Telegraph is the invention that came first.
What is the current produced by a potential difference of 240 Volts through a resistance of .28 ohms
Given:
The potential difference is V = 240 Volts
The resistance is R = 0.28 Ohms
To find the current produced.
Explanation:
The current produced can be calculated by the formula
[tex]I=\text{ }\frac{V}{R}[/tex]On substituting the values, the current produced will be
[tex]\begin{gathered} I=\frac{240}{0.28} \\ =\text{ 857.14 A} \end{gathered}[/tex]Thus, the current produced is 857.14 A
Derive the equation S=ut+½at²
Average velocity = (initial velocity + final velocity )/ 2
AVG v = (u+v)/2
Also,
Distance = avg v x time
s= ([u+v]/2) x t
s= distance travelled by a body in time t
u= initial velocity
a= acceleration
From the equation of motion:
v= u +at
Replacing
s= ( [u+u+at]/2)xt
s= [( 2u+ 2at ( x t ]/2
s= (2ut+at^2)/2
s= ut+1/2at^2
QUESTION 26If the woman in the previous question doubles the constant horizontal force that she exerts on the box to push it on the same horizontal floor,O with a constant speed that is double the speed "vo" in the previous question.with a constant speed that is greater than the speed "vo" in the previous question, but not necessarily twice as great.for a while with a speed that is constant and greater than the speed "vo" in the previous question, then with a speed that increases thereafO for a while with an increasing speed, then with a constant speed thereafter.
When the woman exerts a
Joanne drives her car with a mass of 1000 kg at a speed of 16m/s. what is the
the magnitude of road friction force needed to bring her car to a halt in 14s .
The frictional force will be the product of mass and the deceleration of the car which is 1142.9 N
What is Friction ?Friction is a force that opposes motion. It depends on the surface in contact and independent on the area of the surface.
Given that Joanne drives her car with a mass of 1000 kg at a speed of 16m/s. Before we calculate the the magnitude of road friction force needed to bring her car to a halt in 14s, let us first list out all the necessary parameters
Mass m = 1000 kgInitial velocity u = 16 m/sFinal velocity v = 0 m/sTime t = 14 sAcceleration a = ?Frictional Force F = ?The Frictional Force F = ma
From first equation of motion, v = u - at
Substitute the necessary parameters into the equation
0 = 16 - (a × 14)
14a = 16
a = 16/14
a = 1.143 m/s²
Then Frictional Force F = 1000 × 1.143
F = 1142.9 N
Therefore, the magnitude of road frictional force needed to bring her car to a halt in 14s is 1142.9 N
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The speed of light in an unknown medium is measured to be 2.76 x10% m/s. What is the index of refraction of the medium?
Answer:
Explanation:
The formula for calculating the index of refraction is expressed as
n = c/v
where
n is the index of refraction
v is the speed of light in the medium
c is the speed of light in vacuum and its value is 3 x 10^8 m/s
From the information given,
v = 2.76 x 10^8 m/s
Thus,
n = 3 x 10^8 m/s/2.76 x 10^8 m/s
n = 1.09
the index of refraction of the medium is 1.09
An organ pipe that is closed at one end has a length of 3 m. What is the second longest harmonic wavelength for sound waves in this pipe
Given
Lenght 3m
Procedure
The relationships between the standing wave pattern for a given harmonic and the length-wavelength relationships for closed-end air columns are summarized below.
Harmonic # 3
[tex]\begin{gathered} \lambda=\frac{4}{3}\cdot L \\ \lambda=\frac{4}{3}\cdot3 \\ \lambda=4\text{ m} \end{gathered}[/tex]The second-longest wavelength would be 4m
An il film (n = 1.42) floating on water is84.8 nm thick. What is the wavelengthIN NANOMETERS of the brightest(m = 1) reflected color?(Hint: If you leave the wavelength in nm,the answer will be in nm. No conversionnecessary.)(Unit = nm)
Given:
n = 1.42
Thickness, t = 84.8 nm
m = 1
Let's find the wavelength in Nanometers of the brightest reflected color.
