According to Newton's Third Law of Motion, the force that object A exerts to object B has the same magnitude as the force that object B to object A, but in the opposite direction:
[tex]\vec{F}_{AB}=-\vec{F}_{BA}[/tex]Then, the force exerted by Earth on the Moon has the same magnitude as the force exerted by the Moon on the Earth.
Therefore, the given statement is false.
Which picture below correctly identifies the effort length and lifting length of a lever?Select one:a. Ab. Bc. Cd. D
d.D
Explanation
A lever is a simple machine made of a rigid beam and a fulcrum.
where
the Load is the object which we are lifting,Fulcrum is the point at which the lever is pivoted. and Effort is he force applied to make the object move
Step 1
check for the graph that:
the lifting length goes from the fulcrum to the load
and
the effort length goes from the fulcrum to the applied force,
therefore,
the answer is
d.D
I hope this helps you
Describe
protons.
Location:
Charge:
Mass:
The protons in an atom are classified according to their mass, charge, and location as follows:
Particle: Protons
Mass: 1.67262 × 10⁻²⁷ kg
Charge: Positive charge (+e or +1)
Location: Found in the nucleus of every atom
The proton is a stable subatomic particle with a rest mass of 1.67262 x 10⁻²⁷ kg, or 1,836 times the mass of an electron, with a positive charge that is equal to one electron's charge in magnitude.
All atomic nuclei, with the exception of the hydrogen nucleus, are composed of protons and neutrons, which are electrically neutral particles (that consist of a single proton). A given chemical element's nuclei all contain the same number of protons. This number establishes an element's atomic number and establishes the element's position in the periodic table. An atom is electrically neutral when the number of protons in its nucleus equals the number of electrons in its orbit.
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A 3 kg ball is dropped from a height of 100 m above the surface of Planet Z. If the ball reaches a velocity of 45 m/s in 7 s, what is the ball’s weight on Planet Z? What is the gravitational field strength on Planet Z?
We are given the following information
Mass of ball = 3 kg
Height = 100 m
Final velocity = 45 m/s
Time = 7 s
Recall from the equations of motion
[tex]s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2[/tex]Where u is the initial velocity of the ball that is zero.
[tex]\begin{gathered} s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2 \\ 100=0\cdot7+\frac{1}{2}\cdot g\cdot7^2 \\ 100=\frac{1}{2}\cdot g\cdot49 \\ g=\frac{2\cdot100}{49} \\ g=4.08\; \frac{m}{s^2} \end{gathered}[/tex]So, the gravitational acceleration of the planet Z is 4.08 m/s^2
The weight o the ball is given by
[tex]\begin{gathered} W=m\cdot g \\ W=3\cdot4.08 \\ W=12.24\; N \end{gathered}[/tex]Therefore, the weight of the ball is 12.24 N
Tall pacific coast redwood trees can reach heights of about 100 m. If air drag is negligibly small, how fast is a sequoia come moving when it reaches the ground if it dropped from the top of a 100 m tree?
Given data:
Height of the tree;
[tex]h=100\text{ m}[/tex]Initial velocity;
[tex]u=0\text{ m/s}[/tex]The velocity of sequoia when it reaches the ground is given as,
[tex]v=\sqrt[]{u^2+2gh}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} v=\sqrt[]{(0\text{ m/s})^2+2\times(9.8\text{ m/s}^2)\times(100\text{ m})} \\ \approx44.27\text{ m/s} \end{gathered}[/tex]Therefore, sequoia will reach the ground with a velocity of 44.27 m/s.
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, what is the ratio of the drag force on ball A to the drag force on ball B?
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, the ratio of the drag force on ball A to the drag force on ball B will be F1 : F2 = 1 : 4
When objects travel through fluids (a gas or a liquid), they will undoubtedly encounter resistive forces called drag forces.
