When given a graph of velocity vs time as the following:
The area under the curve represents the distance that the object has moved in the given time interval.
A 2.10-kg block is moving to the right at 2.00 m/s just before it strikes and sticks to a 1.00-kg block initially at rest. What is the total momentum of the two blocks after the collision? Enter a positive answer if the total momentum is toward right and a negative answer if the total momentum is toward left. Kg m/s
Given,
The mass of the moving block, M=2.10 kg
The velocity of the moving block, u=2.00 m/s
The mass of the block that was at rest, m=1.00 kg
The velocity of the second block before the collision, v=0 m/s
From the law of conservation of momentum, the total momentum of a system always remains constant. That is, the total momentum of the two blocks before the collision is equal to the total momentum of the blocks after the collision.
Therefore, the total momentum of the two blocks after the collision is given by,
[tex]p=Mu+mv[/tex]On substituting the known values,
[tex]\begin{gathered} p=2.10\times2.00+1.00\times0 \\ =4.2\text{ kg. m/s} \end{gathered}[/tex]Therefore, the total momentum of the blocks after the collision is 4.2 kg· m/s
A wire with resistance of 8.0 Ω is drawn out through a die such that its new length is twice its original length. Find the resistance of the longer wire assuming that the resistivity and the density of the material of the wire are unchanged during the drawing process.
Explanation
the electrical resistivity is define as
[tex]\begin{gathered} \sigma=R\frac{A}{l} \\ where \\ R\text{ is the electrical resitance} \\ \text{A is the cross seccional area} \\ l\text{ is the length} \end{gathered}[/tex]so,if we isolate R
[tex]\begin{gathered} \sigma= R\frac{A}{l} \\ R=\sigma\frac{l}{A} \end{gathered}[/tex]hence, the ratio of the resitances is
[tex]\frac{R_1}{R_2}=\frac{\sigma\frac{l}{A}}{\sigma\frac{l_2}{A}}[/tex]the volume of the wire is constant , therefore
[tex]A_1l_1=A_2l_2[/tex]if the new length is twice the original
The resistance of the longer wire assuming that the resistivity and the density of the material of the wire are unchanged during the drawing process is twice the resistance of the wire before, i.e. 16 Ω.
What is Resistance?The impediment to current flow in an electrical circuit is measured by resistance. The ohm, a unit of measurement for resistance, is represented by the Greek letter omega. The name of the unit of resistance is derived from Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current, and resistance.
Given:
The resistance of the wire, R = 8 Ω.
Calculate the electrical resistivity by the formula given below,
σ = [tex]R\frac{A}{l}[/tex],
Here σ is the electrical resistivity, A is the area, and l is the length of the wire.
According to Que's statement,
R₁ / R₂ = {σ₁ l₁ / A₁ } / {σ₂l₂ / A₂}
R₂ = 2 *R₁
R₂ = 2*8
R₂ = 16 Ω
Therefore, the resistance of the longer wire assuming that the resistivity and the density of the material of the wire are unchanged during the drawing process is twice the resistance of the wire before, i.e. 16 Ω.
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Three friends each do work to push an object across the room. Katie does 400 J of work while pushing her object across the room in 6 seconds. Alice does 300 J of work while pushing her object across the room in 7 seconds. Alyssa does 350 J of work while pushing her object across the room in 5 seconds. Which friend has the largest power output?
The friend that has the largest power output among the three friends is Alyssa with 70 J/s of output power.
What is power output?
The output power of an object or a device is the rate at which energy is used over a given period of time.
Mathematically, output power is given as;
P = E/t
where;
P is the output powerE is the output energy of the objectt is the time periodThe output power of Katie is calculated as follows;
P = 400 J / 6 s = 66.67 J/s
The output power of Alice is calculated as follows;
P = 300 J / 7 s = 42.86 J/s
The output power of Alyssa is calculated as follows;
P = 350 J / 5 s = 70 J/s
Thus, the output power of each of the friend depends on the amount of energy applied over a given time period.
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Answer: Alyssa
Explanation:
005 (part 1 of 3) A 59 N object is in free fall. What is the magnitude of the net force which acts on the object? Answer in units of N.006 (part 2 of 3) What is the magnitude of the net force when the object encounters 23.4 N of air resistance? Answer in units of N.007 (part 3 of 3) What is the magnitude of the net force when it falls fast enough to encounter an air resistance of 59 N?Answer in units of N.
