Given:
The specific heat capacity of steel is
[tex]c=\text{ 450 J/kg }^{\circ}C[/tex]The mass of the steel is m = 2 kg
The initial temperature of steel is
[tex]T_i=\text{ 10}^{\circ}\text{ C}[/tex]The final temperature of steel is
[tex]T_f\text{ = 32}^{\circ}C[/tex]
The time is t = 30 s
To find the heat flow rate.
Explanation:
The heat flow rate can be calculated by the formula
[tex]\frac{Q}{t}=\frac{mc(T_f-T_i)}{t}[/tex]On substituting the value, the heat flow rate will be
[tex]\begin{gathered} \frac{Q}{t}=\text{ }\frac{2\times450\times(32-10)}{30} \\ =\text{ 660 W} \end{gathered}[/tex]Thus, the heat flow rate is 660 W.
Consider the direction of the net force in the problem. Which of the following is true? A) The object is pulled by a net force in 2 directions.B) You cannot determine this information from the graphs provided. C) The object is pulled by a net force in only 1 Direction.
Answer:
A) The object is pulled by a net force in 2 directions.
Explanation:
The direction of the force is given by the sign of the force. Since the force is negative in the first interval and then it is positive in the third interval, we can say that the object is pulled in two opposite directions. So, the true statement is:
A) The object is pulled by a net force in 2 directions.
Question 7
A race car traveling at 120 km/h accelerates at a rate of 0.8 m/s2 for 20 s. What is the final speed of the race car?
207.3 km/h
213.8 km/h
177.6 km/h
125.8 km/h
Answer:
177.6 km/h
Explanation:
[tex]\boxed{v=u+at}[/tex]
where:
u is initial velocity in meters per second (m/s).v is final velocity in meters per second (m/s).a is acceleration in meters per second per second (m/s²).t is time in seconds (s).Given values:
u = 120 km/ha = 0.2 m/s²t = 20 sConvert the initial velocity from kilometers per hour to meters per second by dividing the value by 3.6:
[tex]\implies u=\dfrac{120}{3.6}=\dfrac{100}{3}\; \sf m/s[/tex]
Subsist the values of u, a and t into the formula and solve for v:
[tex]\begin{aligned} \textsf{Using} \quad v&=u+at\\\\\implies v&=\dfrac{100}{3}+0.8(20)\\\\&=\dfrac{148}{3}\; \sf m/s\end{aligned}[/tex]
Convert back to km/h by multiplying the value by 3.6:
[tex]\implies v=\dfrac{148}{3} \times 3.6=177.6\; \sf km/h[/tex]
Therefore, the final speed of the race car is 177.6 km/h.
Answer:
The final speed of the race car is 177.6 km/h.
Explanation:
The final speed of the race car can be determined using the following formula:
[tex]\boxed{\rm{\:v = u + at\:}}[/tex]where:
v = final speedu = initial speed (which is 120 km/h)a = acceleration (which is 0.8 m/s^2, but needs to be converted to km/h^2)t = time (which is 20 s)To convert the acceleration from m/s² to km/h², we need to multiply it by (60 x 60) / 1000, which is equivalent to 3.6. So:
[tex]\rm{a = 0.8\: m/s^2 \times 3.6 = 2.88\: km/h^2}[/tex]Substituting the values in the formula, we get:
[tex]\rm{v = 120\: km/h + (2.88\: km/h^2 \times 20\: s)}[/tex][tex]\rm{v = 120\: km/h + 57.6\: km/h}[/tex][tex]\rm{v = 177.6\: km/h}[/tex][tex]\therefore[/tex] The final speed of the race car is 177.6 km/h.
[tex]\blue{\overline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]
Using the parenthesis method, convert the local value of the acceleration of gravity (g=9.798 m/s2) to the referent mph/s.
