Answer:
the final temperature of the system of the system is 25.32°C
Explanation:
We are not given specific capacity of water and aluminium, so we use their standard values, also we are not given the density if water so we assume the standard vale of density of water
The aluminium calorimeter has a mass Mc= 120g
Volume of water in calorimeter = 100ml at θc =20°C
Density of water is
1000Kg/m³ = 1g/mL
Then, density = mass/ volume
Mass=density ×volume
Mass=1g/mL×100mL
Mass=100gram
Then, the mass of water is
Mw = 100gram
Mass of brass is Mb = 100gram
The temperature of brass is θb=100°C
The specific heat capacity of water is Cw= 1cal/g°C
The specific heat capacity of aluminum Ca=0.22cal/g°C
We are looking for final temperature θf=?
Given that the specific heat capacity of brass is Cb=0.09Cal/g°C
Using the principle of calorimeter;
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, the calorimeter gained heat and the liquid in the calorimeter gain heat too
Heat gain by calorimeter(Hc) = Mc•Ca•∆θ
Where Mc is mass of calorimeter,
Ca is Specific Heat capacity of Calorimeter
∆θ=(θf-θc)
Hc=Mc•Ca•∆θ
Hc=120•0.22•(θf-20)
Hc=26.4(θf-20)
Hc=26.4θf-528
Also, heat gain by the water
Heat gain by wayer(Hw) = Mw•Cw•∆θ
Where Mw is mass of water,
Cw is Specific Heat capacity of water
∆θ=(θf-θw),
Note that the temperature of the water and the calorimeter are the same at the beginning i.e. θc=θw=20°C
Hw=Mw•Cw•∆θ
Hw=100•1•(θf-20)
Hw=100(θf-20)
Hw=100θf-2000
Also heat loss by the brass is given by
heat loss by brass
Heat loss by brass(Hb)= Mb•Cb•∆θ
Where Mb is mass of brass,
Cb is Specific Heat capacity of brass
∆θ=(θb-θf)
Therefore,
Hb=Mb•Cb•∆θ
Hb=100•0.09•(100-θf)
Hb=9(100-θf)
Hb=900-9θf
Applying the principle of calorimeter
Heat gain = Heat loss
Hc+Hw=Hb
26.4θf-528 + 100θf-2000=900-9θf
26.4θf+100θf+9θf=900+2000+528
135.4θf=3428
Then, θf=3428/133.4
θf=25.32°C
The final temperature of the system is 25.32 degree Celsius.
Given data:
The mass of aluminum cup is, m = 120 g .
The mass of brass piece is, m' = 100 g.
The volume of water in aluminum cup is, V = 100 ml.
The temperature of water is, T = 20 degree Celsius.
The specific heat of brass is, [tex]c''=0.09 \;\rm cal/g ^\circ C[/tex].
The temperature of brass is, T'' = 100 degree Celsius.
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, first calculate heat gain by calorimeter,
H = mc (T' - T)
Here, c is the specific heat of aluminum and its value is, [tex]0.22 \;\rm cal/g^\circ C[/tex]. Solving as,
[tex]H = 120 \times 0.22 (T' - 20)\\H = 26.4T' - 528[/tex]
And, heat gain by water is,
[tex]H'=m'c'(T'-T)[/tex]
Here, c' is the specific heat of water. ( c' =1 )
Solving as,
[tex]H'=100 \times 1 \times (T'-20)\\H' = 100T' -2000[/tex]
Now, heat loss by brass is,
[tex]H'' = m'c'' (T''-T')\\\\H'' = 100 \times 0.09 \times (100-T')\\\\H'' = 900-9T'[/tex]
Applying the principle of calorimeter
Heat gain = Heat loss
H + H' = H''
(26.4T' - 528) + (100T' - 2000) = (900 - 9T')
135.4 T' = 3428
T ' = 25.32 degree Celsius
Thus, we can conclude that the final temperature of the system is 25.32 degree Celsius.
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A stiff wire 35.5 cm long is bent at a right angle in the middle. One section lies along the z-axis and the other is along the line y=2x in the xy-plane. A current of 22.5 A flows in the wire-down the z-axis and out the line in the xy-plane. The wire passes through a uniform magnetic field given by B = (.318 i)T.
Determine the magnitude and the direction of the total force on the wire.
Answer:
Explanation:
Length, l = 35.5 cm = 0.355 m
y = 2 x
Slope, y / x = 2
tan θ = 2
θ = 63.4°
i = 22.5 A
B = 0.318 i Tesla
[tex]\overrightarrow{l}=0.355\widehat{k}+0.355 Cos63.4\widehat{i}+0.355 Sin63.4\widehat{j}[/tex]
[tex]\overrightarrow{l}=0.16\widehat{i} + 0.32\widehat{j}+0.355\widehat{k}[/tex]
[tex]\overrightarrow{B}=0.318\widehat{i}[/tex]
The magnetic force is given by
[tex]\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})[/tex]
[tex]\overrightarrow{F}=22.5\left ( 0.16\widehat{i}+0.32\widehat{j}+0.355\widehat{k} \right )\times 0.318\widehat{i}[/tex]
[tex]\overrightarrow{F}=2.54\widehat{j}-2.29\widehat{k}[/tex]
Magnitude of force
[tex]F=\sqrt{2.54^{2}+2.29^{2}}[/tex]
F = 3.42 N
The angle is Ф
tanФ = -2.29/2.54
Ф = 42° below y axis
The total force experienced by the wire in the magnetic field has a magnitude of 145 N and is directed -30 degrees relative to the z-axis, according to the right-hand rule and Lorentz force law.