Apply the formula:
[tex]t=(m-\frac{1}{2})(\frac{\lambda}{2n})[/tex]Where:
• t = 84.8 nm
,• m = 1
,• n = 1.42
Plug in the values in the formula and solve for the wavelength λ:
[tex]\begin{gathered} 84.8=(1-\frac{1}{2})(\frac{\lambda}{2*1.42}) \\ \\ 84.8=(\frac{1}{2})(\frac{\lambda}{2.84}) \\ \\ 84.8=\frac{\lambda}{2*2.84} \\ \\ 84.8=\frac{\lambda}{5.68} \\ \\ \lambda=84.8*5.68 \\ \\ \lambda=481.67\text{ nm} \end{gathered}[/tex]Therefore, the wavelength is 481.67 nm.
• ANSWER:
481.67 nm
How much heat is needed to bring 25.5 g of water from 29.3 °C to 43.87 °C.Q =m=ΔT= C=solution =
We are given the following information
Mass of water = 25.5 g
Initial temperature of water = 29.3 °C
Final temperature of water = 43.87 °C
The specific heat capacity of water is 4.186 J/g.°C
The amount of heat required is given by
[tex]\begin{gathered} Q=m\cdot c\cdot\Delta T \\ Q=m\cdot c\cdot(T_f-T_i) \end{gathered}[/tex]Let us substitute the given values into the above formula
[tex]\begin{gathered} Q=25.5\times4.186\cdot(43.87-29.3) \\ Q=1,555.25\; J \end{gathered}[/tex]Therefore, we need 1,555.25 Joules of heat to bring 25.5 g of water from 29.3 °C to 43.87 °C.
You need to construct a 400 pF capacitor for a science project. You plan to cut two L x L metal squares and place spacers between them. The thinnest spacers you have are 0.20 mm thick. What is the proper value of L?Express your answer in centimeters.
Consider that the formula for the capacitance of a square parallel plate capacitor is:
[tex]C=\epsilon_o\frac{A}{d}=\epsilon_o\frac{L^2}{d}[/tex]where A=L^2 is the area of each plate, d is the separation between plates and
ε0 is the dielectric permitivity of vacuum ans is equal to 8.82*10^-12 F/m.
If you solve the previous expression for L and replace the given values for d and C, you obtain:
[tex]\begin{gathered} L=\sqrt[]{\frac{dC}{\epsilon_o}} \\ d=0.20mm=0.20\cdot10^{-3}m=2.0\cdot10^{-4}m \\ C=400pF=400\cdot10^{-12}F=4.00\cdot10^{-10}F \\ L=\sqrt[]{\frac{(2.0\cdot10^{-4}m)(4.00\cdot10^{-10}F)}{8.85\cdot10^{-10}\frac{F}{m}}} \\ L\approx0.0095m=0.95cm \end{gathered}[/tex]Hence, the proper value of L to construct the required capacitor is approximately 0.95cm
what is the wavelength of 2.6 million Hz ultrasound as it travels through human tissue ?
The wavelength of ultrasound waves can be given as,
[tex]\lambda=\frac{v}{f}[/tex]The speed of waves in human tissue is 1540 m/s.
Plug in the known values,
[tex]\begin{gathered} \lambda=\frac{1540\text{ m/s}}{(2.6\text{ million Hz)(}\frac{10^6}{1\text{ million}})}(\frac{10^3\text{ mm}}{1\text{ m}}) \\ =0.592\text{ mm} \end{gathered}[/tex]Thus, the wavelength of waves in human tissue is 0.592 m.
The press box at a basketball park is 38.0ft above the ground. A reporter in the press box looks at an angle of 15 degrees below the horizontal to see second base. What is the horizontal distance from the press box to second base?
The press box at a basketball park is 38.0 ft above the ground.
A reporter in the press box looks at an angle of 15 degrees below the horizontal to see the second base.
Let us draw the diagram to better understand the problem.
Here x is the horizontal distance from the press box to the second base.
With respect to angle 15°, the opposite side is 38 ft and the adjacent side is x.
Recall from the trigonometric ratios,
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \tan 15\degree=\frac{38}{x} \\ x=\frac{38}{\tan 15\degree} \\ x=141.8\; ft \end{gathered}[/tex]Therefore, the horizontal distance from the press box to the second base is 141.8 ft.