Given
Gravity = 9.798 m/s2
Procedure
Let's convert m/s2 to mph/s
1 Meters Per Second Squared (m/s2)is equal to
2,24 Miles Per Hour Per Second (mph/s)
Parethesis method:
[tex]9.798\text{ m/s2}\cdot(\frac{2.24\text{ mph/s}}{1\text{ m/s2}})=21.92\text{ mph/s}[/tex]Therefore,
9.798 Meters Per Second Squared (m/s2)is:
21,92 Miles Per Hour Per Second (mph/s)
Force acting on an object or system will NOT change the momentum.
From Newton's second law, the force acting on an object is given by the rate of change of momentum.
That is,
[tex]\begin{gathered} F=\frac{dp}{dt} \\ =\frac{d(mv)}{dt} \end{gathered}[/tex]Where p is the momentum of the object, m is the mass and v is the velocity of the object.
Thus, the force acting on an object will change its momentum.
Therefore the given statement is false.
Amplitude of damped oscillations reduces two folds during one second. Find the time of its five-fold decrease.
The time for five-fold decrease = 2.32 seconds
Explanation:The final amplitude of a damped oscillation is given as:
[tex]A=A_0e^{-kt}[/tex]The amplitude reduces two-folds during one second
That is:
t = 1 second
A = 0.5A₀
[tex]\begin{gathered} 0.5A_0=A_0e^{-kt} \\ \frac{0.5A_0}{A_0}=e^{-kt} \\ 0.5=e^{-k(1)} \\ 0.5=e^{-k} \\ \ln 0.5=-k \\ k=-\ln 0.5 \\ k=0.693 \end{gathered}[/tex]For a five-fold decrease
[tex]\begin{gathered} \frac{A_0}{5}=A_0e^{-kt} \\ 0.2A_0=A_0e^{-kt} \\ \frac{0.2A_0}{A_0}=e^{-kt} \\ 0.2=e^{-kt} \\ \ln 0.2=-kt \\ \ln 0.2=-0.693t \\ -1.609=-0.693t \\ t=\frac{-1.609}{-0.693} \\ t=2.32 \end{gathered}[/tex]The time for five-fold decrease = 2.32 seconds
What is the equivalent resistance in a series circuit if there are three resistors of values 5.00, 2.00, and 3.00?10.001.030.968Ω0.1000
The answer is 10.00
in series circuit , the fo
these are 5 of the hw questions that I'm struggling with, you might have to zoom in a bit.
Given data
*The given time is t = 6 seconds
*The given distance from the base of the cliif is R = 30 m
The formula for the speed is given as
[tex]\begin{gathered} R=ut \\ u=\frac{R}{t} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} u=\frac{30}{7} \\ =4.28\text{ m/s} \end{gathered}[/tex]i need physics help.The charge of an electron is - 1.6 x 10^ -19 C.Show that there are about 3 x 10^18 electrons in 5 x 10^8 nC of charge.
Given:
Charge of an electron = -1.6 x 10⁻¹⁹ C
Let's show that there are about 3 x 10¹⁸ electrons in 5 x 10⁸ nC of charge.
Using the given charge of an electron, to find the number of electrons in 5 x 10⁸ nC of charge, we have:
[tex]\frac{5*10^8*10^{-9}}{1.6*10^{-19}}=3*10^{18}[/tex]Now, let's prove this equation is true.
Simplify the left side:
[tex]\begin{gathered} \frac{5*10^8*10^{-9}}{1.6*10^{-19}} \\ \\ =\frac{5*10^{-1}}{1.6*10^{-19}} \\ \\ =3.125*10^{-18}\text{ electrons} \end{gathered}[/tex]We can see the number of electrons calculated is equivalent to the number of electrons given.
Therefore, we have shown that there are about 3 x 10¹⁸ electrons in 5 x 10⁸ nC of charge.
ANSWER:
Number of electrons = 3.125 x 10¹⁸ electrons
two vectors are defined as follows: A= (-2.2m)x and B = (1.4m)y. (a) Is the magnitude of 1.4 A greater than, less than or equal to the magnitude of 1.2B? (b) Is the x component of 1.4A greater than, less than or the same as the y component of 2.2B?