Explanation:In order to determine the direction of the total force on the wire, you would need to use the right-hand rule. First, you know the current is flowing down the z-axis and out along the y=2x line in the xy-plane. You also know that the magnetic field, B, is given by (.318 i)T which lies along the x-axis.
According to the Lorentz force law, the force exerted on a current-carrying conductor in a magnetic field is given by F = I * (L x B), where I represents the current, L is the length vector of the wire, and x is the cross product. Because the wire is bent at a right angle, the force will have two components: one for each section of the wire.
The force along the z-axis is given by F = I*LB*sin(theta), where theta is the angle between L and B. Here, L = 35.5/2 = 17.75 cm, B = 0.318 T, I = 22.5 A, and theta = 90 degrees (since B is along x-axis and L is along z-axis). Calculating gives Fz = 22.5 A * 17.75 cm * 0.318 T * sin(90) = 126.6 N.
The force along the y-axis is similar, but with the length vector along y=2x in the xy plane and B still along the x-axis, so theta = 180 degrees - arctan(2). Substituting the values we get Fy = - 22.5 A * 17.75 cm * 0.318 T * sin(180 - arctan(2)) = -73.8 N.
Therefore, the total force is given by the vector sum of Fz and Fy, which has magnitude sqrt(Fz^2 + Fy^2) = 145 N, in the direction of atan2(Fy, Fz) = -30 degrees relative to the z-axis.
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A water tank is in the shape of a right circular cone with height 18 ft and radius 12 ft at the top. If it is filled with water to a depth of 15 ft, find the work done in pumping all of the water over the top of the tank.
Answer:
210,600πft-lb
Explanation:
Force is a function F(x) of position x then in moving from
x = a to x= b
Work done = [tex]\int\limits^b_a {Fx} \, dx[/tex]
Consider a water tank conical in shape
we will make small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w
we will have ,
[tex]\frac{w}{12} = \frac{(18-h)}{18} \\w = \frac{2}{3} (18-h)a[/tex]
weight of slice under construction
weight = volume × density × gravitational constant
[tex]weight = \pi \times w^2 \times dh \times 62.4\\= (62.4\pi w^2dh)lb[/tex]
Now we can find work done
[tex]W = \int\limits \, dw\\[/tex]
[tex]\int\limits^{18}_{3} {(62.4\pi } \, dx w^2dh)h\\= 62\pi \frac{4}{9} \int\limits^{18}_{3} {(18-h)^2hdh} \,[/tex]
= [tex]62.4\pi \times\frac{4}{9} (\frac{324}{2} h^2-\frac{36}{3} + \frac{h^4}{4})^1^8_3[/tex]
= 210,600πft-lb
Suppose a student recorded the following data during Step 10 of this experiment,
A 6.2 deg 4.7 deg 3.2 deg 2.4 deg 1.8 deg
Time 2 s 4 s 6 s 8 s 10 s
The student was careful to start taking data when the angular position of the pendulum was zero. Unfortunately he does not have a computer available to fit the data as you will in Step 12, but he knows that the amplitude of oscillations decays exponentially according to A= A0e-(Bt)+ C, where C = 1. Use the data obtained by the student to plot a straight line graph of the appropriate variables on the graph paper, and use it to find values for the parameters A0 and B.
Answer:
B = -0.215 s⁻¹, A₀ = 1.537º
Explanation:
To be able to make this graph we must linearize the data, the best procedure is to calculate the logarithm of the values
A = A₀ [tex]e^{-Bt}[/tex] + C
The constant creates a uniform displacement of the graph if we start the graph at the angle of 1, the constant disappears, this is done by subtracting 1 from each angle, so the equation is
(A-1) = A₀ e^{-Bt}
We do the logarithm
Log (A-1) = log Ao –Bt log e
Make this graph because paper commercially comes in logarithm 10, if we use graph paper if we can calculate directly in base logarithm e, let's perform this calculation
Ln (A-1) = Ln A₀ - Bt
To graph the points we subtract 1 from each angle and calculate the logarithm, the data to be plotted are
θ’= ln (θ -1)
θ(º) t(s) θ'(º)
6.2 2 1.6487
4.7 4 1.3083
3.2 6 0.7885
2.4 8 0.3365
1.8 10 -0.2231
We can use a graph paper and graph on the axis and the primary angle (θ') and on the x-axis time, mark the points and this graph is a straight line, we see that the point for greater time has a linearity deviation , so we will use the first three for the calculations
To find the line described by the equation
y -y₀ = m (x -x₀)
m = (y₂ -y₁) / (x₂ -x₁)
Where m is the slope of the graph e (x₀, y₀) is any point, let's start as the first point of the series
(x₀, y₀) = (2, 1.65)
(x₂, y₂) = (2, 1.65)
(x₁, y₁) = (6, 0.789)
We use the slope equation
m = (1.65 - 0.789) / (2-6)
m = -0.215
The equation is
y - 1.65 = -0.215 (x- 2)
y = -0.215 x +0.4301
We buy the two equations and see that the slope is the constant B
B = -0.215 s⁻¹
The independent term is
b = ln A₀
A₀ = [tex]e^{b}[/tex]
A₀ = [tex]e^{0.43}[/tex]
A₀ = 1.537º
A student sees her physical science professor approaching on the sidewalk that runs by her dorm. She gets a water balloon and waits. When the professor is 2.0s from being directly under the window about 11m from the sidewalk, she drops the balloon. Finish the story.
Answer:
The balloon falls to the ground before the professor gets there. The student is DEFINITELY in for some TROUBLE!
