Answer:
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Explanation:
Given;
initial distance covered by Bolt, d = 200 m
time of this motion, t = 19.3 s
The second distance covered by Bolt, = 100 m
Assuming Bolt maintained the same acceleration for both races.
His acceleration can be determined from the 200 m race.
d = ut + ¹/₂at²
where;
u is his initial velocity = 0
d = ¹/₂at²
[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]
Let the final or maximum velocity for the 100 m race = v
v² = u² + 2ad₂
v² = 2 x 1.074 x 100
v² = 214.8
v = √214.8
v = 14.66 m/s
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At which of the following times does the average angular acceleration equal the instantaneous angular acceleration?
a. 0.50 T
b. 0.67 T
c. 0.71 T
d. all of the above
The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?
Answer:
Acosθ
Explanation:
The x-component of a vector is defined as :
Magnitude * cosine of the angle
Maginitude * cosθ
The magnitude is represented as A
Hence, horizontal, x - component of the vector is :
Acosθ
Furthermore,
The y-component is taken as the sin of the of the angle multiplied by the magnitude
Vertical, y component : Asinθ
(a) If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve on wet concrete where the coefficient of kinetic friction is 0.5 and the coefficient of static friction is 0.7.
(b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate where the coefficient of kinetic friction is 0.3 and the coefficient of static friction is 0.55?
(c) If the truck has four-wheel drive, and the cabinet is wooden, what is it's maximum acceleration (in m/s2)?
Answer:
a) a = 27.44 m / s², b) a = 5.39 m / s², c) a = 156.8 m / s², cabinet maximum acceleration does not change
Explanation:
a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.
Let's apply Newton's second law
we set a regency hiss where the x axis is in the direction of movement of the truck
Y axis y
N- W = 0
N = W = m g
X axis
2fr = m a
the expression for the friction force is
fr = μ N
fr = μ m g
we substitute
2 μ m g = m /2 a
a = 4 μ g
a = 4 0.7 9.8
a = 27.44 m / s²
b) let's look for the maximum acceleration that can be applied to the cabinet
fr = m a
μ N = ma
μ m g = m a
a = μ g
a = 0.55 9.8
a = 5.39 m / s²
as the acceleration of the platform is greater than this acceleration the cabinet must slip
c) the friction force is in the four wheels as well
With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all
applying Newton's second law
4 fr = (m/4) a
16 mg = (m) a
a = 16 g
a = 16 9.8
a = 156.8 m / s²
cabinet maximum acceleration does not change
An object with mass m is located halfway between an object of mass M and an object of mass 3M that are separated by a distance d. What is the magnitude of the force on the object with mass m?A) 8GMm/d^2B) GMm/(4d^2)C) 4GMm/d^2D) GMm/(2d^2)E) 3GMm/2d^2
Answer:
A) 8GMm/d^2
Explanation:
We are given that
[tex]m_1=M[/tex]
[tex]m_2=3M[/tex]
[tex]m_3=m[/tex]
Distance between m1 and m2=d
Distance of object of mass m from m1 and m2=d/2
Gravitational force formula
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
Using the formula
Force acting between m and M is given by
[tex]F_1=\frac{GmM}{d^2/4}[/tex]
Force acting between m and 3M is given by
[tex]F_2=\frac{Gm(3M)}{d^2/4}[/tex]
Now, net force acting on object of mass is given by
[tex]F=F_2-F_1[/tex]
[tex]F=\frac{Gm(3M)}{d^2/4}-\frac{GmM}{d^2/4}[/tex]
[tex]F=\frac{12GmM}{d^2}-\frac{4GmM}{d^2}[/tex]
[tex]F=\frac{12GmM-4GmM}{d^2}[/tex]
[tex]F=\frac{8GmM}{d^2}[/tex]
Hence, the magnitude of the force on the object with mass m=[tex]\frac{8GmM}{d^2}[/tex]
Option A is correct.
Explain why the flow from the battery increases when the switch is closed. Give the label of the concept(s) that you use from the model of electricity. [
Answer:
Due to the applied filed the electrons move in a particular direction.
Explanation:
Initially when the switch is off, the free electrons move here and there in any random directions in the conductor with the random speeds called thermal velocity. So, tat the net flow is almost zero.