We are given the following two vectors
[tex]\begin{gathered} \vec{A}=(1.4\; m)\hat{x} \\ \vec{B}=(-2.2\; m)\hat{y} \end{gathered}[/tex](a) The magnitude of vector A (1.4 m) is less than the magnitude of vector B (2.2 m) because the absolute value of vector A (1.4 m) is less than the absolute value of vector B (-2.2 m)
[tex]\begin{gathered} |\vec{A}|=|1.4|=1.4\; m \\ |\vec{B}|=|-2.2|=2.2\; m \end{gathered}[/tex](b) The x component of vector A (1.4 m) is greater than the y component of vector B (-2.2 m) because the x component of vector A (1.4 m) is positive as compared to the y component of vector B (-2.2 m)
A cart on frictionless rollers approaches a smooth, curved slope h = 0.45 meters high. What minimum speed v is required for the cart to reach the top of the slope?
Given data
*The given height is h = 0.45 m
*The value of the acceleration due to gravity is g = 9.8 m/s^2
The formula for the minimum speed (v) required for the cart to reach the top of the slope is given by the conservation of energy as
[tex]\begin{gathered} U_k=U_p \\ \frac{1}{2}mv^2=mgh \\ v=\sqrt[]{2gh} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{2\times9.8\times0.45} \\ =2.96\text{ m/s} \end{gathered}[/tex]Hence, the minimum speed (v) required for the cart to reach the top of the slope is v = 2.96 m/s
Your shopping cart has a mass of 63.9 kilograms. In order to accelerate the shopping cart down an aisle at 0.22 m/s^2, what force (in newtons, N) would you need to apply to the cart?Answer: ________ N
Given:
The mass of the cart is,
[tex]m=63.9\text{ kg}[/tex]The acceleration of the cart is,
[tex]a=0.22\text{ m/s}^2[/tex]To find:
The force applied to the cart
Explanation:
The force on the cart is,
[tex]\begin{gathered} F=ma \\ =63.9\times0.22 \\ =14.06\text{ N} \end{gathered}[/tex]Hence, the force is 14.06 N.
10 Isabella researched how waves travel through the ground during an earthquake.
She drew a diagram of one, called an S wave, moving through Earth's crust.
Wave motion
Based on her diagram, what kind of wave is an S wave?
A light
B sound
C longitudinal
D transverse
PLEASE REPLY ASAP
A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.20 s for the ball to reach its maximum height.(a) Find the ball's initial velocity. ______m/s upward(b) Find the height it reaches. _______m
ANSWER:
(a) 31.36 m/s
(b) 50.2 m
STEP-BY-STEP EXPLANATION:
Given:
Time (t) = 3.2 s
At the maximum height, the velocity is 0
Final velocity (v) = 0 m/s
(a)
We can calculate the initial velocity as follows:
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \\ \text{ we replacing} \\ \\ -9.8=\frac{0-u}{3.2} \\ \\ u=9.8\cdot3.2 \\ \\ u=31.36\text{ m/s} \end{gathered}[/tex](b)
Now the distance is determined with the following formula:
[tex]\begin{gathered} d=ut+\frac{1}{2}at^2 \\ \\ \text{ We replacing:} \\ \\ d=(31.36)(3.2)+\frac{1}{2}(-9.8)(3.2^2) \\ \\ d=100.352-50.176 \\ \\ d=50.176\approx50.2\text{ m} \end{gathered}[/tex]A speedboat increases its speed from 18.5 m/s to 30.6 m/s in a distance of 226 m. Determine the acceleration of the speedboat?