Given that weight of the object, mg= 59 N
Here, m is the mass and g is the acceleration due to gravity.
(a) The net force acting on the object is
[tex]\begin{gathered} F=mg \\ =59\text{ N} \end{gathered}[/tex](b) Given that object encounters air resistance of 23.4 N which acts in opposite direction of free fall.
Thus, the net force will be
[tex]\begin{gathered} F^{\prime}=\text{ mg-air resistance} \\ =59-23.4 \\ =35.6\text{ N} \end{gathered}[/tex](c) Given that air resistance is 59 N.
Thus the net force will be
[tex]\begin{gathered} F^{\doubleprime}=\text{ 59-59} \\ =0\text{ N} \end{gathered}[/tex]Thus, the net forcce is zero in this case.
What is the net change in internal energy (ΔU) if the amount of work done (W) is the same as the amount of energy transferred in by heat (Q)?
ANSWER:
Zero
STEP-BY-STEP EXPLANATION:
Heat (Q) and work (W) are the two ways to add or remove energy from a system. The processes are very different, however, both can change the internal energy (ΔU) of a system.
It is given by the following equation:
[tex]\begin{gathered} \Delta U=Q-W \\ \\ \text{ So if they are the same:} \\ \\ Q=W \\ \\ \Delta U=Q-Q \\ \\ \Delta U=0 \end{gathered}[/tex]The net change in internal energy (ΔU) is zero
How do I solve this problem? String one is 5kg String 2 is 11 kg
ANSWERS
• Forces on the 5kg block: ,3
,• Forces on the 11kg block: ,2
,• Forces balanced? ,Yes, the system is at rest.
,• Tension in string 1:
,• Tension in string 2: ,107.91 N
EXPLANATION
First we have to draw the forces on each block:
Hence, block 1 has 3 forces acting on it, while block 2 has 2 forces acting on it.
It is said that the system is at rest, which means that the forces on the system are balanced.
By Newton's second law we know that,
[tex]F_{t1}-F_{t2}-F_{g1}=0[/tex]And,
[tex]F_{t2}-F_{g2}=0[/tex]Both equal zero because the blocks are not moving, and therefore there's no acceleration.
From the second equation we can find the tension in string 2,
[tex]\begin{gathered} F_{t2}=F_{g2} \\ F_{t2}=m_2\cdot g \end{gathered}[/tex]m2 = 11kg and g = 9.81m/s²,
[tex]F_{t2}=11\operatorname{kg}\cdot9.81m/s^2=107.91N[/tex]Now that we know that the tension in string 2 is 107.91N, we can find the tension in string 1 replacing this into the first equation and solving for Ft1,
[tex]F_{t1}=F_{t2}+F_{g1}[/tex][tex]undefined[/tex]A circuit breaker is a short piece of metal that melts when a current that is too large passes through it. Is this true or false?
A circuit beaker is a short piece of metal that melts when a current that is too large passes through it. This given statement is false statement.
Which pan of water shows molecules that have received the most heat from a stove?Select one:a. Ab. Bc. Cd. D
Given:
The water in the pan is heated the most.
To find:
Pan of water that has received the most heat from the stove.
Explanation:
When water is heated, the water molecules in the water acquire kinetic energy due to a rise in water temperature. The more the water is heated, the higher the temperature of the water. Thus the kinetic energy of molecules of water will be more.
The higher kinetic energy of water molecules corresponds to the higher speed of the water molecules.
Thus, in option (a), the water molecules are moving very fast, meaning they have high kinetic energy due to high temperature. And the temperature will be maximum if we provide maximum heat to the water.
Thus, option (a) is correct.
Conclusion: The correct option is option (a).