Explanation:
The balloon picks up speed due to gravity and we can calculate the time taken for it to fall to the ground as follows:
Gravity (g) = 9.81 m/s^2
Height or distance (s) = 11 meters
Initial Speed (u) = 0 m/s
[tex]s = u*t + 0.5 * (a*t^2)[/tex]
[tex]11 = 0*t + 0.5 (9.81*t^2)[/tex]
[tex]t= 1.4975 s[/tex]
So we can see that the balloon takes 1.4975 seconds to fall to the ground, and since the professor takes 2 seconds to get to that place, the balloon hits the ground right before the professor gets there.
A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 1.5 in. When twisted by a torque T, the tube develops a maximum normal strain of 170 106 . What is the magnitude of the applied torque T?
Explanation:
Given data:
G = 11.5×10⁶psi
d₂ = 2.0 inch
d₁ = 1.5 inch
ε[tex]_{max}[/tex] = 170 × 10⁻⁶
Y[tex]_{max}[/tex] = 2ε
T/J = τ[tex]_{max}[/tex] /R
[tex]\frac{Td_{2} }{2J}[/tex] = τ[tex]_{max}[/tex] (1)
τ[tex]_{max}[/tex] = G Y
from 1 and 2
[tex]\frac{T d_{2} }{2J} = G Y_{max}[/tex]
T = [tex]\frac{2 G Y_{max}J }{d_{2} }[/tex]
[tex]\frac{2* 11.5*10^{6}*0.006895*10^{6}*340*10*^{-6} *\frac{\pi }{32}[2^{4}-1.5^{4}]*(0.0254)^{4} }{2* 0.0254}[/tex]
= 474.14 Nm
At what minimum speed must a roller coaster be traveling so that passen- gers upside down at the top of the circle (Fig. 5–48) do not fall out? Assume a radius of curvature of 8.6 m.
Answer: [tex]v \approx 18.37 \frac{m}{s}[/tex]
Explanation:
Let assume that system is conservative. From application of the Principle of Energy Conservation, it is noticed that initial linear kinetic energy must be equal to the gravitational energy at the top of the circle. That is to say:
[tex]K_{1} = U_{2}\\[/tex]
[tex]\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (2\cdot R) \\v = \sqrt{4 \cdot g \cdot R}[/tex]
Where [tex]g = 9.81 \frac{m}{s^{2}}[/tex].
[tex]v = \sqrt{4 \cdot (9.81 \frac{m}{s^2})\cdot(8.6 m)} \\v \approx 18.37 \frac{m}{s}[/tex]
A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects beats such that the
frequency of the echo is 900 Hz higher than the frequency the bat is emitting. What is the speed of the bat? The
speed of sound in air is 340 m/s.
Answer:
5.02 m/s
Explanation:
given,
frequency of sound emitted by bat, [tex]f_1[/tex] = 30 kHz = 30000 Hz
beat frequency detected by bat[tex]f_b[/tex] = 900 Hz
Speed of sound, v = 340 m/s
[tex]f_b = f_1 -f_2[/tex]
[tex]900 = 30000-f_2[/tex]
[tex]f_2 = 30900[/tex]
Using Doppler's effect formula
[tex]f_2=f_1(\dfrac{v+v_b}{v-v_b})[/tex]
[tex]v_b = velocity\ of\ bat[/tex]
[tex]30900=30000(\dfrac{340+v_b}{340-v_b})[/tex]
[tex]1.03 =(\dfrac{340+v_b}{340-v_b})[/tex]
[tex]v_b = 5.02 m/s[/tex]
Hence, the speed of bat is equal to 5.02 m/s
A 67 kg person climbs up a uniform 12 kg ladder. The ladder is 5 m long; its lower end rests on a rough horizontal floor (static
friction coefficient 0.39) while its upper end rests against a frictionless vertical wall. The angle between the ladder and the horizontal is 43◦.
Let d denote the climbing person’s distance from the bottom of the ladder (see the above diagram). When the person climbs too far (d > dmax), the ladder slips and falls down (kaboom!). Calculate the maximal distance dmax the person will reach before the ladder slips. The
acceleration of gravity is 9.8 m/s*s. Answer in two decimal places max.
Answer:
1.7 m
Explanation:
Draw a free body diagram of the ladder. There are 5 forces:
Normal force N pushing up at the base of the ladder.
Friction force Nμ pushing right at the base of the ladder.
Weight force mg pushing down a distance d up the ladder.
Weight force Mg pushing down a distance L/2 up the ladder.
Reaction force R pushing left at the top of the ladder.
Sum of forces in the x direction:
∑F = ma
Nμ − R = 0
Sum of forces in the y direction:
∑F = ma
N − mg − Mg = 0
Sum of moments about the base of the ladder:
∑τ = Iα
mg (d cos θ) + Mg (L/2 cos θ) − R (L sin θ) = 0
Use the first equation to substitute for R:
mg (d cos θ) + Mg (L/2 cos θ) − Nμ (L sin θ) = 0
Use the second equation to substitute for N:
mg (d cos θ) + Mg (L/2 cos θ) − (mg + Mg) μ (L sin θ) = 0
Simplify and solve for d:
m (d cos θ) + M (L/2 cos θ) − (m + M) μ (L sin θ) = 0
m (d cos θ) = (m + M) μ (L sin θ) − M (L/2 cos θ)
d = [ (m + M) μ (L sin θ) − M (L/2 cos θ) ] / (m cos θ)
Plug in values and solve:
d = [ (67 kg + 12 kg) (0.39) (5 m sin 43°) − (12 kg) (2.5 m cos 43°) ] / (67 kg cos 43°)
d = 1.70 m
Rounded to two significant figures, the maximum distance is 1.7 m.