The drag force always acts in the opposite direction to fluid flow. If the body’s motion exists in the fluid-like air, it is called aerodynamic drag.
formula to calculate drag force is = F(d) = 1/2 * C * rho*A * [tex]v^{2}[/tex]
C = drag coefficient
A = area of object
rho = density in which object is moving
v = velocity of object
A = area of the object
F1 ( drag force on ball A ) = 1/2 * C * rho * area of ball A * [tex]v^{2}[/tex]
F2 (drag force on ball A ) = 1/2 * C * rho * area of ball B * [tex]v^{2}[/tex]
since , both the balls have same speed and falling in same environment hence , density and speeds are the same , the only difference is in area of both the balls
F1/F2 = area of ball A / area of ball B = 4 * pi * [tex]r1^{2}[/tex] / 4 * pi * [tex]r2^{2}[/tex]
= [tex]r1^{2}[/tex] / [tex]r2^{2}[/tex]
= [tex](\frac{d}{2} )^{2}[/tex]/ [tex](\frac{2d}{2}) ^{2}[/tex]
= 1/4
F1 : F2 = 1 : 4
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44) Find the x coordinate of the center of mass of the bricks shown.
We are asked to determine the x-coordinate of the center of mass of the given bricks. To do that, we will use the following formula:
[tex]\bar{x}=\frac{\Sigma x_im_i}{\Sigma m_i}[/tex]Where:
[tex]\begin{gathered} x_i=\text{ x-coordinate of the center of mass of each brick} \\ m_i=\text{ mass of each brick} \end{gathered}[/tex]Since we have three bricks, the formula expands to:
[tex]\bar{x}=\frac{x_1m_1+x_2m_2_{}+x_3m_3}{m_1+m_2+m_3}[/tex]Since we have three bricks with the same characteristics we will assume the three of them have the same mass:
[tex]\bar{x}=\frac{x_1m_{}+x_2m+x_3m_{}}{m_{}+m_{}+m_{}}[/tex]Taking "m" as a common factor and adding like terms in the denominator we get:
[tex]\bar{x}=\frac{m(x_1+x_2+x_3)}{3m}[/tex]Now we cancel out the "m":
[tex]\bar{x}=\frac{x_1+x_2+x_3}{3}[/tex]Now we determine the x-coordinates of each brick. Each brick is a parallelepiped, therefore, the x-coordinate is in the middle. Since each brick measures L, this means that the x-coordinate of the first brick is:
[tex]x_1=\frac{L}{2}[/tex]For the second brick, we have the L/2 of the separation from the first plus the L/2 of its length, therefore:
[tex]x_2=\frac{L}{2}+\frac{L}{2}=L[/tex]Now, for the third brick we have the L/4 of the separation from the second brick plus the L/2 of the separation of the second brick and the first brick and the L/2 of the length of the third brick, therefore:
[tex]x_3=\frac{L}{2}+\frac{L}{4}+\frac{L}{2}=\frac{5L}{4}[/tex]Now we substitute in the formula for the x-coordinate:
[tex]\bar{x}=\frac{(\frac{L}{2})+(L)+(\frac{5L}{4})}{3}[/tex]Adding like terms in the numerator:
[tex]\bar{x}=\frac{\frac{11L}{2}}{3}[/tex]Simplifying:
[tex]\bar{x}=\frac{11L}{6}[/tex]Therefore, the x-coordinate of the center of mass is located at 11L/6 from the origin.
I got first part correct but dont know how to solve second part: Two new particles with identical positive charge 3 are placed the same 0.0809 m apart. The force between them is measured to be the same as that between the original particles. What is 3 ?
Answer:
5.92 *10^-6 C
Explanation:
For the two charges q3 the force between them is given by
[tex]F=k\frac{q_3\times q_3}{d^2}[/tex]Now we know that
F = 48.1 N, d = 0.0809 m, and k = 8.99 *10^9 kg⋅m^3⋅s^−2⋅C^-2; therefore, the above gives
[tex]48.1=(8.99\times10^9)\frac{q_3\times q_3}{(0.0809)^2}[/tex][tex]\Rightarrow48.1=(8.99\times10^9)\frac{(q_3)^2}{(0.0809)^2}[/tex]Now we solve for q_3.