8. An object of mass 13 kg going to the right witha speed of 26 m/s collides with a(n) 10 kg object at rest.After collision the 13 kg object moves with a speed of 11 m/smaking an angle of 20 degree with the horizontal-right.Calculate the x-component of the velocity of the 10 kgobject after collision. (1 point)A. 07.069 m/sB. O 17.629 m/sC. O 3.745 m/sD. O 11.865 m/sE. O 20.362 m/s
8)Answer: 20.362 m/s
Explanation:
According to the law of conservation of momentum,
Total momentum before collision = total momentum after collision
Total momentum before collision = m1u1 + m2u2
where
m1 is the mass of the object moving to the right
u1 is its velocity
m2 is the mass of the object at rest
u2 is its velocity
From the information given,
m1 = 13
u1 = 26
m2 = 10
u2 = 0
Total momentum before collision = 13 x 26 + 10 x 0 = 338
Total momentum after collision = m1v1cosθ1 + m2v2cosθ2
where
θ1 is the angle made by m1 with the horizontal
θ2 is the angle made by m2 with the horizontal
v1 is the final velocity of m1 after collision
v2 is the final velocity of m2 after collision
From the information given,
v1 = 11
θ1 = 20
Total momentum = 13 x 11cos20 + 10v2cosθ2 = 134.376 + 10v2cosθ2
Thus,
338 = 134.376 + 10v2cosθ2
10v2cosθ2 = 338 - 134.376 = 203.624
v2cosθ2 = 203.624/10
v2cosθ2 = 20.362 m/s
The x component of the velocity of the 10kg ball after collision is 20.362 m/s
A simple machine does 30 J of work with an efficiency of 28%. How much energy was put into the machine?
Take into account that efficiency is given by the following expression:
[tex]\text{ efficiency=(output work/input work)}\cdot100[/tex]In this case, you have:
efficieny = 28%
output work = 30J
input work = ?
Replace the previous values of the parameters into the formula for efficieny, solve for input work and simplify:
28% = (30J/input work)*100%
input work = 30J*(100% / 28%)
input work = 107.14 J
Hence,about 107.14 J was put into the machine
What kinds of nutrients would you expect to find in a slice of pizza?
please help me im stuck on the science.
Answer:
Pizza contains a mix of different nutrients such as, protein, carbohydrates, fats and calcium.
Answer:
Pizza contains a mix of nutrients such as protein, carbohydrates, fats and calcium. Grain (sometimes, but usually not, whole grain) is one of the primary ingredients in the crust.
rock that is attached to a 2.6 m rope is whirled around and covers 3 Radians every 1.2seconds. What is the Tangential velocity of the rock when the rope is released?
Answer: Tangential velocity = 6.5 m/s
Explanation:
The formula for calculating tangential velocity is expressed as
v = rw
where
v represents tangential velocity
r represents the radius of the circular path
w is the angular velocity
From the information given,
r = 2.6
If 3 radians is covered in 1.2 seconds, then
angular velocity, w = distance /time = 3/1.2
w = 2.5 rad/s
Thus,
v = 2.6 x 2.5
v = 6.5 m/s
Tangential velocity = 6.5 m/s
Vector C is 6.28 units long in a105° direction. In unit vectornotation, this would be writtenas:C = [?]î + [?]i coefficient (green)j corfficient (yellow)
We have a vector 6.28 units long in 105° direction
[tex]\begin{gathered} 6.28\cos (105)=-1.62 \\ 6.28\sin (105)=6.03 \end{gathered}[/tex]Therefore the vector C is
[tex]\vec{C}=-1.62i+6.03j[/tex]A particle moves in a straight line, and you are told that the torque acting on it is zero about some unspecified origin. Does this necessarily imply that the total force on the particle is zero? Can you conclude that its angular velocity is constant?
Horizontal force will be there but angular force is zero because it is dependent on angle and angular velocity is constant.
[tex]\tau = rF\sin\theta[/tex]
T= torque
r= radius
In mechanics, every action that tries to preserve, alter, or change a body's motion is referred to as a force. Isaac Newton's Principia Mathematica has three principles of motion that are usually used to explain the concept of force (1687).
Newton's first law states that in the absence of an external force, a body will continue to be in either its resting or evenly moving condition along a straight path. According to the second law, any time an outside force acts on a body, the body accelerates (changing velocity) in the force's direction.
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Data/Research:Mass: _ 1 kgMaterial: CopperInitial temperature (Ti measure in °C): _330°Final temperature (T, measure in ºc): 3300at*C.Specific heat (c value measure in J/kg.°C):Phase Change Temperatures:Melting (melting point)the change from___to___Bolling.(boiling point)the change from___to___Heat of fusion (H, value measure in J/kg):Heat of vaporization (H, value measure in J/kg):
Specific heat (C) for copper is 389 J/kg.°C.
Melting (melting point) the change from soild to liquid at 1084.62°C.