The maximal distance, [tex]d_{max[/tex], the person will reach before the ladder slips will be 3.41 m.
Here, we need to consider the forces acting and torque equilibrium. Let's start by noting down the given data:
Mass of the person, [tex]m_p[/tex] = 67 kgMass of the ladder, [tex]m_l[/tex] = 12 kgLength of the ladder, L = 5 mCoefficient of static friction, μ = 0.39Angle with the horizontal, θ = 43°The forces acting on the ladder are:
The weight of the ladder ([tex]W_l[/tex]) acting at its center of mass, which is at L/2.The weight of the person ([tex]W_p[/tex]) acting at distance d from the bottom.The normal force from the ground (N) and the frictional force from the ground (f).The normal force from the wall ([tex]F_w[/tex]), which is horizontal since the wall is frictionless.Using Newton’s second law for horizontal and vertical equilibrium and setting torques about the base of the ladder:
1. Horizontal forces:
[tex]\[F_w = f\][/tex]2. Vertical forces:
[tex]N=[/tex] [tex]W_l + W_p[/tex]For torque equilibrium about the base:
[tex]\[N \times L \sin(\theta) = W_l \left(\frac{L}{2}\right) \cos(\theta) + W_p d \cos(\theta)\][/tex]Substitute: [tex]\[N = (m_l + m_p)g\][/tex]
And frictional force: f = μN
We solve for d:
[tex]\[0.39(m_l + m_p)g \cdot L \sin(\theta) = m_l g \left(\frac{L}{2}\right) \cos(\theta) + m_p g \cdot d \cos(\theta)\][/tex]Simplify:
[tex]\[d = \frac{0.39(m_l + m_p)L \sin(\theta) - \left(m_l \frac{L}{2}\right) \cos(\theta)}{m_p \cos(\theta)}\][/tex]Substituting values:
d = [tex]\[\frac{0.39(12 + 67) \times 5 \sin(43^\circ) - \left(12 \times \frac{5}{2}\right) \cos(43^\circ)}{67 \cos(43^\circ)}\][/tex]d ≈ 3.41 m11 Initially, a single resistor R 1 is wired to a battery. Then resistorR 2 is added in parallel. Are (a) the potential difference across R 1and (b) the current i 1 through R 1 now more than, less than, or thesame as previously?
Answer:
Explanation:
Since , resistances are joined in parallel , potential difference across either of R₁ or R₂ will be same , and it will be equal to emf of the cell provided no internal resistance exists in the cell.
Since potential difference across R₁ is equal to emf and resistance is R₁ , current through it too will be same as that in the previous case.
The potential difference across resistor R1 remains the same when R2 is added in parallel, and the current through R1 also remains unchanged because the voltage across it does not change.
Explanation:When R2 is added in parallel to R1, the potential difference (voltage) across R1 remains the same because in a parallel circuit, all components have the same potential difference across them. However, the current i1 through R1 remains the same as previously as well since the potential difference across it hasn't changed and Ohm's Law (I = V/R) dictates that the current through a resistor will depend only on the potential difference across it and its resistance, which remains unchanged.
Adding R2 in parallel with R1 has the effect of decreasing the overall resistance of the circuit as found using the equation 1/Rp = 1/R1 + 1/R2. This means that the total current drawn from the battery will increase, but it will be divided among R1 and R2. Since the voltage across R1 remains constant, the current through R1 is unaffected by the addition of R2.
What is the motion of the negative electrons and positive atomic nuclei caused by the external field?
Answer:
The negative electrons moves to the left While the positively charged current moves in the opposite direction to the right when an external field is applied.
The positive atomic nuclei remains almost motionless.
In an external field, electrons and atomic nuclei move differently. In an electric field, electrons tend towards the positive field direction, whereas nuclei tend towards the negative. In a magnetic field, their paths are perpendicular to the field lines.
Explanation:The motion of negative electrons and positive atomic nuclei under the influence of an external field depends on the nature of the external field. This field could be of many types such as electric, magnetic, or electromagnetic. In an electric field, electrons move towards the positive field direction while the nuclei move towards the negative field direction. In a magnetic field, the motion of electrons and nuclei is governed by the right-hand rule, which means they move in a path perpendicular to the magnetic field lines.
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A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential height between the two arms of the manometer is 28 in. determine the density
Answer:
77.88 lbm/ft³
Explanation:
Given,
Specific gravity, SG = 1.25
Density of water, ρ = 62.30 lbm/ft³
density of the fluid =
= S.G x ρ_{water}
= 62.30 x 1.25
= 77.88 lbm/ft³
Density of the fluid is equal to 77.88 lbm/ft³
A constant force of 12 N in the +x direction acts on a 4 kg block as it moves from the origin to the point (6 − 8) m. How much work is done on the block by this force during this displacement?
Answer:
72 Nm²
Explanation:
work = F . d
work = (12i ,0j) . (6i ,-8j )
work = 72 J
The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.
It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j
When a force vector F acts in a nonparallel direction d, the work is given by:
W = |F|*|d|*cos(theta) where theta is the angle between the vectors.
alternatively you can use the dot product of the two vectors to get:
W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²
Final answer:
The work done on the block by the constant force of 12 N in the +x direction is 24 J.
