Answer:
We had this question yesterday, let me check my book quickly
(A) The acceleration of the ball is 9.81 m/s² just before it strikes the ground.
(B) The initial speed of the ball is equal to 1.14 m/s.
(C) The initial speed must be 3.21 m/s if it is to land with a speed of 5 m/s.
What are the equations of motion?The equation of motion is the way to represent the relation between the time, acceleration, initial and final velocity, and distance covered by a moving object.
The three equations of motion are:
[tex]v= u+ at\\v^2 = u^2 +2aS\\S = ut +(1/2) at^2[/tex]
The acceleration of the ball just before it strikes the ground is equal to gravitational acceleration, g = 9.81 m/s².
Given, the final velocity of the ball, v = 4m/s
The height of the table, h = 0.75 m
The initial velocity of the ball, [tex]u = \sqrt{v^2-2gh}[/tex]
[tex]u = \sqrt{(4)^2-2\times 9.8\times 0.75}[/tex]
u = 1.14 m/s
When the final velocity of the ball, v = 5m/s
The initial velocity will be :[tex]u = \sqrt{(5)^2-2\times 9.8\times 0.75}[/tex]
u = 3.21 m/s.
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You throw an object up with an initial velocity of Voy = 11 m/s from a height of y = 25 m. How long, in seconds, does it take for the object to reach the ground? What is the objects final velocity, in meters per second, as it impacts the ground? Find the time, in seconds, if you instead threw the object DOWN with the same velocity Voy
Calculate the ball's greatest height using the vertical motion model, h = -16t2 + vt + s, where v is the beginning velocity in feet/second and s is the height in feet.
What does a ball being thrown upwards accelerate to?A ball is thrown into the air, where it gradually loses speed until it abruptly comes to a rest at the peak of the motion. The body is travelling upward against gravity at the top, hence the acceleration there is 9.8 ms2. For example, g=9.8 ms2 is the formula for the acceleration caused by gravity.
Only at the greatest point of a body being hurled vertically upwards would velocity be zero because of the constant downward acceleration brought on by the gravitational force. As a result, velocity is zero throughout the rest of the motion.
A ball is originally moving upward when it is tossed into the air, for instance.
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A net force of 43.1 N causes a mass to accelerate at a rate of 0.2 m/s2. Determine the mass in kilograms.
The mass of the an object caused to accelerate at 0.2m/s² by a 43.1 Newton force is 215.5 kilograms.
What is the mass of the object?A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
Force, according to Newton's Second Law is expressed as;
F = m × a
Where a is acceleration and m is the mass.
Given the data in the question;
Force applied F = 43.1N = 43.1kgm/s²Acceleration a = 0.2m/s²Mass m = ?Plug the given values into the formula above and solve for m.
F = m × a
43.1kgm/s² = m × 0.2m/s²
m = 43.1kgm/s² / 0.2m/s²
Mass m = 215.5kg
Therefore, the mass of the object is 215.5 kilograms.
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consider the graph at the right. the object whose motion is represented by this graph is…
Here, the slope of the graph is negative and constant.
Hence, acceleration is constant and is in opposite direction to its motion.
So, The object whose motion is represented by this graph is moving with constant acceleration.
Motion is defined in physics as the phenomenon through which an object changes its location with respect to time. Motion is mathematically characterized in terms of displacement, distance, velocity, acceleration, speed, and frame of reference to an observer, with the change in position of the body relative to that frame measured as time passes. Kinematics is the branch of physics that studies forces and their effects on motion, whereas dynamics is the branch that studies forces and their effects on motion.
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Diffraction is:A.the difference in density of the compression and rarefaction parts of a sound wave.B.the change of frequency heard by an observer when sound waves come from a moving source.C.when waves suddenly appear in a medium without a source.D.the apparent bending of sound waves around obstacles.
Diffraction is a phenomenon that occurs when a wave hits an object or it passes through a small gap.
The propagation of the wave will have a circular pattern after the diffraction effect:
Therefore the correct option is D.
A superelastic collision is one in which1) kinetic energy before the collision equals kinetic energy after the collision.2) kinetic energy after the collision is zero.3) kinetic energy before the collision is less than kinetic energy after the collision.4) kinetic energy before the collision is greater than kinetic energy after the collision
A superelastic collision is the the collision in which the potential energy of the system is converted into the kinetic energy such the kinetic energy of the system after the collision is more than the kinetic energy of the system before the collision.