Boiling (boiling point) the change from liquid to gas at 2595°C.
Heat of fusion of copper is 206000 J/kg.
Heat of vaporization of copper is 4730000 J/kg.
is 11/12 greater = or less then 2/3
11/12 is greater than 2/3
Explanation
to solve this we have two options, rewriting them in terms of a common denominator, or the fastest way to compare fractions is to convert them into decimal numbers.
Step 1
a)rewriting in terms of a common factor
we can see that the denominator of the second fraction is 3, so we can multiply by (4/4) in order to ger 12 as denominator , so
[tex]\frac{2}{3}\cdot\frac{4}{4}=\frac{8}{12}[/tex]note that we are not changing or affecting the fraction ( we are just multipliyng by 1)
Step 2
now, we have both fraction with the same denominator, let's compare
When the denominators are the same, the fraction with the lesser numerator is the lesser fraction and the fraction with the greater numerator is the greater fraction
[tex]\frac{11}{12}\text{ and }\frac{8}{12}[/tex]therefore, the greater fraction is 11/12
in other words
11/12 is greater than 2/3
I hope this helps you
A force of 600 N is applied to stretch a spring by 15 cm. What is the spring constant of the spring?
A. 0.025
B. 40 N/m
C. 4000 N/m
D. 9000 N/M
The value of the Spring Constant, k is 40 N/m
The formula for Spring Force(F) is given by
F = - kx
where F is the spring force,
k is the spring constant
x is the displacement
This type of force is Conservative in nature as its magnitude only depends on the initial and final points. It is a restoring force.
The given Information is as follows,
F = Spring Force = 600 N
x = displacement = 15 cm
Now, Spring Constant = k = F/x = 600/15 = 40
Thus, The value of the Spring Constant, k is 40 N/m
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Which of the following best describes acceleration?
Answer:
Increasing speed
Explanation:
summary for story throwing fire by skyler tegland
Answer:
y₀ is the initial position
y is the final position
a is the acceleration, in this case, the acceleration is the gravity so it is equal to -9.8 m/s²
v₀ is the initial speed
So, 115 m is the initial height of an object, 0 m is the final height, and the initial speed is 0 m/s
instantaneous velocity for a displacement function dt=2-2t at any given time
a -2t
b-2
c 2
d 2t
Instantaneous velocity for a displacement function [tex]d_{t}[/tex] = 2-2t is -2 m/s. So the correct option is (b)
What is instantaneous velocity?Instantaneous velocity is defined as the rate of change of position over a very short (near zero) time interval. Measured in SI units m/s. The instantaneous velocity is the magnitude of the instantaneous velocity. Same value as instantaneous velocity, but without direction. Simply put, the velocity of an object at that point in time is called the instantaneous velocity. Hence the definition is given as "velocity of a moving object at a given point in time". It can also be determined by taking the slope of the displacement-time plot or the x-t plot.
If the object's velocity is constant, the instantaneous velocity can be the same as the default velocity.
For the given case,
Displacement function [tex]d_{t}[/tex] = 2-2t
Assume x = 2-2t
Time = t
Since [tex]v_{t}[/tex] = [tex]\frac{d_{} _{x} }{d_{t} }[/tex]
[tex]v_{t}[/tex] = [tex]\int\limits^t_0[/tex]2 [tex]d_{t}[/tex] - [tex]\int\limits^t_0[/tex]-2t [tex]d_{t}[/tex]
[tex]v_{t}[/tex] = -2 m/s
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Blood plasma (at 37.0°C) is to be supplied to a patient at the rate of 2.80 × 10−6 m3/s. If the tube connecting the plasma to the patient’s vein has a radius of 2.00 mm and a length of 52.5 cm, what is the pressure difference between the plasma and the patient’s vein? Viscosity of blood plasma is 1.30 × 10−3 Pa·s.
