Gievn:
[tex]r^{\prime}^{\prime}=\lbrack4.0\operatorname{cm}+(2.5\operatorname{cm}/s^2)t^2\rbrack\text{ i +(5.0cm/s)tj}[/tex]Part A.
Let's find the magnitude of the dot's average velocity between t = 0 and t = 2.0s.
At t = 0:
Substitute 0 for t
[tex]\begin{gathered} r_1=\lbrack4.0\operatorname{cm}+(2.5\operatorname{cm})0^2\rbrack i+(5.0\operatorname{cm})(0)j \\ \\ r_1=4.0\operatorname{cm}i+0\operatorname{cm}j \end{gathered}[/tex]At t = 2:
Substitute 2 for t
[tex]\begin{gathered} r_2=\lbrack4.0\operatorname{cm}+(2.5\operatorname{cm})2^2\rbrack i+(5.0\operatorname{cm})(2)j \\ \\ r_2=\lbrack4.0\operatorname{cm}+10.0\operatorname{cm}\rbrack i+10.0\operatorname{cm}j \\ \\ r_2=14.0\operatorname{cm}i+10.0\operatorname{cm}j \end{gathered}[/tex]To find the magnitude of the dot's average velocity, apply the formula:
[tex]\begin{gathered} V_{ar}=|\frac{r_2-r_1}{t_2-t_1}| \\ \\ V_{ar}=|\frac{(14.0\operatorname{cm}i+10.0\operatorname{cm}j)-(4.0\operatorname{cm}i+0\operatorname{cm}j)}{2-0}| \\ \\ V_{ar}=|\frac{14.0\operatorname{cm}i-4.0\operatorname{cm}i+10.0\operatorname{cm}j-0\operatorname{cm}j}{2-0}| \\ _{} \\ \\ V_{ar}=|\frac{10.0\operatorname{cm}i+10.0\operatorname{cm}j}{2}| \\ \\ V_{ar}=|\frac{10.0\operatorname{cm}i}{2}+\frac{10.0\operatorname{cm}j}{2}| \\ \\ V_{ar}=|5.0\operatorname{cm}i+5.0\operatorname{cm}j| \\ \\ V_{ar}=\sqrt[]{5^2+5^2} \\ \\ V_{ar}=\sqrt[]{25+25} \\ \\ V_{ar}=\sqrt[]{50} \\ \\ V_{ar}=7.1\operatorname{cm}\text{ /s} \end{gathered}[/tex]Therefore, the magnitude of the dot's average velocity is approximately 7.1 cm/s.
Part B.
FInd the direction angle of the dot's average velocity between t = 0s and t = 2.0 s.
To find the direction angle between t = 0s and t = 2.0s, we have:
[tex]\begin{gathered} \tan V_{ar}=\frac{25}{25} \\ \\ \tan V_{ar}=1 \end{gathered}[/tex]Take the inverse tangent of both sides:
[tex]V_{ar}=\tan ^{-1}(1)=45^{\circ}[/tex]Therefore, the direction of the dot's average velocity between t = 0 and t = 2.0 s is 45 degrees.
ANSWER:
• a) 7.1 cm/s
• b) 45 degrees.
Which has the greatest potential energy and the least amount of kinetic energy
Answer:
Last option
Explanation:
The potentital energy depends on the height, so the greater the height, the greter the potential energy.
On the other hand, the kinetic energy depends on the velocity, so when there is no velocity, there is no kinetic energy.
It means that the option with the greatest potential energy and the least amount of kinetic energy is the last option because it has the skater at a higher position and with the less velocity.
If one liquid is poured into another liquid, which would demonstrate transfer of energy in the form of heat?Select one:a. I & IIb. II & IIIc. I & IIId. I, II & III
b. II & III
Explanation:Note that:
• Heat energy is only transferred if there is a difference in temperature
,• Both glass and pan of water in (I) are at 95°, so there will be be no exchange of heat
,• There is a transfer of heat energy in II & III because the temperatures of the glass and pan of water are different
Based on all the points highlighted above, only II & III would demonstrate transfer of energy in the form of heat
i am studying for a test and i need help answering this question please! i don't know what the answer would be!!