When the battery is connected is switch is ON, the random motion of the electrons aligned in a particular direction due to the force applied by the electric filed, so the net flow is not zero it increases and thus the current flow.
An electron in a hydrogen atom is in a p state. Which of the following statements is true?
a.
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).
b.
The electron has an energy of -13.6 eV.
c.
The electron has a total angular momentum of ħ.
d.
The electron has a z-component of angular momentum equal to sqrt(2)* ħ.
Answer:
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).
Explanation:
We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.
P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.
Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node
A 2.90 m segment of wire supplying current to the motor of a submerged submarine carries 1400 A and feels a 2.00 N repulsive force from a parallel wire 4.50 cm away. What is the direction and magnitude (in A) of the current in the other wire? magnitude A direction
Answer:
[tex]I_2=30.9A[/tex]
Explanation:
From the question we are told that:
Wire segment [tex]l_s=2.9m[/tex]
Initial Current [tex]I_1=1400A[/tex]
Force [tex]F=2.00N[/tex]
Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]
[tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]
[tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]
[tex]I_2=30.9A[/tex]
A block of mass M is connected by a string and pulley to a hanging mass m.
The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg.
b. Find the acceleration of the system and tensions on the string.
c. How far will block m drop in the first seconds after the system is released?
d. How long will block M move during the above time?
e. At the time, calculate the velocity of block M
f. Find out the deceleration of block M if the connection string is removal by cutting after the first second. Then, calculate the time taken to contact block M and pulley
How far will block m drop in the first seconds after the system is released?
(b) Use Newton's second law. The net forces on block M are
• ∑ F (horizontal) = T - f = Ma … … … [1]
• ∑ F (vertical) = n - Mg = 0 … … … [2]
where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.
Right away, we see n = Mg, and so f = µn = 0.2Mg.
The net force on block m is
• ∑ F = mg - T = ma … … … [3]
You can eliminate T and solve for a by adding [1] to [3] :
(T - 0.2Mg) + (mg - T ) = Ma + ma
(m - 0.2M) g = (M + m) a
a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)
a = 1.96 m/s²
We can get the tension from [3] :
T = m (g - a)
T = (10 kg) (9.8 m/s² - 1.96 m/s²)
T = 78.4 N
(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of
1/2 (1.96 m/s²) t
(e) Assuming block M starts from rest, its velocity at time t is
(1.96 m/s²) t
(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have
∑ F = -f = Ma
The effect of friction is constant, so that f = 0.2Mg as before, and
-0.2Mg = Ma
a = -0.2g
a = -1.96 m/s²
Then block M slides a distance x such that
0² - (1.96 m/s²) = 2 (-1.96 m/s²) x
x = (1.96 m/s²) / (2 (1.96 m/s²))
x = 0.5 m
(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)
Meanwhile, block m would be in free fall, so after 1 s it would fall a distance
x = 1/2 (-9.8 m/s²) (1 s)
x = 4.9 m
Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation results
Answer:
No, it will not affect the results.
Explanation:
For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.
What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.
Which image illustrates reflection?
A
B
с
D
Answer: I beleive A
Explanation:
Answer:
A
Explanation:
We can see the light being reflected off the mirror.
Please help,it is urgent!)
Answer:
answer is 18.58 because
Answer:
My answer is d.25.1 because
1
An astronaut weighs 202 lb. What is his weight in newtons?
Answer:
978.6084 Newton
Explanation:
Given the following data;
Weight = 220 lbTo find the weight in Newtown;
Conversion:
1 lb = 4.448220 N
220 lb = 220 * 4.448220 = 978.6084 Newton
220 lb = 978.6084 Newton
Therefore, the weight of the astronaut in Newton is 978.6084.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, the weight of an object is given by the formula;
Weight = mg
Where;
m is the mass of the object.g is the acceleration due to gravity.Note:
lb is the symbol for pounds.N is the symbol for Newton.how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?
Assuming no heat lost to the surrounding,
-5⁰C ice → 0⁰C ice
Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(2.0 x 10³) x (0-(-5))
Q = 50000J
0⁰C ice → 0⁰C water
Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg
Q = mLf
Q = 5(3.34 x 10⁵)
Q = 1670000J
0⁰C water → 100⁰C water
Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(4.2 x 10³) x (100-0)
Q = 2100000J
100⁰C water → 100⁰C steam
Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Q = mLv
Q = 5(2.26 x 10⁶)
Q = 11300000J
Total amount of heat required
= 50000 + 1670000 + 2100000 + 11300000
= 15120000J
The velocity-time graph of a body is given. What quantities are represented by (a) slope of the graph and (b) area under the graph?