An ox exerts a forwards force of 7100 N. If the ox has a weight of 8000 N, what is the minimum coefficient of static friction? (HINT: if there was no static friction the ox would slip and not move forward, what friction is required to allow the ox to move without slipping)
Given data
*An ox exerts a forwards force is F = 7100 N
*An ox has weight is N = 8000 N
The formula for the minimum coefficient of static friction is given as
[tex]\begin{gathered} F=\mu_sN \\ \mu_s=\frac{F}{N} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \mu_s=\frac{7100}{8000} \\ =0.887 \end{gathered}[/tex]Hence, the minimum coefficient of static friction is 0.887
A periodic wave has a frequency of 3.2 Hz. What is the wave period?answer in:____ s
Given:
Frequency of wave, f = 3.2 Hz.
Let's find the period of the wave.
To find the wave period, apply the formula:
[tex]T=\frac{1}{f}[/tex]Where:
T is the period in seconds
f is the frequency hertz.
Thus, we have:
[tex]\begin{gathered} T=\frac{1}{3.2} \\ \\ T=0.3125\text{ s} \end{gathered}[/tex]Therefore, the wave period is 0.3125 seconds.
ANSWER:
0.3125 s
A plant that is 4.1 cm tall is 10.3 cm from a converging lens. You observe that the image of this plant is upright and 6.2 cm tall. What is the focal length of the lens?
We know that the magnification is given by:
[tex]M=\frac{h^{\prime}}{h}[/tex]where h is the height of the object and h' is the height of the image, then in this case we have:
[tex]M=\frac{6.2}{4.1}[/tex]On the other hand we also know that the magnification is given by:
[tex]M=-\frac{i}{o}[/tex]where i is the distance of the image to lens and o is the distance of the object to the lens. From this we have:
[tex]\begin{gathered} \frac{6.2}{4.1}=-\frac{i}{10.3} \\ i=-\frac{10.3\cdot6.2}{4.1} \end{gathered}[/tex]Once we have the distance of the image we can use the lens equation to find the focal point:
[tex]\begin{gathered} \frac{1}{10.3}+\frac{1}{-\frac{10.3\cdot6.2}{4.1}}=\frac{1}{f} \\ \frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2}=\frac{1}{f} \\ f=(\frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2})^{-1} \\ f=30.41 \end{gathered}[/tex]Therefore the focal distance of the lens is 30.41 cm and this can be round to 30 cm, hence the answer is D
An astronaut, of mass 95.0 kg, sits on a newton scale as the shuttle takes off. The newton scale reads 5500 N. What is the acceleration of the shuttle?
Given data:
* The mass of the astronaut is m = 95 kg.
* The force read by the Newton scale when the shuttle takes off is F = 5500 N.
Solution:
According to Newton's second law, the force on the astronaut in terms of the acceleration and mass is,
[tex]\begin{gathered} F=\text{ma} \\ a=\frac{F}{m} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} a=\frac{5500}{95} \\ a=57.9ms^{-2} \\ a\approx58ms^{-2} \end{gathered}[/tex]The shuttle and the astronaut are moving with the same acceleration.
Thus, the acceleration of the shuttle is 58 meters per second squared.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
A. the speed of the object
Explanation:
The color wouldn't make any difference to the potential energy and knowing the mass the way the object is shaped or the height of it would make no difference
How long will it take a sample of Polonium- 194 to decay to 1/16 of its original amount the half -life of 0.7 seconds
The time it takes the Polonium-194 to decay to 1/16 of its original amount is 2.8 seconds.
What is half-life?half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay
To calculate the time it takes the sample of Polonium-196 to decay to 1/16 of its original amount, we use the formula below
Formula:
[tex]2^{n/t}[/tex] = R/R'......... Equation 1Where:
n = Total number of time it takes Polonium-194 to decay to 1/16 of its original amountt = Half-life of Polonium-194R = Original amount of Polonium-194R' = Amount of Polonium-194 after decayFrom the question,
Given:
R = 1R = 1/16t = 0.7Substitute these values into equation 1 and solve for n
[tex]2^{n/0.7}[/tex]= 1/(1/16)[tex]2^{n/0.7}[/tex] = 16[tex]2^{n/0.7}[/tex] = 2⁴Equating the base,
n/0.7 = 4n = 0.7×4n = 2.8 seconds.Hence, the time it takes the Polonium-194 to decay is 2.8 seconds.