At the moment shown on the figure we notice that the velocity points down, which means that's is direction is pointing south. Now, the acceleration is not shown but since the object is moving in a circle, we know that the acceleration is a centripetal acceleration which means that its direction always point towards the center of the circle along the string; at this moment this means that the acceleration points west.
Therefore, at the moment shown:
The velocity points south.
The acceleration points west.
Note: The direction of the velocity and the acceleration will change as the object moves at any given point their direction will fulfil:
The velocity is tangent to the circle.
The acceleration points towards the center along the string.
D. O3.015 m/sE. O 9.381 m/s7. An object of mass 18 kg moving with a speedof 23 m/s to the right collides with an objectof mass 17 kg moving with a speed of 16 m/sin the same direction. After collision, the 18 kg object movesto the right with a speed of 3 m/s to the right. Calculatethe velocity of the 17 kg object after collision. (1 point)A.40.894 m/sB. O 44.612 m/sC. O 26.024 m/sD.37.176 m/sAE. O 22.306 m/s
We are given the following information
Before collision:
Mass of the 1st object = 18 kg
Initial speed of the 1st object = 23 m/s
Mass of the 2nd object = 17 kg
Initial speed of the 2nd object = 16 m/s
After collision:
Final speed of the 1st object = 3 m/s
Final speed of the 2nd object = ?
Recall from the law of conservation of momentum, the total momentum before the collision and after the collision must be equal.
[tex]\begin{gathered} momentum\;before=momentum\;after \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \end{gathered}[/tex]Let us substitute the given values and solve for the final speed of the 2nd object (v2).
[tex]\begin{gathered} 18\cdot23+17\cdot16=18\cdot3+17\cdot v_2 \\ 686=54+17\cdot v_2 \\ 17\cdot v_2=686-54 \\ v_2=\frac{632}{17} \\ v_2=37.176\;\;\frac{m}{s} \end{gathered}[/tex]Therefore, the velocity of the 17 kg object after the collision is 37.176 m/s
Option D is the correct answer.
Hello, can you please help me check this? I already have an answer
Given:
The force of attraction is F = 10 N.
Let the mass of each ball is m and the distance between them is r.
To find:
The force of attraction when one of the balls has a mass of 2m and the distance between them is 3r.
Solution:
The force of attraction is given by the formula
[tex]F=\frac{Gmm}{r^2}[/tex]Here, G is the gravitational constant.
The new force of attraction will be
[tex]\begin{gathered} F^{\prime}=\frac{Gm(2m)}{(3r)^2} \\ =\frac{2}{9}\times\frac{Gmm}{r^2} \\ =\frac{2}{9}F \\ =2.22\text{ N} \end{gathered}[/tex]Final Answer: The force of attraction when the mass of one of the balls doubled and the distance between them is tripled will be 2.22 N.
please help, i’ll give brainliest!a cheetah, the fastest animal on earth, can run to a maximum speed of 88.1 ft/s. how fast is this in kph? explain your answer with solution.