Dividing both sides by 8.99 * 10^9 gives
[tex]\frac{48.1}{(8.99\times10^9)}=\frac{(q_3)^2}{(0.0809)^2}[/tex]multiplying both sides by (0.0809)^2 gives
[tex]\frac{48.1}{(8.99\times10^9)}\times\mleft(0.0809\mright)^2=(q_3)^2[/tex]finally, taking the square root of both sides gives
[tex]\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}=\sqrt{(q_3)^2}[/tex][tex]q_3=\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}[/tex]Evaluating the right-hand side gives
[tex]\boxed{q_3=_{}5.92\times10^{-6}C\text{.}}[/tex]Hence, the charge q_3 is 5.92 x 10^-6 C.
A ball is thrown directly downward with an initial speed of 8.45 m/s, from a height of 29.9 m. After what time interval does it strike the ground? s
Given:
The initial speed of the ball is: u = 8.45 m/s.
The ball is thrown from the height: h = 29.9 m
To find:
The time ball takes to strike the ground.
Explanation:
The time taken by the ball to strike the ground can be determined by using the following equation.
[tex]x=ut+\frac{1}{2}at^2[/tex]Here, x = -29.9 m, u = -8.45 m/s and a = -9.8 m/s^2. The negative sign indicates that the ball is falling in the downward direction.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} -29.9=-8.45t-\frac{1}{2}\times9.8t^2 \\ \\ 4.9t^2+8.45t-29.9=0 \end{gathered}[/tex]Solving the above quadratic equation, we get:
t = 1.754 s and t = -3.478 s
But the time is never negative, thus t = 1.754 s.
Final answer:
The ball takes 1.754 seconds to strike the ground.
A steel cable on a bridge has a linear mass density of 15 kg/m. If the cable has been pulled taunt with a tension of 5536 N, what is the speed of a wave on it?
A fireman standing on a 14 m high ladderoperates a water hose with a round nozzle ofdiameter 2.65 inch. The lower end of the hose(14 m below the nozzle) is connected to thepump outlet of diameter 3.49 inch. The gaugepressure of the water at the pump isCalculate the speed of the water jet emerg-ing from the nozzle. Assume that water is anincompressible liquid of density 1000 kg/m3and negligible viscosity. The acceleration ofgravity is 9.8 m/s?Answer in units of m/s.
Given data,
The height, H = 14 m
The diameter, D = 2.65 inch
The gauge pressure, P = 317.84 kPa
We need to calculate the speed of the water jet emerging from the nozzle.
Using Bernoulli's equation,
[tex]\begin{gathered} \frac{1}{2}\rho(v^2_n-v^2_p)=P_{gauge\text{ }}-\rho gh \\ (v^2_n-v^2_p)=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{A_n}{A_p})^2v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{r_n}{r_p_{}})^4v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \end{gathered}[/tex]Further solved as,
[tex]\begin{gathered} v_n=\sqrt[]{\frac{(\frac{2}{\rho})P_{gauge}-2gh}{1-(\frac{r_n}{r_p})^4}} \\ v_n=\sqrt[]{\frac{(\frac{2}{1000})\times317.84\times10^3-2\times9.8\times14}{1-(\frac{1.325_{}}{1.745_{}})^4}} \\ v_n=\sqrt[]{\frac{635-274.4}{0.667}} \\ v_n=\sqrt[]{540.62} \end{gathered}[/tex]Thus, the speed of the water jet is
[tex]v=23.25\text{ m/s}[/tex]what is the smallest amount of time in which the person can accelerate the car from rest to 23 m/s and still keep the coffee cup on the roof. The coefficient of the static friction is 0.21. The maximum acceleration of the car that is allowed so that the cup does not fall is 2.1 m/s^2
Given:
The coefficient of the static friction, μ=0.21
The maximum acceleration of the car so that the cup does not fall, a=2.1 m/s²
The initial velocity of the car, u=0 m/s
The final velocity of the car, v=23 m/s
To find:
The smallest amount of the time in which the car can accelerate so that the coffee cup will still be on the roof.
Explanation:
From the equation of motion,
[tex]v=u+at[/tex]Where t is the smallest amount of time in which the person can accelerate and still keep the cup on the car.