Hence, 3rd option is the correct answer.
n Fig. P9.82, the cylinder Figure P9.82 and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is Pulley wrapped around the cylinder, passes over the pulley, and has a 3.00 kg box Cylinder Box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 2.50 m.
The speed of the box when it has fallen 2.50 m is 4.22 m/s.
What is the speed of the box?
The speed of the box when it has fallen through the given height is calculated as follows;
Apply the principle of conservation of energy to determine the speed of the box.
ΔK.E = ΔP.E
K.Ef - K.Ei = mg(hf - hi)
K.Ef - 0 = mg(hf - 0)
K.Ef = mghf
where;
K.E is the final kinetic energy = rotational + translational kinetic energyhf is the final height of the box¹/₂mv² + ¹/₂I_pω² + ¹/₂I_cω²= mghf
¹/₂mv² + ¹/₂(I_p + I_c)ω² = mghf
where;
I_p is moment of inertia of the pulleyI_c is the moment of inertia of the cylinderω is the angular speed of the boxm is mass of the boxI_p = ¹/₂MR²
where;
M is mass of the pulleyR is the radius of the pulleyI_p = ¹/₂(2)(0.2)² = 0.04 kgm²
I_c = MR²
I_c = (5)(0.4)²
I_c = 0.8 kgm²
¹/₂mv² + ¹/₂(I_p + I_c)(v/r)² = mghf
¹/₂mv² + ¹/₂r²(I_p + I_c)v² = mghf
¹/₂v²[m + 1/r²(I_p + I_c)] = mghf
v²[m + 1/r²(I_p + I_c)] = 2mghf
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
8.25v² = 147
v² = 147/8.25
v² = 17.8
v = √17.8
v = 4.22 m/s
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5. Was you hypothesis supported by the data collected for question 2? Why or why not ?
Answer:
Your hypothesis was not supported by the data collected
Explanation:
On question 2 you said that the point with the greatest kinetic energy would be point 5. However, when you analyze the data, the point with the greatest velocity was point 2 which means that this is the point with the greatest kinetic energy.
Therefore, your hypothesis was not supported by the data collected because based on the picture, it is hard to say where is the lowest point of the roller coaster.
What is the “time constant” for a capacitor, and why is it important?
We know that in a RC circuit the voltage in the capacitor when is charging is given by:
[tex]V_C(t)=V_0(1-e^{-\frac{t}{RC}})[/tex]when this is happening the voltage in the resistor is given by:
[tex]V_R(t)=V_0e^{-\frac{t}{RC}}[/tex]In both equations V0 denotes the voltage given by the source, R is the resistance of the resistor and C is the capacitance of the capacitor.
We notice that in both expressions the product RC appear, this product is what we call the time constant of the capacitor; and it is important since it determines the time intervals in which the voltage, charges and currents chage in a RC circuit. This means that while the capacitor is charging or discharging the variables mentioned will always have the time constant in their expressions.
FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of inertia of the pulley is 2 kg m². (i) Sketch the free body diagram of the 1.5 kg block.(ii) When the mass is released from rest, calculate the angular velocity and number of revolutions of the pulley at t = 4.2 s.
Part (i)
Free body diagram of the 1.5 kg block;
Part (ii)
Only 1 force is acting on the pulley is the weight of the block attached with the sting. The torque acting on the pulley is given as,
[tex]\begin{gathered} \tau=F\times r \\ =Fr\sin \theta \\ =mgr\sin \theta \end{gathered}[/tex]Here, g is the acceleration due to gravity and the θ is the angle between force F and r (as force is acting tangentially hence θ=90°)
Substituting all known values,
[tex]\begin{gathered} \tau=(1.5\text{ kg})\times(9.8\text{ m/s}^2)\times(20\text{ cm})\times\sin (90\degree) \\ =(1.5\text{ kg})\times(9.8\text{ m/s}^2)\times(20\text{ cm})\times(\frac{1\text{ m}}{100\text{ cm}})\times1 \\ =2.94\text{ N}\cdot m \end{gathered}[/tex]In rotational dynamics torque is given as,
[tex]\tau=I\alpha[/tex]Here, I is the moment of inertia of the pulley (I=2 kg.m²) and α is the angular acceleration.