V² = U² + 2aS
30.6² = 18.5² + 2*226a
936.36 = 342.25 + 452a
936.36 - 342.25 = 452a
594.11/452
a= 1.31m/s^2
How do I solve this problem Hint: 1. Draw the forces -rear car: weight downwards, normal force upwards, tension right, drag leftmiddle car: weight downwards, normal force upwards, tension right and left, drag leftlocomotive: weight downwards, normal force upwards, tension left, drag left, thrust/applied right2. The system is moving with a constant speed, so the forces are balanced!3. Start by just looking at the rear car. The tension to the right must balance the drag. This will also be equal to the tension on the middle car. 4. Next, just look at the middle car. The tension to the right must balance the drag and tension to the left. This will also be equal to the tension on the locomotive. 5. Now, just look at the locomotive. The thrust/applied force must balance the drag plus the tension.
The free body diagram of locomotive can be shown as,
Therefore, according to free body diagram, four types of forces are acting on locomotive.
The free body diagram of middle car can be shown as,
Therefore, there are two forces acting on the middle car.
The free body diagram of the last car is shown as,
Therefore, there are three number of forces acting on the last car.
According to free body diagram of locomotive, the tension in cable 2 is equal to net force acting on locomotive. Therefore, the tension in cable 2 is 1000 N.
According to free body diagram of middle car, the tension in cable 1 is equal to tension in cable 2 therefore, the tension in cable 1 is also 1000 N.
The net force generated by locomotive can be given as,
[tex]F=F_{drag}-F_{middle\text{ car}}-F_{\text{rear car}}[/tex]Substitute the known values,
[tex]\begin{gathered} F=1000\text{ N-200 N-200 N} \\ =600\text{ N} \end{gathered}[/tex]Thus, the net force generated by locomotive is 600 N.
If you pick a mass of 11.4 kg from the ground up to 0.84 m above the ground and it takes 2.3 s to do so, how much power is expended?
Power is the rate at which energy is transferred with respect to time.
When you pick a mass m from the ground up to a height h, that mass gains potential energy (U), which is given by:
[tex]U=mgh[/tex]Where g is the gravitational acceleration:
[tex]g=9.81\frac{m}{s^2}[/tex]Find the potential energy that the mass of 11.4kg gains when it is raised 0.84m:
[tex]U=(11.4\operatorname{kg})(9.81\frac{m}{s^2})(0.84m)=93.94056J[/tex]Then, divide the total potential energy by 2.3s to find the power expended in doing so:
[tex]P=\frac{U}{t}=\frac{93.94056J}{2.3s}=40.84372174\ldots W\approx41W[/tex]Therefore, the power expended to raise a mass of 11.4kg up to 0.84m above the ground in 2.3 seconds, is 41 Watts.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Thermal energy can be converted to heat energy.
Option C is correct.
Different Energy Conversion MethodsElectrical energy is a form of chemical energy. Heat energy is a form of thermal energy. Electrical, potential, and other forms of energy can be created from mechanical energy. Heat and light can both be produced with nuclear energy.
What type of energy conversion occurs more during running?Running requires the body to transform potential energy into kinetic energy. A system's internal energy is known as its potential energy. When a system moves horizontally using kinetic energy, potential energy is consumed. Potential energy is held in the form of chemical energy within the human body.
To know more about energy conversion visit:-
https://brainly.com/question/11234965
#SPJ13
the diagram shows four identical magnets that have been dipped into piles of shavings of four different metals. what can be conducted from the diagram above
We can conclude from the experiment that not all the metals can be attracted by a magnet. Since not all the metals could be picked up by the magnet
What is magnetism?A magnet is a material that is able to pick up metallic materials. This implies that the materials can be picked up by the magnets. Metals have magnetic domains in them. These domains get aligned when the magnets have been exposed to a metal.
However, the magnetic domains of all metals do not get aligned when they are exposed to a magnet. As such, not all the metals that we can see are the metals that can be attracted by a magnet.
If we look at the image, we can see that the magnet does attract the cobal and the nickel particles very well but not the aluminium and the copper shavings.
Learn more about magnets:https://brainly.com/question/2841288
#SPJ1
Missing parts;
.