ANSWER:
304.3 Pa
STEP-BY-STEP EXPLANATION:
We have the poiseuille law, which would be the following equation:
[tex]v=\frac{\pi\cdot\Delta P\cdot r^4\cdot t}{8\cdot\eta\cdot L}[/tex]Where,
v = volume of the liquid
r = radius
t: time
n = coefficiente of viscosity
Δp = change of pressure
L : lenght
We solve for Δp, and we would have:
[tex]\begin{gathered} \Delta P=\frac{8\cdot\eta\cdot L\cdot v}{\pi\cdot\cdot r^4\cdot t} \\ \frac{v}{t}=Q \\ \text{ therefore:} \\ \Delta P=\frac{8\cdot\eta\cdot L\cdot Q}{\pi\cdot r^4} \\ \text{ replacing:} \\ L=52.5\text{ cm = 0.525 m} \\ r=2\text{ mm = 0.002 m} \\ \Delta P=\frac{8\cdot1.3\cdot10^3\cdot0.525\cdot2.8\cdot10^{-6}}{3.14\cdot(0.02)^4} \\ \Delta P=304.3\text{ Pa} \end{gathered}[/tex]The pressure difference is 304.3 Pa
If the resistance is 150 Ohm connected to 300 volt line. Calculate the current drawn by the line.
__Given__:
The resistance of the line is,
[tex]R=150\text{ Ohm}[/tex]The potential difference across the line is,
[tex]V=300\text{ Volts}[/tex]The current is given by from Ohm's law as,
[tex]i=\frac{V}{R}[/tex]substituting the values we get,
[tex]\begin{gathered} i=\frac{300\text{ Volts}}{150\text{ Ohm}} \\ =2\text{ A} \end{gathered}[/tex]Hence the current drawn by the line is 2A.
A police officer in hot pursuit drives her car through a circular turn of radius 379 mwith a constant speed of 90.0 km/h. Her mass is 52.0 kg.What are (a) themagnitude and (b) the angle (relative to vertical) of the net force of the officer onthe car seat? (Hint: Consider both horizontal and vertical forces.)
The centripetal force keeps the car in the circle. This force is directed towards the center of the circle. The upward force on the police officer is f = mg and the centripetal force is represented as follows
[tex]F_c=\frac{mv^2}{r}[/tex]The net force will be
[tex]\begin{gathered} F_{net}=F_c+mg \\ F_{net}=\sqrt[]{(\frac{mv^2}{r})^2+(mg)^2} \end{gathered}[/tex]A.
[tex]\begin{gathered} r=379m \\ m=52.0\text{ kg} \\ \text{speed}=90\text{ km/h} \\ \text{speed}=25\text{ m/s} \\ F_{net}=\sqrt[]{(\frac{52\times25^2}{379})^2+(52\times9.8)^2} \\ F_{net}=\sqrt[]{(\frac{32500}{379})^2+(509.6)^2} \\ F_{net}=\sqrt[]{(85.7519788918)^2+259692.16} \\ F_{net}=\sqrt[]{7353.40188386+259692.16} \\ F_{net}=\sqrt[]{267045.561884} \\ F_{net}=516.764512988 \\ F_{net}\approx516.76N \end{gathered}[/tex]B.
[tex]\begin{gathered} \tan \emptyset=\frac{\frac{mv^2}{r}}{mg} \\ \tan \emptyset=\frac{mv^2}{r}\times\frac{1}{mg} \\ \tan \emptyset=\frac{v^2}{rg} \\ \tan \emptyset=\frac{25^2}{379\times9.8} \\ \tan \emptyset=\frac{625}{3714.2} \\ \tan \emptyset=0.16827311399 \\ \emptyset=\tan ^{-1}0.16827311399 \\ \emptyset=9.55185382438 \\ \emptyset=9.55^{\circ} \end{gathered}[/tex]Need to solve step by step this exercise about fluids..
Given:
The pressure in the lower pipe is,
[tex]P_1=120\text{ kPa}[/tex]The speed of water is,
[tex]v_1=1\text{ m/s}[/tex]The radius of the lower pipe is
[tex]\begin{gathered} r_1=12\text{ cm} \\ =0.12\text{ m} \end{gathered}[/tex]the radius of the upper pipe is
[tex]\begin{gathered} r_2=6\text{ cm} \\ =0.06\text{ m} \end{gathered}[/tex]The height of the upper pipe is,
[tex]h_2=2\text{ m}[/tex]The density of water is,
[tex]\rho=1000kg/m^3[/tex]Using the continuity equation,
[tex]\begin{gathered} \pi r^2_{1^{}}v_1=\pir_2^{}^2_{}v_2 \\ v_2=\frac{r^2_{1^{}}v_1}{r^2_{2^{}}} \\ v_2=\frac{0.12\times0.12\times1}{0.06\times0.06} \\ v_2=4\text{ m/s} \end{gathered}[/tex]Hence the speed of water is 4 m/s.