Answer:
96.67 kph
Explanation:
To convert 88.1 ft/s to kph (kilometer per hour), we will use the following relationships:
1 km = 3280.84 ft
1 hour = 3600 seconds
Then, 88.1 ft/s is equivalent to:
[tex]88.1\frac{ft}{s}\times\frac{1\operatorname{km}}{3280.84\text{ ft}}\times\frac{3600s}{1hour}=96.67\frac{\operatorname{km}}{h}[/tex]Therefore, 88.1 ft/s is equal to 96.67 kph
A 4.00 kg object is moving at 5.00 m/s NORTH. It strikes a 6.00 kg object that is moving WEST at 2.00 m/s. The objects undergo a perfectly inelastic (stick together) collision. The kinetic energy lost in the collision is
We are given that an object is moving north at a speed of 5 m/s, lets call this object 1. We have another object moving west at a speed of 2 m/s, this is object 2. A diagram of the problem is the following:
To determine the loss in kinetic energy we need to determine the difference in kinetic energy before the collision and after the collision:
[tex]\Delta K=K_f-K_0[/tex]The final kinetic energy is:
[tex]K_f=\frac{1}{2}(m_1+m_2)v^2_f[/tex]We use the sum of the masses because the objects are stuck together after the collision. The initial kinetic energy is:
[tex]K_0=\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02}[/tex]Substituting we get:
[tex]\Delta K=\frac{1}{2}(m_1+m_2)v^2_f-(\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02})[/tex]The only missing variable is the final velocity. To determine the final velocity we will use the conservation of momentum.
We will use the conservation of momentum in the horizontal direction (west) and the conservation of momentum in the vertical direction (north).
In the horizontal direction we have:
[tex]m_1v_{h1}+m_2v_{h2}=(m_1+m_2)v_{hf}[/tex]Since the object 1 has no velocity in the horizontal direction we have that:
[tex]\begin{gathered} m_1(0)+m_2v_{h2}=(m_1+m_2)v_{hf} \\ m_2v_{h2}=(m_1+m_2)v_{hf} \end{gathered}[/tex]Now we solve for the final horizontal velocity:
[tex]\frac{m_2v_{h2}}{\mleft(m_1+m_2\mright)}=v_{hf}[/tex]Now we substitute the values:
[tex]\frac{(6kg)(2\frac{m}{s})}{(4kg+6kg)}=v_{hf}[/tex]Solving the operations we get:
[tex]1.2\frac{m}{s}=v_{hf}[/tex]Now we use the conservation of momentum in the vertical direction, we get:
[tex]m_1v_{v1}+m_2v_{v2}=(m_1+m_2)v_{vf}[/tex]Since the second object has no vertical velocity we get:
[tex]m_1v_{v1}=(m_1+m_2)v_{vf}[/tex]Now w solve for the final vertical velocity, we get:
[tex]\frac{m_1v_{v1}}{\mleft(m_1+m_2\mright)}=v_{vf}[/tex]Now we substitute the values:
[tex]\frac{(4kg)(5\frac{m}{s})}{(4kg+6kg)}=v_{vf}[/tex]Now we solve the operations:
[tex]2\frac{m}{s}=v_{vf}[/tex]Now we determine the magnitude of the final velocity using the following formula:
[tex]v_f=\sqrt[]{v^2_{hf}+v^2_{vf}}[/tex]Substituting the values:
[tex]v_f=\sqrt[]{(1.2\frac{m}{s})^2+(2\frac{m}{s})^2}[/tex]Solving the operations:
[tex]v_f=2.53\frac{m}{s}[/tex]Now we substitute this in the formula for the kinetic energy and we get:
[tex]\Delta K=\frac{1}{2}(4kg+6kg)(2.53\frac{m}{s})^2-(\frac{1}{2}(4kg)(5\frac{m}{s})^2+\frac{1}{2}(6kg)(2\frac{m}{s})^2)[/tex]Solving the operations:
[tex]\Delta K=32J-62J=-30J[/tex]Therefore, there was a loss of 30J of kinetic energy.
What acceleration takes an object to 8m/s from 20m/s over a period of 3s?is the answer -4?
In order to calculate the acceleration, let's use the following formula:
[tex]a=\frac{v_2-v_1}{\Delta t}[/tex]Where v2 is the final speed, v1 is the initial speed and Delta t is the period of time.