On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]On substituting the known values,
[tex]\begin{gathered} t=\frac{23-0}{2.1} \\ =10.95\text{ s} \end{gathered}[/tex]Final answer:
Thus the smallest amount of the time in which the person can accelerate the car at the given rate and still keep the cup on the roof of the car is 10.95 s
A puck is moving on an air hockey table. Relative to an x,y coordinate system at time t =0s, the x
components of the puck's initial velocity and acceleration are Vix=1.0 m/s and ax=2.0 m/s². The y
components of the puck's initial velocity and acceleration are Viy=2.0 m/s and ay=2.0 m/s². Find the
magnitude and direction of the puck's velocity at a time of t=0.50 s. Specify the direction relative to
the x axis. HELPP!!!
The supplied puck is moving at a speed of v0x=+3.4m/s on an air hockey table at time t=0.
What is the meaning of velocity?The direction of the movement of the body or the object is defined by its velocity. Most of the time, speed is a scalar quantity. In its purest form, velocity is a vector quantity. It measures how quickly a distance changes. It is the rate at which displacement is changing.
What does the term "tangential velocity" refer to?Any object traveling in a circular motion has a linear speed known as tangential velocity. On a turntable, a point in the center moves less distance in a full rotation than a point near the outside edge.
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Two cars in opposite directions were going at 32 mph before a collision. They had a head on inelastic collision, i.e. the two cars stuck together afterward. The common speed of the combined piece right after the collision is 20 mph. The mass of Car 1 was 2,000 lb. Car 2 was heavier. The mass of Car 2 was ____ lb.
The mass of Car 2 was 3000 lb.
We need to apply the concept of conservation of momentum.
The velocity of both cars= 32mph
Combined velocity = 20mph
Mass of Car 1= 2000 lb
According to the conservation of momentum
M1V1+ M2V2= (M1+M2)
2000x32- (-M2x 32)=20(2000+M2)
64000+32M2=40000 +20M2
24000= 8M2
M2= 3000lb
Therefore the mass of Car 2 is 3000lb.
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A robotic arm lifts a barrel of radioactive waste, as shown in the figure.
If the maximum torque delivered by the arm about the axis O is 3.00 x 10° N•m and the distance r is 3.00 m, what is the maximum mass m of the barrel?
The maximum mass of the barrel, with maximum torque of 3.0×10° N.m delivered about the axis is 0.102 kg.
What is mass?Mass is the quantity of matter a body contains. The S.I unit of mass is kilogram (kg). Mass can also be defined as the ratio of the force to the acceleration of a body.
To calculate the maximum mass of the barrel, we use the formula below.
Formula:
m = τ/dg........... Equation 1Where:
m = Maximum Mass of the barrel τ = maximum Torque deliveredd = Distanceg = Acceleration due to gravity.From the question,
Given:
τ = 3.00×10° N.md = 3.00 mg = 9.8 m/s²Substitute the values above into equation 1
m = (3.00×10°/3×9.8)m = 1/9.8 kg.m = 0.102 kg.Hence, the maximum mass of the barrel is 0.102 kg.
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The critical angle for a certain liquid-air surface is 20°. What is the index of refraction of this liquid?
ANSWER
[tex]\begin{equation*} 2.92 \end{equation*}[/tex]EXPLANATION
To find the index of refraction of the liquid, we have to apply the formula for critical angle:
[tex]\theta=\sin^{-1}(\frac{n_r}{n_i})[/tex]where nr = refractive index of air = 1
ni = refractive index of liquid
Hence, by substituting the given values into the equation, we have that the index of refraction of the liquid is:
[tex]\begin{gathered} 20=\sin^{-1}(\frac{1}{n_i}) \\ \sin20=\frac{1}{n_i} \\ n_i=\frac{1}{\sin20} \\ n_i=2.92 \end{gathered}[/tex]That is the answer.
Four wires running through the corners of a square with sides of length 16.166 cm carry equal currents, 3.684 A. Calculate the magnetic field at the center of the square.