Explanation:
The work done on the block can be calculated by multiplying the force applied to the block by the displacement of the block in the direction of the force. In this case, the force applied is 12 N in the +x direction and the displacement is (6 - 8) m = -2 m. Since the force and displacement are in opposite directions, the work done will be negative. The formula for work is given by:
Work = Force * Displacement * cos(theta)
where theta is the angle between the force and the displacement.
In this case, the angle between the force and the displacement is 180 degrees, so cos(theta) = -1. Substituting the values into the formula:
Work = 12 N * -2 m * (-1) = 24 J
A solid nonconducting sphere of radius R carries a uniform charge density throughout its volume. At a radial distance r1 = R/4 from the center, the electric field has a magnitude E0. What is the magnitude of the electric field at a radial distance r2 = 2R?
Answer:
Explanation:
Given that,
Sphere of radius R
At radial distance ¼R it has an electric field of Eo
Magnitude of electric field E at r=2R
Solution
electric field is given as
E=KQ/r²
Now at R,
E=KQ/R²
When, at r=R/4 inside the sphere is given as
Eo=kQ/R³ • R/4
Eo=kQ/4R²
When, r =2R
Then,
E=kQ/(2R)²
E=kQ/4R²
Comparing this to Eo
Then, E=Eo
Then, the electric field at r=2R is equal to the electric field at r=R/4
What is the measure of the ability of a force to rotate or accelerate an object around an axis?
A. Centripetal Force
B. Level Arm
C. Axis of Rotation
D. Torque
D. Torque
Explanation:
Torque is a pattern of the force that can make a victim to revolve on an axis. The torque's direction vector based on the force's direction on the axis. The SI unit for torque is the Newton-meter. Torque can be both static or dynamic.
A static is one that does not generate an angular acceleration.The Torque's magnitude vector ζ for a torque generated by a given force F is
ζ = F .r sin(θ)
where,
r is the width of the moment arm
θ is the angle within the force vector and the moment arm.
In rotational kinematics, torque is measured as
ζ = Iα
Where,
α is the angular acceleration
I is the rotational inertia
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Suppose you want to operate an ideal refrigerator with a cold temperature of − 13.5 °C , and you would like it to have a coefficient of performance of at least 7.75. What is the maximum hot reservoir temperature for such a refrigerator?
Answer:
Th = 50°C
Explanation:
See the attachment below.
K = 7.75
Tc = 13.5°C = 273+13.5 = 286.5K
Answer:
292.984 K or 19.98 °C
Explanation:
Using,
η = Tc/(Th-Tc)................ Equation 1
Where η = coefficient of performance, Tc = cold temperature of the refrigerator, Th = hot temperature of the refrigerator.
Make Th the subject of the equation
η (Th-Tc) = Tc
η Th-η Tc = Tc
Th = (Tc+ηTc)/η ....................... Equation 2
Given: Tc = -13.5°C+273 = 259.5 °C, η = 7.75
Substitute into equation 2
Th = (259.5+7.75×259.5)/7.75
Th = (259.5+2011.125)/7.75
Th = 2270.625/7.75
Th = 292.984 °C or 19.98 °C
Hence the maximum hot reservoir temperature = 292.984 °C or 19.98 °C
A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12.0 A through the body at 25.0 V for a very short time, usually about 3.00 ms.
(a) What power does the defibrillator deliver to the body?
(b) How much energy is transferred?
To solve this problem we will apply the concepts related to Power, such as the product of voltage and current. Likewise, Power is defined as the amount of energy per unit of time, therefore, using this relationship we will solve the second part.
PART A)
We know that Power is,
[tex]P = VI[/tex]
Substitute [tex]25.0V[/tex] for V and 12.0 A for I in the expression,
[tex]P=(25.0V)(12.0A)[/tex]
[tex]P = 300W[/tex]
Therefore the power deliver to the body is 300W
PART B) Converting the time to SI,
[tex]t = 3ms = 0.003s[/tex]
Therefore if we have that the expression for energy is,
[tex]P = \frac{E}{t} \rightarrow E = Pt[/tex]
Here,
E = Energy,
P = Power,
t = Time,
Replacing,
[tex]E = (300W)(0.003s)[/tex]
[tex]E = 0.9J[/tex]
Therefore the energy transferred is equal to 0.9J
This question involves the concepts of electrical power and energy.
(a) The power delivered by the defibrillator is "300 W".
(b) The energy trasferred is "0.9 J".
(a) POWERThe electrical power delivered by the defibrillator can be given by the following formula:
[tex]P=IV[/tex]
where,
P = electrical power = ?I = electric current = 12 AV = voltage = 25 VTherefore,
[tex]P=(12\ A)(25\ V)[/tex]
P = 300 W
(b) Energy
The transferred energy can be given by the following formula:
[tex]P=\frac{E}{t}\\\\E=Pt[/tex]
where,
E = Energy = ?P = Power = 300 Wt = time = 3 ms = 3 x 10⁻³ sTherefore,
[tex]E=(300\ W)(3\ x\ 10^{-3}\ s)\\[/tex]
E = 0.9 J
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A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the tube has magnitude f. You attach the upper end of a lightweight vertical spring of force constant k to the cap at the top of the tube, and attach the lower end of the spring to the top of the cylinder. Initially the cylinder is at rest and the spring is relaxed. You then release the cylinder. What vertical distance will the cylinder descend before it comes momentarily to rest? Express your answer in terms of the variables m, f, and constants g, k.
The cylinder will descend a distance of f/k before coming momentarily to rest.
Explanation:
When the cylinder is released, it will initially accelerate downwards due to the force of gravity. As it accelerates, the spring attached to it will begin to stretch, exerting an upward force on the cylinder. The cylinder will continue to descend until the force exerted by the spring is equal in magnitude to the force of kinetic friction between the cylinder and the walls of the tube. At this point, the net force on the cylinder will be zero, and it will come momentarily to rest.