So, for v2 = 8, v1 = 20 and Delta t = 3, we have:
[tex]a=\frac{8-20}{3}=\frac{-12}{3}=-4\text{ m/s2}[/tex]Which situation is an example of transferring heat by means of convection?Question 5 options:the sun warming the earthsteaming hot cup of teayou wear pajamas in order to get warmyou accidentally grabbing the hot pan from the oven
The weather patterns outside is an example of transferring heat by means of convection.
The correct answer is B
What is it Convection?When heat is transmitted from the a solid surface to the a nonsolid, such air or water, the process is known as convection. Convection is the process through which heat is transported over an interface and involves fluid motion in relation to a solid surface.
By which method convection moves heat?Convection. The exchange of heat through convection, or the movement of a gas or liquid, between two substances. Free convection occurs when warm air and water rises and therefore is replaced by cooler air or water, which causes air or water to move away from of the heated body.
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Charge A(qA=1C) and charge B(qB=2C) are 2 m apart. Charge C(qC=2C) is located between them, and the force on charge C is zero. How far from charge A is charge C?
Given,
Charge A and charge B are 2 m apart.
Let Charge A and Charge C are x m apart
Since force at C is zero.
Thus the force on C due to A and due to B is same
So,
[tex]\begin{gathered} F_{AC}=F_{BC} \\ \Rightarrow k\frac{1C\times2C}{x^2}=k\frac{2C\times2C}{(2-x)^2} \\ \Rightarrow2(2-x)^2=4x^2 \\ \Rightarrow x^2+4x-4=0 \\ \Rightarrow x=\frac{-4\pm\sqrt[]{4^2-4\times1(-4)}}{2\times1} \\ \Rightarrow x=-2\pm2\sqrt[]{2} \\ \Rightarrow x=0.82,-4.82 \end{gathered}[/tex]Since distance cannot be negative.
Thus the distanec between A and C is 0.82 m
Higher frequency sound waves produce quicker images in sonography True or false
The correct answer is TRUE. Since high frequency waves have a higher range, they able to create a better image and create higher quality image. However, the downside to this is that they are not able to create images that can penetrate objects.
If a 97 kg object is subjected to 115 N of force for 8.3 s, what is the change in velocity?
Answer:
The change in velocity was about [tex]9.84~\text{m/s}[/tex]
Explanation:
Step 1: Understand which formulas to use
The problem has the quantities: mass, force, time and change in velocity, and they are all already in their standard form/
We know that change in velocity divided by time equals to acceleration:
[tex]a=\frac{v-v_{0}}{t}[/tex]
Another formula that involves acceleration, mass and force is:
[tex]F=m\times a[/tex]
Step 2: Do the calculations:
The time is said to be [tex]8.3~\text{s}[/tex], so the acceleration will be:
[tex]a=\frac{v-v_{0}}{8.3}[/tex]
Substitute this value and the value for force, mass into the other formula:
[tex]F=m\times a\\115=97\times \frac{v-v_{0}}{8.3}[/tex]
[tex]v-v_{0}[/tex] denotes the change in velocity, so we will have to make it the subject.
[tex]115=97\times \frac{v-v_{0}}{8.3}\\\\\frac{115}{97}=\frac{v-v_{0}}{8.3}\\\\\frac{115}{97}\times 8.3 =v-v_{0}\\\\v-v_{0}\approx 9.84[/tex]
you have three substances in front of you at room temperature. One is solid phase one is in a liquid phase and one is in the gaseous phase. Which one has the strongest attractive forces between molecules
Consider that in a solid phase the molecules are closer toguether in comparisson with the other phases (gas or liquid). It means that the attractive forces between the molecules in a solid substance are strongest, related to the other phases.
Hence, the substance in the solid phase has the strongest attractive forces between molecules.
A gluon is a boson that holds what types of particles together to form baryons?Select one:a.photonb.bosonc.quarkd.lepton
To find
A gluon is a boson that holds what types of particles together to form baryons?
Explanation
Baryons are made up of three types of quarks according to the standard model.