For practical reasons, we can consider each side of the square as an infinite wire. This can be seen on the following drawing:
This way, the field on the center will be the sum of the contribution of each wire. We can calculate the contribution of a single wire as:
[tex]B=\frac{\mu_0i}{2\pi d}=\frac{4\pi *10^{-7}*3.684}{2\pi(\frac{16.166*10^{-2}}{2})}=9.115*10^{-6}T[/tex]Then, the total field will be this, multiplied by the number of wires:
[tex]B_t=4*9.115*10^{-6}=36.46\mu T[/tex]Then, the resulting field will be Bt=36.46uT
A baseball player pitches a fastball toward home plate at a speed of 41.0 m/s. The batter swings, connects with the ball of mass 195 g, and hits it so that the ball leaves the bat with a speed of 37.0 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.A. What is the impulse delivered to the ball by the bat? Enter a positive value if the impulse is in the direction the bat pushes the ball and enter a negative value if the impulse is in the opposite direction the bat pushes the ball. (kg m/s)B. If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat? (kN)
Given:
Initial velocity, vi = 41.0 m/s
Mass of ball, m = 195 g = 0.195 kg
Final velocity, vf = 37.0 m/s
Assuming the ball is moving horizontally just before and after collision with the bat, let's solve for the following:
• (A). What is the impulse delivered to the ball by the bat?
To find the impulse, apply the change in momentum formula:
[tex]\Delta p=p_f-p_i[/tex]Where:
pi is the initial momentum = -mvi
pf is the final momentum = mvf
Thus, we have:
[tex]\begin{gathered} \Delta p=mv_f-(-mv_i) \\ \\ \Delta p=mv_f+mv_i \\ \\ \Delta p=m(v_f+v_i) \\ \\ \Delta p=0.195(37.0+41.0) \\ \\ \Delta p=15.21\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Impulse can be said to equal change in momentum.
Therefore, the impulse delivered to the ball by the bat is 15.21 kg.m/s away from the bat.
• (B). If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat?
Apply the formula:
[tex]\text{ Impulse = Force }\ast\text{ time}[/tex]Rewrite the formula for force:
[tex]\text{ Force=}\frac{impulse}{time}[/tex]Where:
time = 3.00 m/s
impulse = 15.21 kg.m/s
Hence, we have:
[tex]\begin{gathered} \text{ F=}\frac{15.21}{3} \\ \\ F\text{ = 5.07 kN} \end{gathered}[/tex]Therefore, the magnitude of the average force exerted on the ball by the bat is 5.07 kN away from the bat.
ANSWER:
(A). 15.21 kg.m/s away from the bat
(B). 5.07 kN.
Calculate the potential energy of a 2 kg ball that is about to be dropped of a height of 25 m
Given:
The mass of the ball, m=2 kg
The height from which the ball is about to be dropped, h=25 m
To find:
The potential energy of the ball.
Explanation:
The type of potential energy that is stored in this ball is called gravitational potential energy. The gravitational potential energy is the energy that an object possesses due to its height. The gravitational potential energy is directly proportional to the height at which the object is situated.
The potential energy of the ball is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=2\times9.8\times25 \\ =490\text{ J} \end{gathered}[/tex]Final answer:
The potential energy of the ball is 490 J
PLEASE HELP
Which is not an accurate statement about Earth's gravitational pull?
A) Earth's gravitational pull helps keep it in orbit.
B) Earth's gravitational pull is the same as Jupiter's gravitational pull.
C) Earth's gravitational pull is 9.8 m/s2.
D) Earth's gravity helps keep people from floating outside of the planet.
Newton’s Second Law states “The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.” Explain how your observations in both a and b support this Law.
Newton’s Second Law states “The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.”
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Force = mass × acceleration
Assuming the force constant the acceleration is inversely proportional to the mass of the object.
Thus, acceleration is directly proportional to force and inversely proportional to the mass of the body.
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A sled of mass 26 kg has an 18 kg child on it. If big brother is pulling with a 30 N force to the right and 10 N up, and big sister is pushing with a 40 N force to the right and 16 N down, what is the normal force?
Given data:
* The mass of the sled is m_1 = 26 kg.
* The mass of the child is m_2 = 18 kg.
* The force in the upwards direction by the big brother is F_1 = 10 N.