To find the distance the cylinder descends before coming to rest, we can equate the force exerted by the spring with the force of kinetic friction:
kx = f
Where k is the force constant of the spring and x is the distance the cylinder has descended. Solving for x gives:
x = f/k
Therefore, the vertical distance the cylinder descends before it comes momentarily to rest is given by x = f/k.
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A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 10.0 m: (a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s; (b) he is then lifted at the constant speed of 4.70 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage
Answer:
(a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s - 6106 J
(b) he is then lifted at the constant speed of 4.70 m/s - 5488 J
(c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage - 4869 J
Explanation:
knowing
d = 10 m
m = 56 kg
The work done by the applied force to pull the spelunker is given by
Wa + Wg = Kf - Ki
Wg = -mgd
First
Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]
Wa = (56*9.8*10) + (0.5*56*[tex]4.7^{2}[/tex])
Wa = 6106 J
Second
Kf = Ki
Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]
Wa = 56*9.8*10
Wa = 5488 J
Third
Kf = 0
Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]
Wa = (56*9.8*10) - (0.5*56*[tex]4.7^{2}[/tex])
Wa = 4869 J
The maximum acceleration a pilot can stand without blacking out is about 7.0 g. In an endurance test for a jet plane's pilot, what is the maximum speed she can tolerate if she is spun in a horizontal circle of diameter 85 m?
The pilot's maximum endurance speed in a circle of 85 m diameter without blacking out can be calculated using the centripetal acceleration formula, taking into account the acceleration limit of 7 g, where 1 g equals 9.80 m/s².
Explanation:The question requires the application of concepts from physics, specifically the idea of centripetal force and acceleration when a body moves in a circular path.
Given that the maximum acceleration a pilot can endure without blacking out is 7 g, and knowing that 1 g is equivalent to 9.80 m/s², we can calculate the maximum speed under these conditions using the formula for centripetal acceleration, which is `a = v² / r`, where `a` is the acceleration, `v` is the speed, and `r` is the radius of the circle.
The diameter of the circle is given as 85 meters, which means the radius `r` will be half of that, 42.5 meters. The maximum acceleration of 7 g translates to 7 × 9.80 m/s², which is 68.6 m/s². We can rearrange the formula to solve for `v`, leading to `v = √(a × r)`.
Substituting the values provided, we get `v = √(68.6 m/s² × 42.5 m)
v = √2923.175 m²/s² ≈ 54.1 m/s
Therefore, the maximum speed the pilot can tolerate while spinning in the horizontal circle is approximately 54.1 m/s, which is equivalent to about 192 km/h.
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ]. What is the amplitude of the magnetic field of this wave? A) 0.06 μT B) 0.23 μT C) 1.10 μT D) 1.20 μT
Answer:
Option D is correct.
Explanation:
Bmax = Emax / c
The general form for electromagnetic wave equation is
E = jEmax ×cos(kx-wt)
We were given
(360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ].
So from the equation above
Emax = 360V/m
Bmax = 360/(3×10⁸) = 1.2 ×10‐⁶ T.
Answer
Option D
Amplitude of Magnetic field = B = 1.2×10⁻⁶ T
Explanation:
The relationship between electric field and magnetic field of an electromagnetic wave is given by
B = E/c
Where B is the amplitude of magnetic field and E is the amplitude of electric field and c is the speed of light
The amplitude of electric field is given as 360 V/m
B = (360 V/m)/(3×10⁸ m/s)
B = 1.2×10⁻⁶ V.s/m²
Since 1 Tesla is equal to 1 V.s/m²
B = 1.2×10⁻⁶ T
Therefore, option D is correct
A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field \vec{E} = (100 N/C) \hat{i}E → = ( 100 N / C ) i ^. Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10.0 \text{ cm}R = 10.0 cm centered at the origin. (ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
Φ = 2.26 10⁶ N m² / C
Explanation:
The electric flow is
Ф = E .ds = [tex]q_{int}[/tex] / ε₀
Since we have two components for this flow. The uniform electric field and the flux created by the point charge.
The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface
The flow created by the point load at the origin is
Φ = q_{int} /ε₀
Let's calculate
Φ = 2.00 10⁻⁶ /8.85 10¹²
Φ = 2.26 10⁶ N m² / C
"Sort the following objects as part of the system or not."An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Answer:
0.080625 m/s
Explanation:
Horizontally, the total momentum must be conserved, meaning the momentum before and after the throw must be the same.
The total momentum before the throw is 0, and so is after the throw:
mv + MV = 0
where m, M are the masses of the ball and the quarterback, respectively. v and V are the velocities of the ball and the quarterback, respectively.
0.43*15 + 80V = 0
80V = -6.45
V = -6.45 / 80 = -0.080625 m/s
So he will be moving in the opposite direction with the ball at the rate of 0.080625 m/s
Final answer:
To determine the quarterback's movement after throwing the football, one would need to use conservation of momentum principles. Initial velocity, time in the air, and the maximum height of the football once thrown can be calculated using the equations of projectile motion.
Explanation:
The scenario presented in the question can be analyzed using the principles of conservation of momentum. Since the situation described involves horizontal motion and the exchange of momentum between the quarterback and the football, we will apply the concepts of momentum conservation and the horizontal aspect of projectile motion.