The angular acceleration is given as,
[tex]\alpha=\frac{\tau}{I}[/tex]Substituting all known values,
[tex]\begin{gathered} \alpha=\frac{2.94\text{ N.m}}{2\text{ kg.m}^2} \\ =1.47\text{ rad/s}^2 \end{gathered}[/tex]The angular velocity is given as,
[tex]\omega=\alpha t[/tex]Here, t is the time.
Substituting all known values,
[tex]\begin{gathered} \omega=(1.47\text{ rad/s}^2)\times(4.2\text{ s}) \\ =6.174\text{ rad/s} \end{gathered}[/tex]Therefore, the angular velocity of the pulley is 6.174 rad/s.
The angular displacement of the pulley in 4.2 s is given as,
[tex]\Theta=\omega t[/tex]Substituting all known values,
[tex]\begin{gathered} \Theta=(6.174\text{ rad/s})\times(4.2\text{ s}) \\ =25.9308\text{ rad} \end{gathered}[/tex]The number of revolutions of the
Given two 2.00μC charges on the horizontal axis are positioned at x=0.8m, and the
other at x=-0.8m, and a test charge q = 1.28x10-18C at the origin.
(a) What is the net force exerted on q by the two 2.00μC charges? [5]
(b) What is the electric field at the origin due to 2.00μC charges? [5]
(c) What is the electric potential at the origin due to the two 2.00μC charges
Answer:Question 1
Given q1=2µC
q2=2µC
q= 1.2×10^-18C at origin
Net force exerted by two charges on q
F_1= force on q due to q1
F_2= force on q due to q2
F_net= F_(1-) F_2
= (Kqq_1)/r^2 - (kqq_2)/r^2 Then q_1=q_2=〖2×10〗^(-6)
F_net=0N
b) The electric field at charge q
E_net= E_1- E_2
= (kq_1)/r^2 - (kq_2)/r^2
Then q_1=q_2
E)_net= 0 N/C
c) The electric potential at origin due to two charge
V_net= V_(1 )- V_2
= (kq_1)/r - (kq_2)/r
Then q1= q2
V_net= 0 V
Explanation:
A jogger jogs from one end to the other of a straight track in 1.17 min and then back to the starting point in 1.67 min. What is the jogger’s average speed in jogging to the far end of the track (assuming the track is 100 m long) in m/s?
ANSWER:
1.21 m/s
STEP-BY-STEP EXPLANATION:
Given:
One way time = 1.17 min
Return time = 1.67 min
1 minute is 60 seconds, therefore:
One way time = 1.17 min = 1.17 * 60 = 70.2 sec
Return time = 1.67 min * 60 = 100.2 sec
We calculate the speed for each journey, knowing that the distance traveled is 100 meters, like this:
[tex]\begin{gathered} v=\frac{d}{t} \\ \\ v_1=\frac{d}{t_1}=\frac{100}{70.2}=1.4245\text{ m/s} \\ \\ v_2=\frac{d}{t_2}=\frac{100}{100.2}=0.998\text{ m\/s} \end{gathered}[/tex]Therefore, the average speed would be:
[tex]\begin{gathered} v=\frac{v_1+v_2}{2}=\frac{1.4245+0.998}{2}=\frac{2.4225}{2} \\ \\ v=1.21\text{ m/s} \end{gathered}[/tex]The average speed is 1.21 m/s
A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°. How far is the ball from the football player when it lands? How much farther would the ball go if he kicked it with the same speed, but at a 45° angle? Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta) \\ v_y=v\sin (\theta) \end{gathered}[/tex]Where v = 16 m/s and θ = 63°
[tex]\begin{gathered} v_x=16\cos (63\degree)=7.26\; \frac{m}{s} \\ v_y=16\sin (63\degree)=14.26\; \frac{m}{s} \end{gathered}[/tex]How far is the ball from the football player when it lands?
The range of the ball is given by
[tex]x=v_x\times t[/tex]Where t is the time the ball remains in the air.
The time (t) can be found as
[tex]y=v_yt+\frac{1}{2}at^2[/tex]y = 0 when the ball is in the air.