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A truck covers 40.0 m in 9.45 s while uniformly slowing down to a final velocity of 2.10 m/s.(a) Find the truck's original speed._____ m/s(b) Find its acceleration._____ m/s2
(a)
In order to find the original speed, let's use the formula below to find an expression for the acceleration:
[tex]\begin{gathered} V=V_0+a\cdot t\\ \\ 2.1=V_0+a\cdot9.45\\ \\ a=\frac{2.1-V_0}{9.45} \end{gathered}[/tex]Now, we can use the following formula to find the initial speed:
[tex]\begin{gathered} \Delta S=V_0t+\frac{at^2}{2}\\ \\ 40=V_0\cdot9.45+\frac{\frac{(2.1-V_0)}{9.45}\cdot9.45^2}{2}\\ \\ 40=9.45V_0+4.725(2.1-V_0)\\ \\ 40=9.45V_0+9.9225-4.725V_0\\ \\ 4.725V_0=40-9.9225\\ \\ V_0=\frac{30.0775}{4.725}\\ \\ V_0=6.3656\text{ m/s} \end{gathered}[/tex](b)
Now, calculating the acceleration, we have:
[tex]\begin{gathered} a=\frac{2.1-6.3656}{9.45}\\ \\ a=-0.4514\text{ m/s^^b2} \end{gathered}[/tex]A rope is vibrating so as to form the standing wave pattern shown at the right. 1.How many nodes are present in the rope?2.What harmonic is shown in the pattern?3.A guitar string has a fundamental frequency of 120 Hz. What is the frequency of the third harmonic of the string?
We will have the following:
1. There are 5 nodes.
2. The harmonic shown is the 4th harmonic.
3. We will detemrine the 3rd harmonic as follows:
[tex]\lambda_3=\frac{3}{2}(120Hz)\Rightarrow\lambda_3=180Hz[/tex]Compare the power of one motorcylce that travels twice as fast as a second indetical motorcycle.A.Same amount of powerB.Twice the powerC.4 times the powerD.Half the power
Given:
The power of one motorcycle travels twice as fast as a second motorcycle.
To compare the power of motorcycles.
Explanation:
Let the speed of one motorcycle be v.
The speed of the second motorcycle will be 2v.
Power is calculated by the formula
[tex]P=\text{ Force}\times velocity[/tex]The power of the first motorcycle will be
[tex]P_1=Fv_1[/tex]The power of the second motorcycle will be
[tex]\begin{gathered} P_2=Fv_2 \\ =F\times2v_1 \\ =2Fv_1 \\ =2P_1 \end{gathered}[/tex]Thus, the power of the second motorcycle is twice the power of the first motorcycle.
Given a material of specific heat c in Cal/gramC^o and mass 6 grams. If the material is heated by absorbing 7 calories of heat then which of these expressions yields the change in temperature of the material in Celsius degrees? A)7 divided by (c - 6) B)7 times (c - 6) C)7 divided by (6 times c)
Given:
The mass is m = 6 grams.
The heat absorbed is Q = 7 calories.
The unit of specific heat, c of the material is cal/gram degree Celsius
To find the change in temperature of the material in degrees Celsius.
Explanation:
The formula to calculate the temperature change is
[tex]\begin{gathered} Q\text{ = mc}\Delta T \\ \Delta T=\frac{Q}{mc} \end{gathered}[/tex]Substituting the values, the change in temperature will be
[tex]\Delta T\text{ = }\frac{7}{6\times c}[/tex]Final Answer: The change in temperature will be 7 divided by (6 times c).
What causes water to freeze solid when placed in a freezer? Takes heat away from the waterAdds heat to the waterPuts electricity in the waterDoes not let light shine on the water
Takes heat away from the water.
Water freezes solid when placed in a freezer.
If heat is added water melts.
Electricity in the water doesn't freeze it.
Light doesn't freeze water.
So, the correct answer is:
Takes heat away from the water
Prove that the area of the parallelogram is equal to | A × B |
The area of a paralellogram with base a and height h is given by:
[tex]A=h\cdot a[/tex]If two adjacent sides of a parallelogram have lengths a and b and are separated by an angle φ, then the base of the parallelogram is a and the height is given by b*sin(φ). Then, the area of the parallelogram is given by:
[tex]A=a\cdot b\cdot\sin (\phi)[/tex]On the other hand, the cross product of two vectors is defined as:
[tex]\vec{a}\times\vec{b}=a\cdot b\cdot\sin (\phi)\hat{n}[/tex]Where the unitary vector is directed toward the direction perpendicular to a and b according to the right hand rule.