* The force in the downwards direction by the big sister is F_2 = 16 N.
Solution:
The net mass on the sled along with the child is,
[tex]\begin{gathered} m=m_1+m_2 \\ m=26+18 \\ m=44\text{ kg} \end{gathered}[/tex]The net weight of the sled along with the child is,
[tex]\begin{gathered} w=mg \\ w=44\times9.8 \\ w=431.2\text{ N} \end{gathered}[/tex]The weight of the sled along the child is acting on the sled in the downwards direction.
Thus, the normal force acting on the sled (taking upward force as negative and downward force as positive) is,
[tex]\begin{gathered} N=w+F_2-F_1 \\ N=431.2+16-10 \\ N=437.2\text{ newton} \end{gathered}[/tex]Thus, the normal force acting on the sled is 437.2 N.
a student drops a pebble from the edge of a vertical cliff. the pebble hits the ground 4 s after it was dropped. what is the height of the cliff? a. 20 m b. 40 m c. 60 m d. 80 m
The object's speed shortly before it lands on the earth is 40 m/s.
What is an example of velocity?The speed at which something moves in a specific direction is known as its velocity. as the speed of a car driving north on a highway or the pace at which a rocket takes off. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.
The parameters are as follows: the pebble's time, t = 4 s; the object's velocity right before impact;
The kinematic equation is as follows;
v = in which
v = 0+10 (4)
The object's speed right before impact with the earth is v = 40 m/s2, where g is the acceleration caused by gravity and an is a constant of 10 m/s2. As a result,
the object's final velocity before impact is 40 m/s.
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Physics
Hello how to solve this
2.15 mL of water will spill from the beaker when a 400mL glass beaker at room temperature is filled to the brim with cold water.
What is temperature and how come 2 mL of water spills off the beaker?Temperature as studied always is the measure for the degree of hotness or coldness.Here in the question is given a glass beaker of 400 mL which is at room temperature .Now the beaker is warmed up to 30 degree celsius, and then some of amount of water spilled.Change in volume = beta x v1 x change in temperature = 210 x 10^-4 x 400 x ( 30 - 4.4) = 2.15 mL.Hence the amount of water that will spill from the beaker would be 2.15 mL.To know more about temperature visit:
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You have a concave mirror with a focal length of 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object. Where should you place the object?Select one:a.9.09 cmb.12.09 cmc.7.12 cmd.5.8 cm
ANSWER:
a. 9.09cm
STEP-BY-STEP EXPLANATION:
Given:
Focal length = 100 cm
Distance image = -10 cm
We use the following formula to calculate the value of the distance of the object:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]We substitute each value and calculate the distance of the object just like this:
[tex]\begin{gathered} \frac{1}{d_o}+\frac{1}{-10}=\frac{1}{100} \\ \\ \frac{1}{d_{o}}=\frac{1}{100}+\frac{1}{10} \\ \\ \frac{1}{d_o}=\frac{11}{100} \\ \\ d_o=\frac{100}{11} \\ \\ d_o=9.09\text{ cm} \end{gathered}[/tex]Therefore, the correct answer is option a. 9.09cm
A ski jumper competing for an Olympic gold metal wants to jump
a horizontal distance of 149 meters. The takeoff point of the ski
jump is at a height of 38.0 meters. With what horizontal velocity
must he leave the jump in order to travel 149 meters?
19.25 m/s is horizontal velocity must he leave the jump in order to travel 149 meters .
How fast is horizontal moving?Standard definitions of horizontal velocity include miles per hour and meters per second, which are horizontal displacement times time. The distance an object has traveled since its origin is simply referred to as displacement.
How can one calculate vertical velocity using horizontal velocity?V * cos() equals the horizontal velocity component Vx. V * sin() is equal to the vertical component of velocity, Vy.
Time before landing = sqrt ( 2 x height / gravity ), sqrt ( 2 x 38/ 9.81) = 7.75
distance / time = avg speed
149/ 7.75 ≅ 19.25 m/s
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An intrepid hiker reaches a large crevasse in his hiking route. He sees a nice landing ledge 60.0 cm below his position but it is across a 2.3 m gap. He spends 1.2 s accelerating horizontally at 5.92 m/s2 [right] in an attempt to launch himself to the safe landing on the far side of the gap. Does he make it?