Applying Bernoulli's principle we get,
[tex]\begin{gathered} P_1+\frac{1}{2}\rho(v_1)^2+\rho g\times0_{}=P_2+\frac{1}{2}\rho(v_2)^2+\rhogh_2 \\ 120\times10^3+\frac{1}{2}\times1000\times1^2+0=P_2+\frac{1}{2}\times1000\times4^2+1000\times9.8\times2 \\ P_2=120500-27600 \\ P_2=92.9\text{ kPa} \end{gathered}[/tex]Hence, the pressure is 92.9 kPa.
A 150 N crate is being pulled up a perfectly smooth ramp that slopes upward at 15 degrees by a pull that is directly at 30 degrees above the surface of the ramp. What is the magnitude of the pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s?
The magnitude of the pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s is 44.83 N.
What is the magnitude of the force required?
The magnitude of the force pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s is calculated by applying Newton's second law of motion.
F(net) = ma
where;
F(net) is the net force on the cratem is the mass of the cratea is the acceleration of the crateat constant velocity, the acceleration of the crate = 0
F(net) = 0
Fx - Fgx = 0
where;
Fx is the horizontal component of the applied forceFgx is the horizontal component of the weight of the crateFcos(30) - 150sin(15) = 0
F = 150 sin(15) / cos30
F = 44.83 N
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URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
An ice cube is placed on a metal block that is at a room temperature of 25°C Which statement best describes the result of this interaction?
A. Heat is transferred from the ice cube to the metal block, and the ice melts
B. Heat is transferred from the air to the ice cube, and the ice melts.
C. Heat is transferred from the air and metal to the ice cube, and the ice melts
D. Heat is transferred from the ice cube to the air, and the ice melts
Answer:
C. Heat is transferred from the air and metal to the ice cube, and the ice melts
Monica is loading a box that weighs 96 N into a truck. She pushes the box up a ramp using a force of 32 N. What is the mechanical advantage of the ramp?a. 3.0b. 128 Nc. 0.33d. 64 N
The mechanical advantage is given by:
[tex]MA=\frac{\text{ output force}}{\text{ input force}}[/tex]In this case the output force is 96 N and the input force is 32 N, then we have:
[tex]MA=\frac{96}{32}=3[/tex]Therefore, the mechanical advantage is 3
As a 600 N woman sits on the floor, the floor exerts a force on her ofa. 6 Nb. 60 NC. 1200 Nd. 600 N
the floor exerts a force of 600 N on her
Explanation
Newton's third law states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction,in this case the wowan exerts a force on the floor, it is her weigth, in the other hand, the floor exerts a force on the woman,this is called Normal force, the normal force is the force that surfaces exert to prevent solid objects from passing through each other. Normal force is a contact force.
Step 1
a) diagram:
as the woman is at equilibrium , the sum of the forces acting on it equals, zero, hence
[tex]\begin{gathered} Normal\text{ force-weigth= 0} \\ Normal\text{ force = weigth} \\ Normal\text{ force=600 N} \end{gathered}[/tex]therefore, the floor exerts a force of 600 N on her
so, the answer is
d) 600 N
I hope this helps you
A woman standing before a cliff claps her hands, 3.6s later she hears the echo. How far away is the cliff? The speed of sound in air at ordinary temperature is 353m/s.
Since we know that the time it takes to the cliff will be half the time it takes for the echo to be heard, the distance is:
[tex]t=\frac{3.6s}{2}\Rightarrow t=1.8s[/tex][tex]d=\frac{353m}{s}\cdot1.8s\Rightarrow d=635.4m[/tex]So, the cliff is 635.4 meters away.
On slide 2f and 2g, what is the value of ΣF if the object is in equilibrium?a)0N b)It is impossible to tell
From the given figure, let's determine sum of forces (ΣF) if the object is in equilibrum.
When an object is said to be in equilibrum, the forces are balanced. This means that the object is in a state of equilibrum.
If an object is in a state of equilibrum, the sum of all forces is zero.
Therefore, we have:
ΣFx = 0 N
ΣFy = 0 N
ΣF = √(ΣFx)² + (ΣFy)²= 0 N
Therefore, the value of ΣF if the object is in equilibrum is 0N.
ANSWER:
a) 0N