The modulus of the cross product of a and b is:
[tex]|\vec{a}\times\vec{b}|=a\cdot b\cdot\sin (\phi)[/tex]We can see that both the area of the parallelogram and the modulus of the cross product have the same expressions. Therefore:
[tex]A=|\vec{a}\times\vec{b}|[/tex]State briefly rules of drawing vector on coordinate systems
In order to draw a vector in the coordinate system, we can follow these steps:
1. Draw the initial point of the vector.
2. Draw the end point of the vector.
3. Create an arrow starting at the initial point and ending at the end point.
For example, let's use a vector that starts at (0, 0) and ends at (3, 4):
As shown in Fig. B9, a block of mass m2=2.1 kg sits on an inclined plane with angle 35 degrees .The coefficients of static and kinetic friction between the block and the plane are unknown.The block is attached to a rope, which is hung over a frictionless and massless pulley and attached to a hanging mass m1.(a) The block my is on the verge of sliding, when the hanging mass is increased to m1=1.6kg.01. find the coefficient of static friction.As shown in Fig. Bob, an identical block (with mass m2) is placed on top of the block on the inclined plane. The hanging mass m1 remains unchanged. The two blocks on the inclined plane are sliding down the incline with the same acceleration a=0.31 m/s^2Find02.the coefficient of kinetic friction.03.what is the minimum value of the coefficient of static friction between the two identical blocks (m2) so that they move together?
Question 1.
The free body diagrams of the situation are shown below:
The forces applied on m1 are in only the y direction then we only have one equation of motion to get it we use Newton's Second law. Then we have:
[tex]T-W_1=m_1a[/tex]The forces applied on m2 are in the x and y direction, this means that we have two equations of motion. We don't expect this block to move on the y-direction (which we defined); this means that the acceleration in this direction has to be zero. Furthermore, since the blocks m1 and m2 are attach thorugh the rope their acceleration has to be the same; this means that the acceleration of block m2 in the x direction has to be a. Applying this and Newton's second law we have for block 2:
[tex]\begin{gathered} (W_2)_x+f_f-T=m_2a \\ N-(W_2)_y=0 \end{gathered}[/tex]where (W2)x and (W2)y denote the component of the weight in the x and y direction, respectively. Now, since block m2 is on the verge of sliding this means that the system is in equilibrium, that is, the acceleration of the system is equal to zero. Then we have that the equations above take the form:
[tex]\begin{gathered} T-W_1=0 \\ (W_2)_x+f_f-T=0 \\ N-(W_2)_y=0 \end{gathered}[/tex]To determine the coefficient of static friction we firs need to determine the force of fricction. To do this we first solve the first equation for T:
[tex]T=W_1[/tex]Now we plug this value in the second equation and solve for the force of friction:
[tex]\begin{gathered} (W_2)_x+f_f-W_1=0 \\ f_f=W_1-(W_2)_x \end{gathered}[/tex]Now we need to remember that the force of friction can be obtain by:
[tex]f_f=\mu_{}N[/tex]where mu is the coefficient of friction (this rule applies for static and kinetic friction). Then we have:
[tex]\begin{gathered} \mu N=W_1-(W_2)_x_{} \\ \mu=\frac{W_1-(W_2)_x}{N} \end{gathered}[/tex]To find the normal force we use the third equation of motion:
[tex]N=(W_2)_y[/tex]Hence we have:
[tex]\mu=\frac{W_1-(W_2)_x}{(W_2)_y}[/tex]Now we need to remember that the components of the weight on an inclined plane are given by:
[tex]\begin{gathered} W_x=W\sin \theta \\ W_y=W\cos \theta \end{gathered}[/tex]where theta is the angle of the plane.
Plugging this on the expression for the coefficient of friction:
[tex]\mu=\frac{W_1-W_2\sin \theta}{W_2\cos \theta}[/tex]Finally we plug the values given to find the coefficient:
[tex]\begin{gathered} \mu=\frac{(1.6)(9.8)-(2.1)(9.8)\sin 35}{(2.1)(9.8)\cos 35} \\ \mu=0.23 \end{gathered}[/tex]Therefore the coefficient of static friction is 0.23
Question 2
For this question we are going to look at the two block in the inclined plane as one block with mass of 4.2 kg and we are going to name the two blocks as m2.