The hiker made it to a safe landing on the other side of the gap after travelling horizontally at 2.49 m.
What is the time motion from the vertical height?
The time taken for the hiker to fall from the given height is calculated as follows;
h = vt + ¹/₂gt²
where;
v is the vertical velocity = 0t is the time of motiong is acceleration due to gravityh is the height of fallh = ¹/₂gt²
t = √(2h/g)
t = √[(2 x 0.6) / (9.8)]
t = 0.35 seconds
The horizontal velocity of the hiker during the period of acceleration is calculated as follows;
Vₓ = at
Vₓ = (5.92 m/s²) x (1.2 s)
Vₓ = 7.104 m/s
The horizontal distance travelled during the time period of 0.35 seconds;
X = Vₓt
X = 7.104 x 0.35
X = 2.49 m
Thus, the hiker made it to a safe landing on the other side of the gap which is 2.3 m wide and smaller to his horizontal displacement of 2.49 m.
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Describe the mathematical relationship between the distance (d) and the attractive force (F) between protons and electrons.
The attractive force and the distance are inversely proportional.
[tex]F\propto\frac{1}{r}[/tex]This relation means that the attractive force decreases as the distance increases, and the attractive force increases as the distance decrease.
A car traveling at 11.6 meters per second crashes into a barrier and stops in 0.287 meters. What force must be exerted on a child of mass 21.2 kilograms to stop him or her in the same time as the car? Answer must be in 3 significant digits.
The equation to obtain the final speed of car is,
[tex]v^2=u^2+2as[/tex]Substitute the known values,
[tex]\begin{gathered} (0m/s)^2=(11.6m/s)^2+2a(0.287\text{ m)} \\ a=\frac{-134.56m^2s^{-2}}{2(0.287\text{ m)}} \\ \approx-234.4m/s^2 \end{gathered}[/tex]The negative sign of acceleration indicates that the car is deaccelerating.
The force required to stop the car is,
[tex]F=ma[/tex]Substitute the magnitude of known values,
[tex]\begin{gathered} F=(21.2kg)(234.4m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =4969.28\text{ N} \\ \approx4970\text{ N} \end{gathered}[/tex]Thus, the force required to stop the car is 4970 N.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." Explain the error in the student's claim. Provide an example of each quantity to support your answer.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." The error here is that acceleration is said to be done when either speed of that object changes or direction of that object changes. Hence , acceleration is not a scaler quantity.
Scalar quantities are quantities that are described only by a magnitude. They do not have a direction of action.
A vector quantity is defined as the physical quantity that has both directions as well as magnitude.
Acceleration is said to be occurred in two cases :
when the object changes its speed
or when the object changes its direction
since , acceleration depends upon both direction and magnitude ,hence it is a vector quantity not a scaler quantity.
for example : a stone attached to a string moving in a circular motion at a constant speed will be considered in accelerated motion because it is constantly changing its direction.
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A wooden sphere of mass 4.0 kg is completely immersed in water. A pushing force of 20. N is
applied.
21°
20 N
4.2 ms²
19⁰
At the moment shown in the diagram, the sphere is stationary and it experiences an
acceleration upwards and to the right as shown.
Calculate the size of the upwards force due to the water (upthrust) acting on the sphere.
The size of the upwards force due to the water (upthrust) acting on the sphere is 12.64 N.
What is upthrust?
Buoyancy or upthrust, is an upward force exerted by a fluid on an object immersed in the fluid due to the weight of the object
Thus, upthrust is the upward force acting on an object immersed in a liquid.
Fu - Df = F(net_u)
where;
Fu is the upward forceDf is the downward force applied on the objectF(net_u) is the net upward forceFu - F x sin(21) = ma x sin(19)
where;
m is the mass of the wooden spherea is the upward acceleration of the wooden sphereFu - 20 x sin(21) = (4 x 4.2) x sin(19)
Fu - 7.17 = 5.47
Fu = 12.64 N
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