The acceleration is due to gravity (-9.8 m/s²)
[tex]\begin{gathered} 0=14.26t+\frac{1}{2}(-9.8)t^2 \\ 0=14.26t-4.9t^2 \\ 0=t(14.26-4.9t) \\ 0=14.26-4.9t \\ 4.9t=14.26 \\ t=\frac{14.26}{4.9} \\ t=2.91\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=7.26\times2.91=21.13\; m[/tex]Therefore, the ball will land 21.13 meters far from the football player.
How much farther would the ball go if he kicked it with the same speed, but at a 45° angle?
We need to repeat the above calculations
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta)=16\cos (45\degree)=11.31\; \frac{m}{s} \\ v_y=v\sin (\theta)=16\sin (45\degree)=11.31\; \frac{m}{s} \end{gathered}[/tex]The time (t) is given by
[tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=11.31_{}t+\frac{1}{2}(-9.8)t^2 \\ 0=11.31_{}t-4.9t^2 \\ 0=11.31_{}-4.9t \\ 4.9t=11.31_{} \\ t=\frac{11.31_{}}{4.9} \\ t=2.31\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=11.31\times2.31=26.13\; m[/tex]Therefore, the ball will land 26.13 meters far from the football player.
Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
The ball kicked at 16 m/s and at a 63° angle takes 2.91 s to land.
The ball kicked at 9 m/s and at a 45° angle will take
[tex]v_y=9\sin (45\degree)=6.36\; \frac{m}{s}[/tex][tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=6.36t+\frac{1}{2}(-9.8)t^2 \\ 0=6.36t-4.9t^2 \\ 0=6.36-4.9t \\ 4.9t=6.36 \\ t=\frac{6.36}{4.9} \\ t=1.30\; s \end{gathered}[/tex]So, the ball kicked at 9 m/s and at a 45° angle takes 1.30 s to land.
Therefore, the ball kicked at 9 m/s and at a 45° angle will land first.
An ant can crawl 72 inches every 6 minutes. The table below shows the distance the ant can travel for different amounts of time at this rate.According to the table, if the ant moves at a constant speed, what is its speed? (Remember that speed is a unit rate).
Given,
The ant crawls 72 inches every 6 minutes.
And the table of the time and distance covered by the ant.
From the table, the values of distance are,
d₁=6 inches
d₂=12 inches
d₃=18 inches
And the values of corresponding time are,
t₁=0.5 min
t₂=1 min
t₃=1.5 min
The speed of an object is given by the ratio of the distance covered by the object to the time it takes the object to cover the distance.
Thus the speed of the ant is given by,
[tex]v=\frac{d_1}{t_1}=\frac{d_2}{t_2}=\frac{d_3}{t_3}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\frac{6}{0.5}=\frac{12}{1}=\frac{18}{1.5} \\ =12\text{ imches/min} \end{gathered}[/tex]Thus the speed of the ant is 12 inches/min
It takes 225 kJ of work to accelerate a car from 20.1 m/s to 28.1 m/s. What is the car's mass?
Answer:
THE REMAINIG WILL BE 75
Explanation:
HOPE IT HELPS YOU
Select all that apply
Which conditions are necessary for the diffusion of a substance to occur across a membrane?
a concentration gradient
a supply of energy
membrane permeability
Answer:
a concentration gradient.
membrane permeability.
Explanation:
diffusion occurs down a concentration gradient.
and is the free net movement of particles therefore, does not require energy.
membrane permeability is necessary to ensure that the particles can pass through the cell membrane.
describe how moral relativism was influenced by einstein theories of relativity and subsequently the trend toward the idea there are no absolutes?
The special and general theories of relativity and Albert Einstein's audacious theory that light is a particle are his most famous works as a physicist and Nobel winner. The most well-known scientist of the 20th century is perhaps him.
In March 1879, he was born in Ulm, Württemberg. He had a great interest in nature and the capacity to comprehend challenging mathematical ideas even as a young man in Munich. He had an unremarkable high school experience, doing exceptionally well in arithmetic but completely failing the classics, which were then thought to be crucial for anybody planning to attend college. He detested school's dreary regimentation and uncreative atmosphere.
The second study established a lot of information regarding the nature of molecules and explained Brownian motion, which is the random jostling of molecules floating in a fluid. 16 years later, this study helped him win the physics Nobel Prize.