The free body diagram in this case is:
Applying Newton's second law in each mass lead to the system of equations:
[tex]\begin{gathered} T-W_1=m_1a \\ (W_2)_x-f_f-T=m_2a \\ N-(W_2)_y=0 \end{gathered}[/tex]From the first equation of motion we have:
[tex]T=W_1+m_1a[/tex]Plugging this in the second equation:
[tex](W_2)_x-f_f-W_1-m_1a=m_2a[/tex]but we know that:
[tex]f_f=\mu_{}N[/tex]and:
[tex]N=(W_2)_y[/tex]Then we have:
[tex](W_2)_x-\mu(W_2)_y-W_1-m_1a=m_2a[/tex]Solving for the coefficient and using the expression for the wight on an inclined plane we have:
[tex]\begin{gathered} (W_2)_x-\mu(W_2)_y-W_1-m_1a=m_2a \\ \mu=\frac{W_2\sin \theta-W_1-m_1a-m_2a}{W_2\cos \theta} \end{gathered}[/tex]Plugging the values given we have:
[tex]\begin{gathered} \mu=\frac{(4.2)(9.8)\sin 35-(1.6)(9.8)-(1.6)(0.31)-(4.2)(0.31)}{(4.2)(9.8)\cos 35} \\ \mu=0.18 \end{gathered}[/tex]Therefore the coefficient of kinetic friction is 0.18.
Question 3.
To find the coefficient of static friction between the two blocks on the inclined plane we need that the upper block to be stationary with respect to the lower block, that means that the friction has to be equal to the weight in the x-direction, that is:
[tex]\begin{gathered} f_f=W_2\sin 35 \\ \mu W_2\cos 35=W_2\sin 35 \\ \mu=\frac{\sin 35}{\cos 35} \\ \mu=0.7 \end{gathered}[/tex]Therefore the coefficient between the blocks has to be at least 0.7
If a sloth is traveling at 0.067 mps or 0.15 mph how long does it take the sloth to travel an 11.5 meter tree
Answer:
Explanation:
Given:
V = 0.067 m/s
D = 11.5 m
___________
t - ?
t = D / V
t = 11.5 / 0.067 ≈ 172 s or t ≈ 2.87 min or t ≈ 2 min 52 s
A 25.0 kg bag of dirt falls 15.0 m at a construction site. Assuming that all of the heat produced is retained by the dirt, how much will its temperature increase? (cdirt = 0.20 cal/g•°C)
Given:
• Mass of bag, m =25.0 kg
,• Height, h = 15.0 m
,• c = 0.20 cal/g•°C
Let's find by how much the temperature will increase.
Apply the Law of Conservation of Energy:
[tex]mc\Delta T=mgh[/tex]Where:
• m is the mass
,• c is the specific heat capacity
,• g is acceleration due to gravity
,• h is the height.
,• ΔT is the temperature change.
Thus, we have:
[tex]\begin{gathered} mc\Delta T=mgh \\ \\ Eliminate\text{ m on both sides:} \\ c\Delta T=gh \end{gathered}[/tex]Now, plug in the values and solve for ΔT:
[tex]\begin{gathered} \Delta T=\frac{gh}{c} \\ \\ \Delta T=\frac{9.8*15.0}{0.20\times10^3\times4.2} \\ \\ \Delta T=\frac{147}{840} \\ \\ \Delta T=0.175^o\text{ C} \end{gathered}[/tex]Therefore, the temperature change is 0.175° C.
• ANSWER:
0.175° C
this is a 3 part questionplease see picture for part A73) The human brain consumes about 22 W of power under normal conditions, though more power may be required during exams.(a) For what amount of time can one Snickers bar (see the note following Problem 48) power the normally functioning brain(one bar provides 280 calories)? (b) At what rate must you lift a 3.6-kg container of milk (one gallon) if the power output of your arm is to be 22 W? (C) How much time does it take to lift the milk container through a distance of 1.0 m at this rate?