The diagram shows four identical magnets that have been dipped into piles of shavings of four different metals.
What conclusion can be made about a magnet’s ability to attract metals based on what is shown in this diagram?
answer choices
Magnets do not attract all metals
Magnets only attract aluminum and copper.
Magnets attract all metals.
Steel is the only metal that magnets attract.
Chase goes to a restaurant and the subtotal on the bill was xx dollars. A tax of 7% is applied to the bill. Chase decides to leave a tip of 15% on the entire bill (including the tax). Write an expression in terms of xx that represents the total amount that Chase paid.
From the information given, the subtotal on the bill was $x. A tax of 7% was applied on this amount. Recall, percentage is expresed in terms of 100. This means that the amount of tax charged is
7/100 * x = 0.07x
Amount of bill including tax = initial amount + tax
Amount of bill including tax = x + 0.07x = 1.07x
Chase decides to leave a tip of 15% on the entire bill (including the tax). This means that the amount paid as tip is
15/100 * 1.07x = 0.1605x
Total amount paid = amount of bill including tax + amount of tip
Total amount paid = 1.07x + 0.1605x = 1.2305x
The expression for the total amount paid in terms of x is 1.2305x
2 pucks on an air-hockey table. Puck A has a mass of 0.0380 kg and is moving along the x axis with a velocity of + 6.29 m/s. It takes a collision with puck B, which has a mass of 0.0760 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angle shown in the drawing, pluck A at an angle of 65 degrees to the x axis and pluc B at an angle of -37 degrees to the x axis. Find the speed of pluck A and pluck B.
Given,
The mass of puck A, m₁=0.0380 kg
The mass of puck B, m₂=0.0760 kg
The velocity of puck A before the collision, u=+6.29 m/s
The angle made by puck A with the x-axis, θ₁=65°
The angle made by puck B, θ₂=-37°
The momentum is conserved in both directions simultaneously and independently. That is, the sum x-components of the momentum before the collision and after the collision are equal. The same goes for the y-axis.
Considering the x-direction,
[tex]m_1u=m_1v_1\cos \theta_1+m_2v_2\cos \theta_2[/tex]Where v₁ is the velocity of puck A and v₂ is the velocity of puck B after the collision.
On substituting the known values,
[tex]\begin{gathered} 0.0380\times6.29=0.0380\times v_1\times\cos 65^{\circ}+0.0760\times v_2\times\cos (-37)^{\circ} \\ \Rightarrow0.24=0.016v_1+0.061v_2\text{ }\rightarrow\text{ (i)} \end{gathered}[/tex]Considering the y-direction,
[tex]0=m_1v_1\sin \theta_1+m_2v_2\sin \theta_2[/tex]On substituting the known values,
[tex]\begin{gathered} 0=0.0380\times v_1\times\sin 65^{\circ}+0.0760\times v_2\times\sin (-37)^{\circ} \\ 0=0.034v_1-0.046v_2\text{ }\rightarrow\text{ (ii)} \end{gathered}[/tex]On solving equations (i) and (ii),
[tex]\begin{gathered} v_1=3.93\text{ m/s} \\ v_2=2.90\text{ m/s} \end{gathered}[/tex]Thus the speed of pluck A is 3.93 m/s and the speed of pluck B is 2.90 m/s
An elevator starts from rest with a constant
upward acceleration and moves 1 m in the first
1.9 s. A passenger in the elevator is holding a
7.4 kg bundle at the end of a vertical cord.
What is the tension in the cord as the elevator accelerates? The acceleration of gravity
is 9.8 m/s
Answer in units of N
The tension in the cord as the elevator accelerates is 76.59 N.
As the elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.9 s, if the acceleration of the elevator is a, then:
1m = 1/2 a(1.9)²
⇒ a = 2/(1.9²) = 0.55 m/s².
Mass of the bundle; m = 7.4 kg.
Acceleration due to gravity; g = 9.8 m/s².