Conclusion
The correct option is
c. quark
determine the total force and the absolute pressure on the bottom oa a swimming pool 28.0m by 8.5 whose uniform depth is 1.8m. what will be the pressure against the side of the side of the pool near the bottom
First, consider that the preassure of a column of water is given by:
[tex]P=\rho gh[/tex]where:
ρ: water density = 1000kg/m^3
g: gravitational acceleration constant = 9.8m/s^2
h: depth of the pool = 1.8m
Replace the previous values of the parameters into the formula for P:
[tex]P=(1000\frac{kg}{m^3})(9.8\frac{m}{s^2})(1.8m)=17640\frac{N}{m^2}[/tex]Now, consider that the surface area of the pool is:
[tex]A=(28.0m)(8.5m)=238m^2[/tex]Then, the total force on the bottom is:
[tex]F=P\cdot A=(17640\frac{N}{m^2})(238m^2)=4,198,320N[/tex]Now, consider that the absolute pressure depends if you take into account the atmospheric pressure or not.
If you consider the atmospheric pressure, then, the absolute pressure is the sum of the pressure due to the water of the pool and the pressure of the atmospheric pressure. Then:
absolute pressure = 101,325N/m^2 + 17,640N/m^2 = 118,965N/m^2
If the atmospheric pressure is not taken into account the absolute pressure is 17,640N/m^2.
Pressure in any direction (i.e. on the wall) very near the bottom will be nearly the same as at the bottom.
Block 1 of mass mi is at rest on a plane that makes an angle of 0 with the horizontal. Thecoefficient of kinetic friction between the block and the plane is p. The block is attachedto block 2 of mass me that hangs freely by an inextensible, light string that passes over africtionless, massless pulley. When the blocks are released, block 1 moves up the inclinewhile block 2 moves down. See Figure 1 for the diagram of the two blocks.m,m2Figure 1: Two blocks connected by a string, with one block resting on an inclined plane.By drawing the free body diagram for each mass and applying Newton's laws, calculate:i) the magnitude of the acceleration of the blocks.ii) the tension in the string connecting the two blocks.
The force diagram of the given system is shown below:
where,
N: normal force = Wy
Wy: component of the weight W perpendicular to the incline = Wcosθ
Wx: component of the weight W parallel to the incline = Wsinθ
W: weight of block 1 = m1*g
W2: weight of block 2 = m2*g
T: tension on the cord
Fr: friction force = μN
i) The acceleration of the two blocks is the same.
The sum of forces on the first block along the direction of the incline is (equation 1):
[tex]\begin{gathered} T-F_r-W_x=m_1\cdot a \\ T-\mu N-m_1g\sin \theta=m_1\cdot a \end{gathered}[/tex]The sum of forces along the direction perpendicular to the incline is (equation 2):
[tex]\begin{gathered} W_y+N=0 \\ N=W_y=m_1g\cos \theta \end{gathered}[/tex]The sum of forces on the second block is (equation 3):
[tex]m_2g-T=m_2\cdot a[/tex]Now, add the first and third equation, solve for a, and replace the expression for N, as follow:
[tex]\begin{gathered} T-\mu N-m_1g\sin \theta+m_2g-T=m_1a+m_2a \\ a(m_1+m_2)=m_2g-\mu N-m_1g\sin \theta \\ a=\frac{m_2g-\mu m_1g\cos\theta-m_1g\sin\theta}{m_1+m_2} \end{gathered}[/tex]The previous expression is the result for the acceleration of the two blocks.
ii) And the tension is:
[tex]\begin{gathered} T=m_2g-m_2a \\ T=m_2g-m_2(\frac{m_2g-\mu m_1g\cos\theta-m_1g\sin\theta}{m_1+m_2}) \end{gathered}[/tex]Which one of the following circuits has the largest resistance?
Given:
Five different circuits are shown here.