Given,
The power consumed by a brain, P=22 W
The energy per bar, E=280 calories=280×4184=1171.52 kJ
The mass of a milk container, m=3.6 kg
The power output of the arm, P₀=22 W
The distance through which the container needs to be lifted, d=1.0 m
a)
The power is given by,
[tex]P=\frac{E}{t}[/tex]Where t is the time.
On substituting the known values in the above equation,
[tex]\begin{gathered} 22=\frac{1171.52\times10^3}{t} \\ \Rightarrow t=\frac{1171.52\times10^3}{22} \\ =53250\text{ s} \end{gathered}[/tex]That is,
[tex]\frac{53250}{3600}=14.79\text{ hr}[/tex]Therefore one snicker bar can power the brain for 14.79 hr
b)
The power output can also be calculated using the formula,
[tex]\begin{gathered} P=F\times v \\ =mg\times v \end{gathered}[/tex]Where F is the force applied by the container on the arm, v is the rate at which the container must be lifted, and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 22=3.6\times9.8\times v \\ v=\frac{22}{3.6\times9.8} \\ =0.62\text{ m/s} \end{gathered}[/tex]Thus the rate at which the milk container must be lifted is 0.62 m/s
c)
The rate at which the container must be lifted is given by,
[tex]v=\frac{d}{t}[/tex]Where t is the time it takes to lift the container at the calculated rate.
On substituting the known values,
[tex]\begin{gathered} 0.62=\frac{1}{t} \\ \Rightarrow t=\frac{1}{0.62} \\ =1.61\text{ s} \end{gathered}[/tex]Thus it takes 1.61 s to lift the container through 1 m at the given rate.
A block having a mass of m = 19.5 kg is suspended via two cables as shown in the figure. The angles shown in the figure are as follows: a = 15° and B = 35°. We will label the tension in Cable 1 as T1, and the tension in Cable 2 as T2. Solve for T1 and T2
We will have the next diagram
Then we can sum the forces in x and sum the forces in y
Forces in x
[tex]\sum ^{}_{}F_x=-T_1\sin (\alpha)+T_2\cos (\beta)=0[/tex][tex]\sum ^{}_{}F_x=-T_1\sin (15)+T_2\cos (35)=0[/tex]Forces in y
[tex]\sum ^{}_{}F_y=T_1\cos (\alpha)+T_2\sin (\beta)=mg[/tex][tex]\sum ^{}_{}F_y=T_1\cos (15)+T_2\sin (35)=19.5(9.8)[/tex]We simplify the equations found and we found the next system of equation
[tex]\begin{gathered} -T_1\sin (15)+T_2\cos (35)=0 \\ T_1\cos (15)+T_2\sin (35)=191.1 \end{gathered}[/tex]then we isolate the T2 of the first equation
[tex]T_2=\frac{T_1\sin(15)}{\cos(35)}[/tex]We substitute the equation above in the second equation
[tex]T_1\cos (15)+(\frac{T_1\sin(15)}{\cos(35)})\sin (35)=191.1[/tex]we simplify
[tex]T_1(\cos (15)+\frac{\sin (15)\sin (35)}{\cos (35)})=191.1[/tex][tex]T_1(1.147)=191.1[/tex]We isolate the T1
[tex]T_1=\frac{191.1}{1.147^{}}=166.6N[/tex]then we can substitute the value we found in T1 in the euation with T2 isolate
[tex]T_2=\frac{(166.6)_{}\sin (15)}{\cos (35)}=52.54N[/tex]
Which of the following descriptions best describes all of the factors that need to be considered when
determining an object's terminal velocity? (1 point)
O air density and the object's drag coefficient
O the object's weight and area it presents
O the object's weight and area it presents, as well as air density
O the object's weight, length, and width
All of the factors that need to be considered when determining an object's terminal velocity are the object's weight and area it presents, as well as air density.
option C is the correct answer
What is terminal velocity?Terminal velocity is obtained when the speed of a moving object is no longer increasing or decreasing. That is the object's acceleration (or deceleration) is zero.
Mathematically, the formula for terminal velocity is given as;
V = √(2mg)/(ρAC)
where;
m is the mass of the falling objectg is the acceleration due to gravityρ is the density of the fluid through which the object is fallingA is the projected area of the objectC is the drag coefficientThus, the variables to consider in determining terminal velocity of an object incudes the area, density of air, mass, etc.
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