So, the tension in the cord as the elevator accelerates is = mg + ma
= m(g+a)
= 7.4 ( 9.8 + 0.55) N.
= 76.59 N.
Hence, the tension in the cord is 76.59 N.
Learn more about acceleration here:
brainly.com/question/12550364
#SPJ1
A hypothesis is:an educated guess about a problem or question.the unit of measurement for volume.the official result of an experiment.None of the choices are correct.
Answer:
an educated guess about a problem or question.
Explanation:
An hypothesis is usually concerned with predicting the outcome of an experiment. It is an educated or well structured guess as to what the outcome would be and it is put to test. Thus, the correct option is
an educated guess about a problem or question.
A particularly scary roller coaster contains a loop-the-loop inwhich the car and rider are completely upside down. If the radiusof the loop is 22.7 m with what minimum speed must the cartraverse the loop so that the rider does not fall out while upsidedown at the top? Assume the rider is not strapped to the car.
The forces on the car at the top of the loop are the normal force and the weight, they both points downward. Also the accereation at that point points downward then Newton's second law is:
[tex]-F_N-F_g=-ma[/tex]Now, since this is a circular motion the acceleration is a centripetal one and that is given as:
[tex]a=\frac{v^2}{R}[/tex]where v is the velocity and R is the radius of the circle. Plugging this in Newton's second law we have:
[tex]-F_N-F_g=-m\frac{v^2}{R}[/tex]If the car has the minimum speed to reamin in contact then it would be on the verge of losing contact , this means that Fn=0 at the toop of the loop. Then we have that the equation above:
[tex]\begin{gathered} -F_N-F_g=-m\frac{v^2}{R} \\ 0-mg=-m\frac{v^2}{R} \\ v=\sqrt[]{gR} \end{gathered}[/tex]Plugging the values of the acceleration of gravity and the radius of the loop we have:
[tex]\begin{gathered} v=\sqrt[]{(9.8)(22.7)} \\ v=14.92 \end{gathered}[/tex]Therefore the minimum velocity is 14.92 meters per second.
Matt has a mass of 37 kg and skis down a hill with no friction or air resistance. The hill has a slop of 24°. A. What is the normal force acting on himB. What is his acceleration down hill?C. Now assume there is friction. If the coefficient of kinetic friction between him and the hill is .31, what is his acceleration down the hill?
Given:
Mass of object = 37 kg
Slope = 24 degrees
Let's solve the following questions.
A. Normal force acting on him.
To find the normal force acting on Matt, apply the formula below:
[tex]mg\cos (\theta)-N=0[/tex]Where N is the force.
m is the mass = 37 kg
g is the gravitational acceleration = 9.8 m/s^2
Θ = 24 degrees.
Thus, we have:
[tex]\begin{gathered} 37\ast9.8\cos (24)-N=0 \\ \\ 37\ast9.8(0.9135)-N=0 \\ \\ 331.25-N=0 \end{gathered}[/tex]Add N to both sides:
[tex]\begin{gathered} 331.25-N+N=0+N \\ \\ 331.25=N \\ \\ N=331.25N \end{gathered}[/tex]Therefore, the normal force acting on Matt is 331.25 N
B. Acceleration down the hill.
The acceleration down the hill will be the opposite side(side opposite the angle).
To find the acceleration down the hill, apply the formula below:
[tex]a=g\sin \theta[/tex]Thus, we have:
[tex]\begin{gathered} a=9.8\sin 24 \\ \\ a=9.8(0.4067) \\ \\ a=3.99m/s^2 \end{gathered}[/tex]Therefore, the acceleration down the hill is 3.99 m/s
C. Given:
Coefficient of friction between Matt and the hill is = 0.31
Let's find the acceleration assuming there is friction.