However, his third work, "On the Electrodynamics of Moving Bodies," was left out of the award's wording. The third article was the one that would have the biggest impact on contemporary physics. It included Einstein's Special Theory of Relativity, which greatly simplified how we think about how radiation, like light, interacts with matter. Speaking about one body moving and another being motionless has no real significance, according to Einstein. Only in connection to one other can bodies be conceived of as moving;
This specifically implies that, regardless of the frame of reference, electromagnetic radiation's (such as light's) speed remains constant. Even well-known scientists struggled to comprehend this theory because of Einstein's insightful and audacious viewpoint. But over time, when the predictions made by his theory were repeatedly verified, the Special Theory of Relativity finally transformed how scientists thought about matter, space, time, and everything that interacts with them.
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Number 1. Part b: what are the final kinetic energy of the system
Given that there is a cart of mass, m = 0.12 kg moving with initial speed of, u1 = 0.45 m/s and it collides with another cart of mass, m = 0.12 kg with initial speed, u2 = 0 m/s
We have to find the initial and final kinetic energy.
(a) Initial kinetic energy,
[tex]\begin{gathered} K\mathrm{}E.1=\frac{1}{2}mv^2 \\ =\frac{1}{2}\times0.12\times(0.45)^2 \\ =0.012\text{ J} \end{gathered}[/tex]According to the conservation of linear momentum,
[tex]mu1+mu2=2mv[/tex]Here, v is the final speed.
[tex]\begin{gathered} 0.12\times0.45=2\times0.12\times v \\ v=\frac{0.45}{2} \\ =0.225\text{ m/s} \end{gathered}[/tex]Here, the final speed is 0.225 m/s.
(b) The formula to find kinetic energy is
[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(2m)v^2[/tex]Substituting the values, we get
[tex]\begin{gathered} K\mathrm{}E\mathrm{}=0.12\times(0.225)^2 \\ =6.075\times10^{-3}\text{ J} \end{gathered}[/tex]Hence the kinetic energy is 6.075 x 10^(-3) J.
A 150 m long train entered the 450 m long bridge. From the entrance locomotives only after the last wagon passed 10 minutes. At what speed the train was going?
The train was going at the speed of 1m/s on the bridge.
The length L of the bridge is 450m.
The length B of the train is 150m.
After entering the bridge,
The last wagon comes out after it has passed ten minutes.
So, the speed of the train can be found out by using the formula,
Speed = Total Distance/total time
Here, total time is 10 minutes,
We know,
One minute = 60 seconds
10 minutes = 10×60 seconds.
Total time = 600 seconds.
Total distance travelled by the train is (L+B) when it comes out from the bridge,
So, putting all the values,
Speed = (450+150)/600
Speed = 600/600
Speed = 1m/s.
So, the speed of the train is 1m/s.
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this is a 2 part question28) A 1100-kg car coasts on a horizontal road with a speed of 19 m/s. After crossing anunpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net work done on the car positive, negative, or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section.
ANSWER:
(a) negative
(b) -120.3 N
STEP-BY-STEP EXPLANATION:
(a)
The net work done on the car is negative because its speed decreases from 19 m/s to 12 m/s
(b)
The acceleration is found from the equationof motion that:
[tex]v^2-u^2=2\cdot a\cdot s[/tex]Replacing and solving for a:
[tex]\begin{gathered} 12-19=2\cdot a\cdot32 \\ -7=64a \\ a=-\frac{7}{64}\frac{m}{s^2} \end{gathered}[/tex]Therefore, the force would be:
[tex]\begin{gathered} F=m\cdot a \\ F=1100\cdot-\frac{7}{64} \\ F=-120.3\text{ N} \end{gathered}[/tex]Suppose you walk 16 m straight east and then 22 m straight south. At what angle, in degrees South of East, is a line connecting your starting point to your final position?
53.9 angle, in degrees South of East and 27.20 m is a line connecting your starting point to your final position.
What is initial position and final position?The distance in decimeters between the starting point and the ending position is in a straight line. The distance between an object's original position and its ultimate position is known as displacement.