Answer:
a) acceleration
b) displacement
Explanation:
The velocity-time graph is a graph of velocity versus time. The velocity (m/s) would be on the Y-axis while time (s) would be on the X-axis.
a) The slope of a graph is given by: change in Y-axis/change in X-axis = ΔY/ΔX
In a velocity-time graph, ΔY = change in velocity and ΔX = change in time.
Hence, the slope of a velocity-time graph becomes: change in velocity/change in time.
Also, acceleration = change in velocity/change in time.
Hence, the slope of a velocity-time graph = acceleration.
b) Assuming that the area under a velocity-time graph is a rectangle, the area is given as:
Area of a rectangle = length x breadth
= velocity x time (m/s x s)
Also, displacement = velocity x time (m)
Hence, the area under a velocity-time graph of a body would give the displacement of the body.
You are stranded in a stationary boat. Your friend is on a dock, but the boat is just beyond his reach. There is a 5 kg anchor in the boat. You'd like to get the boat to move closer to the dock so your friend can rescue you. Select from the following list what effect each change will have on the position of the boat relative to the dock. A. The boat will move closer to the dock. B. The boat will move away from the dock. C. The position of the boat relative to the dock will not change.
Answer:
running away and launching the anchor that will give a greater speed towards the dock v₄.
Explanation:
To try to bring the boat closer to the dock, several cases can be carried out.
* move inside the ship so that the center of mass changes and since moving away you have a speed v, the ship will approach the dock at a speed v₂,
* Throw the anchor in the opposite direction to the dock so that using the conservation of the moment the boat moves towards it, it moves at a speed v₃
* A combination of the two processes running away and launching the anchor that will give a greater speed towards the dock v₄.
In all cases, the friction must be zero.
All other movements move the ship away from the dock
A 10 kg box is at static equilibrium and the downward pull of gravity acting on the box is 98 Newton’s what is the minimum force that would require to just pick up the box
Explanation:
static equilibrium means its on the floor or something
so slightly greater than 98 newtons in the upward direction
if a body covers 100m in 5 second from rest find the acceleration produced by a body in 10 second
Answer:
a=10m/s^2
Explanation:
acceleration= final velocity+ initial velocity/time taken
v-u/t=a
100-0/5=a
100/5=a
a=20m/s^2
case2
100-0/10=a
100/10=a
a=10m/s^2
Don't forget to write the units.
Hope this helps
please mark me as brainliest.
The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?
Explanation:
The period T of a simple pendulum is given by
[tex]T = 2 \pi \sqrt{\dfrac{l}{g}}[/tex]
Doubling the length of the pendulum gives us a new period T'
[tex]T' = 2 \pi \sqrt{\dfrac{l'}{g}} = 2 \pi \sqrt{\dfrac{2l}{g}}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2} \left(2 \pi \sqrt{\dfrac{l}{g}} \right)[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2}\:T = \sqrt{2}(3.5\:\text{s})= 4.95\:\text{s}[/tex]
Write submultipels units of time
What
is the the improvetanse of water in our body and how do
Or
open Meet and enter this code
Answer:
Or open Meet and enter this code to the source of definition and how to use it to make
A wheel 30 cm in diameter accelerates uniformly from 245 rpm to 380 rpm in 6.1 s . Part A How far will a point on the edge of the wheel have traveled in this time
Answer:
A point on the edge of the wheel will travel 199.563 radians at the given time.
Explanation:
Given;
initial angular velocity of the wheel; [tex]\omega _i = 245 \ rev/\min = 245\ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 25.66 \ rad/s[/tex]
final angular velocity of the wheel;
[tex]\omega _f = 380 \ rev/\min = 380 \ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 39.80 \ rad/s[/tex]
radius of the wheel, d/2 = (30 cm ) / 2 = 15 cm = 0.15 m
time of motion, t = 6.1 s
The angular distance traveled by the edge of the wheel is calculated as;
[tex]\theta = (\frac{\omega_f + \omega_i}{2} )t\\\\\theta = (\frac{39.8 + 25.66}{2} )\times 6.1\\\\\theta = 199.653 \ radian[/tex]
Therefore, a point on the edge of the wheel will travel 199.563 radians at the given time.