To find the acceleration, we have the formula:
[tex]Fg=m\ast a_g[/tex]Where:
Fg is the force due to gravity
m is the mass of Matt
Thus, we have:
[tex]Fg=37\ast3.99=147.5N[/tex]Also, let's find the force due to fricton using the formula:
[tex]F_f=uN[/tex]Where:
u is the coeficient of friction = 0.31
N is the normal force
We have:
[tex]F_f=0.31\ast147.5=45.7N[/tex]Thus, we have the formula:
[tex]F_g-F_f=m\ast a[/tex]Let's solve for a:
[tex]\begin{gathered} 147.5-45.7=37\ast a \\ \\ 101.8=37\ast a \\ \\ a=\frac{101.8}{37} \\ \\ a=2.75m/s^2 \end{gathered}[/tex]Therefore, the acceleration assuming there is friction is 2.75 m/s^2
ANSWER:
[tex]\begin{gathered} A\text{. 331.25 N} \\ \\ B.3.99m/s^2 \\ \\ \text{ C. 2.75 m/s}^2 \end{gathered}[/tex]How to find the compressive stress in a circular tube
The compressive stress is 177.93 MN/m²
Explanations:1 inch = 0.0254 meters
The outside diameter, D = 2.5 in
D = 2.5 x 0.0254
D = 0.0635 m
The inner diameter, d = 1.5 in
d = 1.5 x 0.0254
d = 0.0381 m
The area of circular tube is calculated as:
[tex]\begin{gathered} A\text{ = }\frac{\pi{}}{4}(D^2-d^2) \\ A\text{ = }\frac{3.142}{4}(0.0635^2-0.0381^2) \\ A\text{ = }0.002m^2 \end{gathered}[/tex]The Area of the circular tube = 0.002 m²
The compressive load = 80 kips
1 kips = 4448.22 N
The compressive load = 4448.22 x 80 N
The compressive load = 355857.6N
[tex]\begin{gathered} \text{Compressive stress = }\frac{Compressive\text{ load}}{\text{Area}} \\ \text{Compressive stress = }\frac{355857.6}{0.002} \\ \text{Compressive stress = }177928800N/m^2 \\ \text{Compressive stress = }177.93MN/m^2 \end{gathered}[/tex]The freezing point of a solution with a nonvolatile solute is higher than that of the pure solvent.O TrueO False
The solute is the substance that is dissolved in the solution and it is generally smaller in quantity.
The freezing point does not depend on the kind of the solute or the size of the solute, it changes when the concentration of the solute changes.
The freezing point of the solution is lower than that of the pure solvent.
Thus, The freezing point of a solution with a nonvolatile solute is lower than that of the pure solvent.
Thus, the above statement is False.
4. The force between two charged balls is 6.0 × 10–6 N. If the distance is doubled and the charge on one ball is doubled, what is the new force between the two charged balls? a. 3.0 × 10–6 N b. 6.0 × 10–6 N c. 3.0 × 10–3 N d. 6.0 × 10–3 N
Given:
The force between two charged balls is,
[tex]F=6.0\times10^{-6}\text{ N}[/tex]The distance between the balls is doubled and the charge on one ball is doubled.
To find:
The new force between the charged balls
Explanation:
Let, the charges are,
[tex]q_1\text{ and q}_2[/tex]The distance between the charges be d, then the force between the charged balls is,
[tex]\begin{gathered} F=k\frac{q_1q_2}{d^2}=6.0\times10^{-6}\text{ N} \\ k=Coulomb^{\prime}s\text{ constant} \end{gathered}[/tex]Now, the distance is
[tex]2d[/tex]and the first charge became,
[tex]2q_1[/tex]The new force is,
[tex]\begin{gathered} F_{new}=k\frac{2q_1q_2}{(2d)^2} \\ =k\frac{q_1q_2}{2d^2} \\ =\frac{F}{2} \\ =\frac{6.0\times10^{-6}}{2} \\ =3.0\times10^{-6}\text{ N} \end{gathered}[/tex]Hence, the new force is,
[tex]\begin{equation*} 3.0\times10^{-6}\text{ N} \end{equation*}[/tex]The average speed of blood in the aorta is 0.348 m/s, and the radius of the aorta is 1.00 cm. There are about 2.00 × 109 capillaries with an average radius of 6.26 μm. What is the approximate average speed of the blood flow in the capillaries?