Conservation of Momentum
For part of the scenario where we determine the quarterback's movement after throwing the ball, we use the conservation of momentum which states that the total momentum of a system is conserved if there are no external forces acting on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity. Just before the quarterback throws the ball, both he and the ball are part of the same system, and their combined momentum is zero if he jumps straight up in the air. After throwing the ball, the quarterback will have a backward velocity to conserve the total momentum of the system; we calculate this by equating the momentum of the thrown ball and the momentum of the moving quarterback.
Projectile Motion
In a situation where the football is thrown in a certain manner, the motion of the football can be separated into horizontal and vertical components. The initial velocity of the football can be calculated based on the given range, height, and angle of throw using kinematic equations. The time it takes for the ball to reach the receiver and its maximum height can also be determined through these equations.
Many spacecraft have visited Mars over the years. Mars is smaller than the earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8 m
Incomplete question.The complete one is here
Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?
Answer:
[tex]T=5900s=99min[/tex]
Explanation:
Given
[tex]r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\[/tex]
To find
orbital period of a spacecraft T
Solution
An the initial calculating is computing the angular velocity of satellite :
[tex]w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s[/tex]
Computing T
[tex]T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min[/tex]
If an astronaut can throw a certain wrench 15.0 m vertically upward on earth, how high could he throw it on our moon if he gives it the same starting speed in both places?
Final answer:
The maximum height a wrench can be thrown on the Moon with the same starting speed as on Earth is 6 times higher.
Explanation:
To determine how many times higher an astronaut could jump on the Moon compared to Earth, we need to compare the gravitational accelerations of both locations. The gravitational acceleration on the Moon is about 1/6 of the acceleration on Earth, denoted as g.
When an object is thrown vertically upward, its maximum height is determined by the initial velocity and the gravitational acceleration. Since the astronaut gives the wrench the same starting speed in both places, the maximum height it can reach on the Moon will be 6 times higher than on Earth. This is because the weaker gravitational acceleration on the Moon allows the object to reach a greater height before being pulled back down.
Therefore, if the astronaut can throw the wrench 15.0 m vertically upward on Earth, they could throw it about 90.0 m (15.0 m * 6) on the Moon with the same starting speed.
A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards th
The angular velocity of the merry-go-round increases as a child walks towards the center, due to the conservation of angular momentum and a decrease in the system's moment of inertia.
Explanation:Angular Momentum and Angular Velocity on a Merry-Go-RoundWhen a child walks from the outer edge of a rotating merry-go-round towards the center, the system's angular momentum remains constant due to the conservation of angular momentum. However, the moment of inertia of the system decreases as the child moves closer to the center. Since angular momentum is the product of the moment of inertia and angular velocity (L = I∙ω), and it is conserved, the angular velocity of the merry-go-round must increase to compensate for the decrease in moment of inertia.
This concept can be compared to a figure skater pulling in their arms to spin faster. As the skater's arms come closer to the body, their moment of inertia decreases, causing an increase in their spinning speed, or angular velocity, with no external torques acting on the system.
The same principle applies to the merry-go-round: as the child moves inward, the system's angular velocity increases to conserve angular momentum. This can be observed in playgrounds and is a practical example of conservation laws in rotational motion.
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As the child walks towards the center of the rotating merry-go-round, the system's angular momentum remains constant due to the conservation of angular momentum. The moment of inertia decreases, causing the angular velocity to increase.
A child is standing on the edge of a merry-go-round that is rotating with frequency f.
As the child walks towards the center of the merry-go-round, the angular momentum of the system (child plus merry-go-round) remains constant.
This is due to the conservation of angular momentum, which states that if no external torque acts on the system, the total angular momentum remains unchanged.
Angular momentum (L) is given by the formula:
L = I * ωwhere I is the moment of inertia and ω is the angular velocity.
As the child moves towards the center, the moment of inertia (I) decreases because it depends on the distance from the axis of rotation. To keep angular momentum constant, the angular velocity (ω) must increase.Suppose the initial angular momentum is Lo = Io * ωo. As the child walks inward, the new moment of inertia In becomes smaller, and thus the new angular velocity ωn must be greater to satisfy Lo = In * ωn.complete question:
A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center of the rotating merry-go-round?
A 60-N box rests on a rough horizontal surface with a coefficient of static friction of 0.5. A horizontal force of 23 N acts on the box but the box is observed to be at rest. What is the value of the static friction force
Answer:23 N
Explanation:
Given
Weight of box [tex]W=60\ N[/tex]
Coefficient of static friction is [tex]\mu _s=0.5[/tex]
Applied force [tex]F=23\ N[/tex]
When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.
Thus Static friction [tex]F_s[/tex]=applied force
[tex]F_s=23\ N[/tex]
Although it maximum value can go up to [tex](F_s)_{max}=\mu _sN[/tex]
[tex]F_s=0.5\times 60[/tex]
[tex]F_s=30\ N[/tex]
The speed of a 3.5-N hockey puck sliding across a level icy surface decreases at a rate of 0.45 m/s2. The coefficient of kinetic friction between the puck and the ice is _____. Round your answer to the nearest hundredth.
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.05.
Explanation:
Given that,
Weight of the hockey puck, W = mg = 3.5 N
Acceleration of the hockey puck, [tex]a=0.45\ m/s^2[/tex]
The frictional force opposing the motion of the puck is given by
[tex]F=\mu N[/tex]
This force is balanced by the force due to motion of the hockey puck. So,
[tex]\mu N=ma\\\\\mu mg=ma\\\\\mu=\dfrac{a}{g}\\\\\mu=\dfrac{0.45}{9.8}\\\\\mu=0.045[/tex]
or
[tex]\mu=0.05[/tex]
So, the coefficient of kinetic friction between the puck and the ice is 0.05. Hence, this is the required solution.
