equation, we observe that the maximum range is achevied when
[tex]\theta=45^{\circ}[/tex]
For this angle, the range is
[tex]d=\frac{v^2}{g}[/tex]
For the spore in this problem, the initial velocity is
v = 1.11 m/s
Therefore, the maximum range is
[tex]d=\frac{(1.11)^2}{9.8}=0.126 m = 12.6 cm[/tex]
Kepler deduced this law of motion from observations of Mars. What information confirms his conclusion that the orbit of Mars is elliptical?
Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.
This led to Kepler's law of orbital motion.
First Law: Planetary orbits are elliptical with the sun at a focus.
Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.
Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.
It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.
An ice cube and a rubber ball are both placed at one end of a warm cookie sheet, and the sheet is then tipped up. The ice cube slides down with virtually no friction, and the ball rolls down without slipping. The ball and the ice cube have the same inertia. Which one reaches the bottom first?
The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically, it is defined as the product of the coefficient of friction and normal reaction.
The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.
Because the ice cube slides down with virtually no friction. Without friction, an object can easily slide with the more speed.
The ball is spherical in nature and a spherical surface is in more contact with the surface during the inclined motion.
Hence, the ice ball reaches the bottom first, when the sheet is then tipped up.
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