We are given the following information
Average speed of blood in the aorta = 0.348 m/s
Radius of the aorta = 1 cm = 0.01 m
Number of capillaries = 2.00 × 10^9
Radius of capillaries = 6.26 μm
We are asked to find the average speed of the blood flow in the capillaries.
The incoming volume flow rate of blood in the aorta must be equal to the outgoing volume flow rate in the capillaries times the number of capillaries.
[tex]Q_a=2\times10^9\cdot Q_c[/tex]The volume flow rate can be written as the product of area and speed
[tex]A_a\cdot v_a=2\times10^9\cdot A_c\cdot v_c[/tex]Recall that area is pi into the square of the radius.
[tex]\begin{gathered} \pi(0.01)^2\cdot0.348=2\times10^9\cdot\pi(6.26\times10^{-6})^2\cdot v_c \\ (0.01)^2\cdot0.348=2\times10^9\cdot(6.26\times10^{-6})^2\cdot v_c \\ v_c=\frac{(0.01)^2\cdot0.348}{2\times10^9\cdot(6.26\times10^{-6})^2} \\ v_c=0.44\times10^{-3}\; \; \frac{m}{s} \end{gathered}[/tex]Therefore, the average speed of the blood flow in the capillaries is 0.44×10^-3 m/s
A 12V battery with 0.3 ohm internal resistance is connected to a 24V voltage source charging it.What is the power delivered by the voltage source?
-2880 Watts
Explanation
Step 1
Let
[tex]\begin{gathered} V=24V \\ Internal\text{ resistance of battery=}0.3\Omega \\ \end{gathered}[/tex]
Now, we need aplly the formula
[tex]I(current\text{ in circuit)=}\frac{e.m.f}{\text{total resistance}}=\frac{24\text{v}}{0.3\Omega}=80\text{ Ampers}[/tex]Step 2
now, we need to find the powe delivered, to do that, let's use this formula:
The power delivered by the voltage source is: PV = – I * V =
[tex]\begin{gathered} P=-I\cdot V \\ P=-80\text{ A}\cdot(36)V \\ P=-2880\text{W} \end{gathered}[/tex]hence the answer is -2880 Watt
I hope this helps you
ff
Calculate the acceleration of the elevator for each 5 second interval
Given,
The weight of the student, W=500 N
Thus the mass of the student is given by,
[tex]m=\frac{W}{g}[/tex]Where g is the acceleration due to gravity,
On substituting the known values,
[tex]\begin{gathered} m=\frac{500}{9.8} \\ =51.02\text{ kg} \end{gathered}[/tex]For the first 5 seconds, the net force acting on the student is 0 N. Thus scale reads only his weight. As the net force is zero the acceleration of the elevator is also zero.
In the next 5 intervals, the net force acting on the student is F=200 N, as seen from the diagram.
Thus the acceleration of the elevator is given by the equation,
[tex]F=ma[/tex]Where a is the acceleration of the elevator.
On substituting the known values,
[tex]\begin{gathered} 200=51.02\times a \\ \Rightarrow a=\frac{200}{51.02} \\ =3.92m/s^2 \end{gathered}[/tex]Thus the acceleration in this interval is 3.92 m/s²
During the interval, 10s-15s, the net force acting on the student is zero as seen from the graph. Thus the acceleration of the elevator is also zero.
During the interval, 15 s-20 s, the net force on the student is F=-200 N as seen from the diagram.
Thus the acceleration is,
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \\ a=\frac{-200}{51.02} \\ =-3.92m/s^2 \end{gathered}[/tex]Thus the accelerating in this interval is -3.92 m/s²That is the elevator is accelerating downwards.