Briefing:You walk 16 m straight east and then 22 m straight south. This forms a right angled triangle with a horizontal distance of 16 m, a vertical distance of 22 m and the hypotenuse is the distance between the ending and starting point. Let x represent the distance between the ending and starting point. Using Pythagoras theorem:
x² = 16² + 22²
x² = 256 + 484
x² = 740
Taking square root of both sides:
√x² = √740
x = √740
x = 27.20 m = distance between the ending and starting point.
Now use trigonometry:
sinθ=B/R
sinθ=22/27.20
sinθ= 0.808
θ = 53.9 degree. This is your angle.
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Steve takes his car out for a joy ride and travels 400 meters north. He then travels 100 meters east and picks up his buddy Frank. They then stop at a 7-11 which is 200 meters south from Frank’s house. If the total trip takes 10 minutes, determine the average velocity of Steve’s car. (Draw a picture!) NEED HELPP ASAPPPP !!!!
The average velocity of Steve's car is 1.17m/s.
How to calculate average velocity?Average velocity is the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.
It can be calculated by dividing the total distance of a moving body by the time taken.
According to this question, Steve takes his car out for a joy ride and travels 400 meters north. He then travels 100 meters east and picks up his buddy Frank. They then stop at a 7-11 which is 200 meters south from Frank’s house.
The total distance traveled by Steve is 400m + 100m + 200m = 700m.
Average velocity = 700m ÷ 600s
Average velocity = 1.17m/s
Therefore, 1.17m/s is the average velocity of the car.
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Information givenknown: mass of Christine=60 kgmass of cart= 22 kg mass of hailey=69The two girls on the cart to the left pushing off of Conner take .3833 s to reach a distance of 0.3m. Conner reaches the same distance in 0.2333s. What is the mass of Conner?
We will have the following:
We use conservation of momentum to solve, that is:
[tex]\begin{gathered} (60kg+22kg+69kg)(0.3m/0.3833s)=(m+22kg)(0.3m/0.233s) \\ \\ \Rightarrow118.1841899kg\ast m/s=(m+22kg)(\frac{300}{233}m/s) \\ \\ \Rightarrow m+22kg=91.78968976kg\Rightarrow m=69.78968976... \\ \\ \Rightarrow m\approx69.8 \end{gathered}[/tex]So, Conner's mass is approximately 69.8 kg.
Which atoms were cations?
Alkali and alkaline earth metals invariably produce cations.
What kind of compounds are cations?Calcium (Ca2+), potassium (K+), and hydrogen (H+) are a few examples of cations.
What does common cation mean?Positively charged ions are referred to as cations. Less electrons than protons make up cations. An ion can be made up of a single atom of an element (a monatomic ion, monatomic cation, or monatomic anion) or many atoms that are chemically connected to one another (a polyatomic ion or polyatomic cation or anion)
Alkali and alkaline earth metals always create cations, whereas halogens always produce anions. Most nonmetals normally create anions, while the majority of other metals typically produce cations (such as iron, silver, and nickel) (e.g. oxygen, carbon, sulfur)
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A boat heads East across a South flowing river. The river is 246 m wide and flows at a rate of 6.4 m/s. The boat sails across at 5.3 m/s..
How long will it take the boak to cross the river?
If a boat heads East across a South flowing river. The river is 246 m wide and flows at a rate of 6.4 m/s. The boat sails across at 5.3 m/s, then the time taken by the boat to cross the river would be 46.41 seconds.
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.
As given in the problem a boat heads East across a South flowing river. The river is 246 m wide and flows at a rate of 6.4 m/s. The boat sails across at 5.3 m/s.
Time taken by the boat to cross the river = 246/5.3
=46.41 seconds
Thus, the time taken by the boat to cross the river would be 46.41 seconds.
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Determine the net force (in N) necessary to give an object with a mass of 2.10 kg an acceleration of 5.20 m/s².N
Newton's second law states that the force is equal to the rate of change of momentum; for a constant mass, force equals mass times acceleration. In mathematical terms this means that:
[tex]F=ma[/tex]where m is the mass and a is the acceleration.