To find:
The circuit with the largest resistance
Explanation:
The resistance in a circuit is,
[tex]R=\frac{V}{I}[/tex]or the first circuit, the resistance is,
[tex]\begin{gathered} R_1=\frac{2}{10} \\ =0.2\text{ ohm} \end{gathered}[/tex]For the second circuit, the resistance is,
[tex]\begin{gathered} R_2=\frac{8}{40} \\ =0.2\text{ ohm} \end{gathered}[/tex]For the third circuit, the resistance is,
[tex]\begin{gathered} R_3=\frac{10}{10} \\ =1\text{ ohm} \end{gathered}[/tex]For the fourth circuit, the resistance is,
[tex]\begin{gathered} R_4=\frac{15}{75} \\ =0.2\text{ ohm} \end{gathered}[/tex]For the fifth circuit, the resistance is,
[tex]\begin{gathered} R_5=\frac{20}{5} \\ =4\text{ ohm} \end{gathered}[/tex]Hence, the fifth circuit has the greatest resistance.
Help on this question for 12 points! Please include a deep explanation of how you got your answer!
If they continue at the same speed the total distance traveled after 90 seconds by student 1 and student 2 would be 60 meters and 30 meters respectively.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.
The speed of the student = Total distance /Total time
As given in the problem we have to calculate the total distance traveled after 90 seconds by student 1 and student 2,
Speed of student 1 = 40 / 60 = 2 / 3 m /s
Distance traveled by the student 1 = 2/3 ×90 = 60 meters
Speed of student 2 = 20 / 60 = 1 / 3 m /s
Distance traveled by the student 2 = 1/3 ×90 = 30 meters
Thus, If they continue at the same speed the total distance traveled after 90 seconds by student 1 and student 2 would be 60 meters and 30 meters respectively.
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You are swimming in the ocean and observe that a wave passes by every 2 seconds and thewaves are 2 m away from each other. What is the wave speed of the waves?
The waves move at a pace of 1 m/s.
What is the wave speed equation?Using wavelength and frequency, the wave speed formula is given by: v = f λ where v denotes the wave's speed and λ is wavelength . The wave's frequency is f.
Is a wave's speed constant?A change in the medium is truly necessary to modify the wave's speed since the wave's speed is a feature of the medium. The wave speed is constant if the medium does not alter during the wave's motion.
calculation:-
Wave speed is calculated as frequency times wavelength.
frequency equals 1/time, or 1/2 second.
wavelength is 2 m, and the wave speed is (1/2)*2 m/s, or 1 m/s.
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How can we increase the gravitational potential energy of the bucket of pennies as seen in the image below.Question 3 options:A) take books awayB) add more booksC) take the pennies awayD) remove the basket the books are resting on
We are given a bucket with pennies. We are asked to increase the gravitational potential energy. The gravitational potential energy is the energy that is related to the position of the object and is determined by the following formula:
[tex]U=mgh[/tex]This means that the gravitational potential energy depends on the height and the mass. Therefore, to increase the gravitational potential energy we must increase the height of the bucket. Therefore, we need to add more books. Therefore, the right option is B.
An electric device draws 6.50 A at 240 V. part aIf the voltage drops by 16 %, what will be the current, assuming nothing else changes?Express your answer using two significant figures.part bIf the resistance of the device were reduced by 16 % , what current would be drawn at 240 V?
Given:
The current is I = 6.5 A
The voltage is V = 240 V
To find the value of current
(a) if the voltage is reduced by 16%
(b) if resistance is reduced by 16%
Explanation:
According to Ohm's law,
[tex]V\propto\text{ I}[/tex](a) So, if the voltage reduces by 16%, the current will also reduce by 16%.