Compare waves on a pond and electromagnetic waves. 1. A wave on a pond is a mechanical wave which doesn’t require a medium to travel. 2. A wave on a pond is an electromagnetic wave which doesn’t require a medium to travel. 3. A wave on a pond is an electromagnetic wave which requires a medium to travel 4. A wave on a pond is a mechanical wave which requires a medium to travel.
Answer:
The correct answer is the number 4. A wave on a pond is a mechanical wave which requires a medium to travel.
Explanation:
Mechanical waves are those that need a material medium to propagate. The waves of the sea and the waves that we produce on a guitar string, the sound, are examples of mechanical waves. Electromagnetic waves are energetic pulses capable of propagating in a vacuum. This way, a wave on a pond is a mechanical wave which requires a medium to travel.
Final answer:
A wave on a pond is a mechanical wave requiring water to propagate, while electromagnetic waves, such as light, don't require a medium and can travel through a vacuum. Therefore, statement 4 is correct. This reflects fundamental differences in how mechanical and electromagnetic waves propagate and their need for a medium.
Explanation:
To compare waves on a pond with electromagnetic waves, we should understand that a wave on a pond is a mechanical wave which requires a medium to travel (option 4). This is because mechanical waves are disturbances that propagate through a material medium and the pond water serves as this medium. On the other hand, electromagnetic waves do not need a medium; they can travel through a vacuum because they consist of oscillating electric and magnetic fields.
So, the correct statement would be that a wave on a pond is a mechanical wave which requires a medium to travel. Therefore, the waves on a pond are similar to sound waves since both are mechanical and need a medium. In contrast, light waves are an example of electromagnetic waves, which means they can travel through the vacuum of space without any medium.
Furthermore, all electromagnetic waves move at the same speed in a vacuum, which is known as the speed of light. This is a fundamental difference between mechanical and electromagnetic waves since the speed of mechanical waves depends on the medium in which they are traveling. The discovery that electromagnetic waves do not require a medium led to the abandonment of the aether theory, which was invented to provide a medium for light and other electromagnetic wave propagation.
Show that one mole of any gas occupies a volume of 22.4 L at 1 atm and 0°C i.e., at standard temperature and pressure (STP). [Given: R = 0.0821 L.atm/mol.K] on T 1 . 1 (b) A tank of volume 0.3 m contains 2 moles of oxygen gas at 20°С. Find the average K.E. per atom and also find the rms speed of the atoms. [Given: k = 1.38 x 10-23 J/K; R= 8.317 J/mol, M= 16 g/mol]
Answer:
a) The statement is true, b) [tex]K_{tr.atom} = 6.073 \times 10^{-21} \frac{J}{atom}[/tex],[tex]v_{rms} \approx 477.904 \frac{m}{s}[/tex]
Explanation:
a) The standard conditions are 273.15 K and 1 atm. Let consider that gas behaves ideally, whose equation of state is:
[tex]P \cdot V = n \cdot R_{u} \cdot T[/tex]
The volume is cleared out:
[tex]V = \frac{n \cdot R_{u} \cdot T}{P}[/tex]
By replacing terms:
[tex]V = \frac{(1 mole)\cdot (0.082 \frac{L \cdot atm}{mole \cdot K} )\cdot (273.15 K)}{1 atm}\\V = 22.399 L[/tex]
Therefore, the statement is true.
b) The average kinetic energy per atom is:
[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{R_{u}\cdot T}{N_{A}}[/tex]
Where [tex]N_{A}[/tex] is the Avogadro constante and is equal to [tex]6.022 \times 10^{23} \frac{atoms}{mole}[/tex].
[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{(8.317 \frac{J}{mole\cdot K} )\cdot (293.15 K)}{6.022 \times 10^{23} \frac{atoms}{mole} } \\K_{tr,atom} = 6.073\times 10^{-21} \frac{J}{atom}[/tex]
The rms speed is determined by the following formula:
[tex]v_{rms} = \sqrt{\frac{3 \cdot k \cdot T}{n \cdot M_{O} \cdot \frac{1}{N_{A}} } }\\v_{rms} = \sqrt{\frac{3\cdot(1.38 \times 10^{-23} \frac{J}{K} )\cdot(293.15 K)}{(2 moles) \cdot (16 \frac{g}{mole} )(\frac{1 kg}{1000 g} ) \cdot (\frac{1 mole}{6.022 \times 10^{23} atoms} )}} \\v_{rms} \approx 477.904 \frac{m}{s}[/tex]
Two objects, A with charge +Q and B with charge +4Q, are separated by a distance r. The magnitude of the force exerted on the second object by the first is F. If the first object is moved to a distance 2r from the second object, what is the magnitude of the electric force on the second object?
Final answer:
The magnitude of the electric force on the second object remains the same when the first object is moved to a distance 2r from it.
Explanation:
To find the magnitude of the electric force on the second object (B) when the first object (A) is moved to a distance 2r from B, we can use Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Given that the charge of A is Q and the charge of B is 4Q, the force exerted on B when A is at distance r is F. Now, when A is moved to a distance 2r from B, the new distance between A and B is 2r. Using Coulomb's law and the proportionalities mentioned earlier, the magnitude of the new electric force on B can be calculated as:
F' = (k * Q * 4Q) / (2r)^2
Where k is the constant in Coulomb's law. Simplifying the equation gives:
F' = (4 * F) / 4 = F
Therefore, the magnitude of the electric force on the second object B remains the same when the first object A is moved to a distance 2r from B.