In this case the mass is 2.10 kg and the acceleration is 5.20 m/s²; plugging these values in Newton's second law we have that:
[tex]\begin{gathered} F=(2.10)(5.20) \\ F=10.92 \end{gathered}[/tex]Therefore, the force needed is 10.92 N
Three blocks are sitting on a horizontal, frictionless table. They
are pushed from the left by an applied force F = 10 N, as
shown. How much force does block 3 exert on block 2, if
m₂ = 1 kg, m₂ = 2 kg, m₂ = 3 kg ?
a) 3 N
b) 5 N
c) 6 N
d) 8 N
Newton's laws of motion, which are composed of three basic principles, describe the interaction between an object's motion and the forces acting on it.
These laws can be summarized as follows: A body remains at rest or in motion in a straight line at a constant speed unless moved upon by a force.
The first law states that an object's motion cannot be changed until a force acts on it.
The second law states that an object's force is calculated by dividing its mass by its acceleration.
The third law states that when two objects come into contact, they apply pressures that are equal in size and direction to one another.
Let a represent the system's acceleration.
T1 = M1a .....(1)
T2 − T1 = M2a ....(2)
F2 − T2 = M3a ......(3)
When we combine (1), (2), and (3), we obtain
(M1 + M2 + M3)a = F
or (1 + 2 + 3) a= 6 a = 1 m/s²
Now , T2 = (M1 + M2)a = (1 + 2)(1) = 3N
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what is constant angular speed
Answer:
in optical storage, constant angular velocity (CAV) is a qualifier for the rated speed of any disc containing information, and may also be applied to the writing speed of recordable discs. A drive or disc operating in CAV mode maintains a constant angular velocity, contrasted with a constant linear velocity (CLV)
A 0.0012 C amount of charge is produced on a Van der Graaf with 1,093 J of energy. A spark of this charge is produced in 0.005 s. What is the resistance of the air gap of the spark? (Voltage can be calculated by using energy divided by charge.)
Given,
The charge generated, q=0.0012 C
The energy of the Van der Graaf generator, E=1093 J
The time it takes for the generator to produce the given amount of charge, t=.005 s
The current is given by the time rate of the flow of charges. That is,
[tex]I=\frac{q}{t}\text{ }\rightarrow\text{ (i)}[/tex]The voltage is calculated using the formula,
[tex]V=\frac{E}{q}\text{ }\rightarrow\text{ (ii)}[/tex]From Ohm's law, the voltage is given by,
[tex]\begin{gathered} V=IR \\ R=\frac{V}{I}\text{ }\rightarrow\text{ (iii)} \end{gathered}[/tex]On substituting equation (i) and (ii) in equation (iii),
[tex]\begin{gathered} R=\frac{\frac{E}{q}}{\frac{q}{t}} \\ =\frac{Et}{q^2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} R=\frac{1093\times0.005}{0.0012^2} \\ =3.8\times10^6\text{ }\Omega \end{gathered}[/tex]Thus the resistance of the air gap is 3.8×10⁶ Ω
In a physics lab a student discovers that the magnitude of the magnetic field in a specific location near a long wire is 21.919 microTesla. If the wire carries a current of 35.483 A, what is the distance from the wire to that location ?
We will have the following:
First, we have that the permeability of free space is:
[tex]\mu_0=4\pi\ast10^{-7}Tm/A[/tex]Then:
[tex]\begin{gathered} B=\frac{\mu_0I}{2r\pi}\Rightarrow r=\frac{\mu_0I}{2B\pi} \\ \\ \Rightarrow r=\frac{(4\pi\ast10^{-7}Tm/A)(35.483A)}{2\pi(2.1919\ast10^{-5}T)}\Rightarrow r=0.3237647703...m \\ \\ \Rightarrow r\approx0.32m \end{gathered}[/tex]So, the distance is approximately 0.32 m.
How to do calculate this the error? What are the rules?(Mark scheme Answer is B)
Here,
potential difference(V)= (2.00±0.02) V;
current (I)=( 5.3 ± 0.1) mA;
Resistance (X) will be given by
[tex]\begin{gathered} X=\frac{V}{I}; \\ X=\text{ }\frac{2.00±0.02}{5.3±0.1}\text{ }\times1000\begin{cases}V={(2.00±0.02)} \\ I=({5.3±0.1\text{ \rparen mA= \lparen5.3}\pm0.1)\times10^{-3\text{ }}}A\end{cases} \\ \\ \therefore X=(\text{ 377.358 }\pm0.001) \\ \text{ } \\ \end{gathered}[/tex]