The value of the current will be
[tex]\begin{gathered} I_a=I-(I\times\frac{16}{100}) \\ =6.5-(6.5\times\frac{16}{100}) \\ =5.46\text{ A} \end{gathered}[/tex](b) The value of resistance initially is
[tex]\begin{gathered} V=IR \\ R=\frac{V}{I} \\ =\frac{240}{6.5} \\ =36.92\text{ }\Omega \end{gathered}[/tex]If the resistance is reduced by 16%, then the current can be calculated as
[tex]\begin{gathered} I_b=\frac{V}{R^{\prime}} \\ =\frac{V}{R-(R\times\frac{16}{100})} \\ =\frac{240}{36.92-(36.92\times\frac{16}{100})} \\ =7.74\text{ A} \end{gathered}[/tex]Thus, the current value decreases to 5.46 A when voltage decreases but the current value increases to 7.74 A when resistance decreases.
13.An outfielder throws a baseball at a speed of 110 ft/sat an angle of 25.0_ above the horizontal to homeplate 290 ft away. Will the ball reach the catcher onthe fly or will it bounce first?
We will have the following:
[tex]Range=\frac{v^2_i\sin (2\theta)}{g}[/tex][tex]R=\frac{(110ft/s)^2\sin(2(25))}{(32.17ft/s^2)}\Rightarrow R=288.1298651\ldots ft[/tex]From this, we can see that the ball will bounce first.
On the desk, students have a straight, graduated cylinder that contains water up to a height h = 10cm and a Berzelius glass that contains oil. They add a volume of oil to the cylinder, equal to that of water. What hydrostatic pressure is exerted on the bottom of the vessel? See the adjacent figure (g = 10N / kg). The density of the oil is 0,8 g/ cm3 and the density of the water is 1g/cm3.
Pressure (p) is given by
[tex]\begin{gathered} p=h\rho g \\ h=\text{ height of liquid column ;} \\ \rho=\text{ density of liquid ;} \\ g=\text{ acceleration due to gravity;} \end{gathered}[/tex]Here, pressure has two parts -- one due to water column and second due to oil column .
Therefore
[tex]\begin{gathered} p=h_1\rho_1g\text{ +h}_2\rho_2g\text{ }\begin{cases}h\placeholder{⬚}_1={h_2} \\ g={10\text{ N/kg}}\end{cases} \\ \\ p=\text{ \lparen10 }\times0.8\times10)\text{ +\lparen10}\times1\times10)\begin{cases}\rho_1={0.8} \\ \rho_2={1}\end{cases} \\ p=180\text{ N/ m}^2 \end{gathered}[/tex]Final answer is :-
[tex]pressure\text{ = 180 N/m}^2\text{ = 180 Pascal}[/tex]Which of the following statements best describes a wave? A. A wave is energy carrierB. A wave cannot travel in a vacuum C. A wave is a disturbance in space and time D. A wave is a disturbance in space and time that carries energy
ANSWER
D. A wave is a disturbance in space and time that carries energy
EXPLANATION
We want to identify the option that best describes a wave.
A wave can be described as a disturbance in a medium that transports energy without causing a movement of particles. Waves can take the form of electric or magnetic intensity, electric potential, or temperature variations.
There are certain waves that can travel in a vacuum. These waves are called Electromagnetic waves.
Hence, the correct option is option D.
Metalloids have __________ and ___________ properties of metals and nonmetals.
5. They act as ______________ .
According to the research, the correct terms to complete the statements about Metalloids are:
Metalloids have unique and intermediate properties of metals and nonmetals.They act as semiconductors.What are Metalloids?They are very varied elements that have three or more electrons in their valence shell ad Boron (B), Silicon (Si) that conduct electricity more effectively than non-metals, being considered as semiconductors.
In this sense, these have intermediate properties between metals and non-metals in terms of bonds and ionization capacity, being unique among themselves in shape, appearance and color.
Therefore, we can conclude that metalloids are electrical semiconductors whose intermediate properties allow them to act as metals in some situations and as non-metals in others.
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What can quantum computers do more efficiently than regular computers?a. Monitor medical equipment for greater efficiency.b. Automate routine hospital administrative tasks.c. Provide accurate warning of extreme weather events.d. Map disaster areas after major adverse events.
Assuming that a quantum computer could be programed to do any of the scenarios in the given options, then the quantum computer would be able to do every single possible task orders of magnitude more efficiently, due to the nature of its processing "method".
So, all the options will be true for a quantum computer. Assuming, that it can be programmed to do anything like the scenarios described.
URGENNTT
What is the car's velocity between 11h and 15h equation : d f-d i/t f -ti
The velocity of car between the time 11 hr and 15 hr will be
4 m/s².
What is Average velocity?It is defined as the rate of change of displacement with respect to time.
Mathematically -
v = dx/dt
dx = v dt
∫dx = ∫v dt
Δx = v Δt
v = Δx/Δt
Given is an displacement - time graph of a car moving in positive x - direction.
We know that the average velocity is given by -
v = Δx/Δt
From the graph, we can write -
Displacement = Δx = x[15] - x[11]
Time = Δt = t[15] - t[11]
Substituting the values in the formula, we get -
v = (20 - 4)/(15 - 11)
v = 16/4
v = 4 m/s²
The car's velocity will be 4 m/s²
Therefore, the velocity of car between the time 11 hr and 15 hr will be
4 m/s².
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At 3:00 P.M., a bank robber is spotted driving north on I-15 at milepost 124. His speed is 133.0 mi/h. At 3:37 P.M., he is spotted at milepost 181 doing 101.0 mi/h. (Assume a straight highway). Assume north to be positive.What is the bank robber’s displacement during this time interval? Enter a positive value if the displacement is toward north and enter a negative value if the displacement is toward south.
The displacement of an object is defined as a change in the position of the object. If the final position is x_f and the initial position is x_i, the displacement is given by:
[tex]\Delta x=x_f-x_i[/tex]Since the driver is spotted at the milepost 124 and then at the milepost 181, then the final position is 181 miles, and the initial position is 124 miles.
Substitute those values into the equation to find the displacement:
[tex]\Delta x=181mi-124mi=57mi[/tex]Then, the change in position was 57 miles towards the North.
Therefore, during that time interval, the bank robber's displacement was 57 miles towards the North. Then, the answer is: 57.
A/An _____ is the flow of positive charges that move from higher to lower potential energy, whereas a/an _____ is described as a flow of charged particles.conventional current, electric currentelectric current, conventional currentconventional current, electric circuitelectric circuit, conventional current
ANSWER:
conventional current, electric current
STEP-BY-STEP EXPLANATION:
We have that a conventional current is described as the flow of positive charges that move from a higher potential energy to a lower one.
And the electric current is described as a flow of charged particles.
Therefore:
A/An conventional current is the flow of positive charges that move from higher to lower potential energy, whereas a/an electric current is described as a flow of charged particles.
10. Object AB has a mass of 10 kg and a velocity of 2 m/s. It splits into two parts, part A has a momentum of 13 Ns and object B has a velocity of 3.5 m/s. What is the mass of object A and B?
The mass of object A and B are 7.82 kg and 2.18 kg respectively.
Given data:
The mass of object AB is M=10 kg.
The velocity of object AB is V=2 m/s.
The final momentum of object A is p=13 Ns.
The final velocity of object B is v=3.2 m/s.
Applying the conservation of momentum before and after the collision,
[tex]\begin{gathered} MV=p+m_Bv \\ (10)(2)=13+(m_B)3.2 \\ m_B=2.18kg \end{gathered}[/tex]Here, mB is the mass of object B.
The mass of object A can be calculated as,
[tex]\begin{gathered} m_A+m_B=M \\ m_A+2.18=10 \\ m_A=7.82kg \end{gathered}[/tex]Here, mA is the mass of object A.
Thus, the mass of object A and B are 7.82 kg and 2.18 kg respectively.