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval
Answer:
The number of revolutions is 44.6.
Explanation:
We can find the revolutions of the wheel with the following equation:
[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]
Where:
[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s
t: is the time = 8 s
α: is the angular acceleration
We can find the angular acceleration with the initial and final angular velocities:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s
[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]
Hence, the number of revolutions is:
[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]
Therefore, the number of revolutions is 44.6.
I hope it helps you!
A proton has been accelerated from rest through a potential difference of -1350 V. What is the proton's kinetic energy, in electron volts? What is the proton's kinetic energy, in joules? What is the proton's speed?
Answer:
1 eV = 1.60 * 10^-19 J work done in accelerating electron thru 1 V
KE (total energy) = 1350 ^ 1 eV (note proton goes from + to -)
KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules
1/2 m v^2 = KE = 2.16 * 10^-16 J
v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11
v = 5.09 * 10^5 m/s
The proton's kinetic energy, in joules is 2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.
What is velocity?
When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
1 eV = 1.60 * [tex]10^{-19} J[/tex] work done in accelerating electron throw 1 V
K.E (total energy) = 1350 ^ 1 eV (note proton goes from + to -)
K.E = 1.60 * [tex]10^{-19}[/tex]J * 1350 = 2.16 * [tex]10^{-16}[/tex] Joules
1/2 m v² = KE = 2.16 *[tex]10^{-16}[/tex] J
Velocity of proton is,
v² = 4.32 * 10[tex]e^{-16}[/tex] / 1.67 * [tex]10{-27}[/tex] = 2.59 * [tex]10^{11}[/tex]
v = 5.09 * [tex]10^{5}[/tex]m/s
The proton's kinetic energy, in joules is 2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.
To learn more about velocity refer the link:
brainly.com/question/18084516
#SPJ2
If the potential (relative to infinity) due to a point charge is V at a distance R from this charge, the distance at which the potential (relative to infinity) is 2V is
A. 4R
B. 2R
C. R/2.
D. R/4
Answer:
R/2
Explanation:
The potential at a distance r is given by :
[tex]V=\dfrac{kq}{r}[/tex]
Where
k is electrostatic constant
q is the charge
The potential (relative to infinity) due to a point charge is V at a distance R from this charge. So,
[tex]\dfrac{V_1}{V_2}=\dfrac{r_2}{r_1}[/tex]
Put all the values,
[tex]\dfrac{V}{2V}=\dfrac{r_2}{R}\\\\\dfrac{1}{2}=\dfrac{r_2}{R}\\\\r_2=\dfrac{R}{2}[/tex]
So, the distance at which the potential (relative to infinity) is 2V is R/2.describe the cause of earth's magnetism ?
Una pelota se lanza verticalmente hacia arriba desde la azotea de un edificio con una velocidad inicial de 35 m/s. Si se detiene en el aire a 200 m del suelo, ¿Cuál es la altura del edificio?
a. 138,8 m
b. 51.2 m
c. 71,2 m
d. 45,0 m
In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length
Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
First step :
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
next :
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x = [tex]\sqrt{\frac{2*23.27}{5029} }[/tex] = 0.0962 m
A tank is full of water. Find the work (in J) required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1,000 kg/m3 as the density of water. Round your answer to the nearest whole number.)
Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.
I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)
Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.
The 2.0-kg block feels
• the downward pull of its own weight, (2.0 kg) g
• the upward normal force of the surface, magnitude n₁
• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction
• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction
• the applied force, mag. F, pointing in the positive horizontal direction
Meanwhile the 3.0-kg block feels
• its own weight, (3.0 kg) g, pointing downward
• normal force, mag. n₂, pointing upward
• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction
• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)
Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:
• net vertical force:
n₂ - (3.0 kg) g = 0 ==> n₂ = (3.0 kg) g ==> f₂ = 0.30 (3.0 kg) g
• net horizontal force:
c₂ - f₂ = 0 ==> c₂ = 0.30 (3.0 kg) g ≈ 8.8 N
Which one of the following statements concerning resistors in "parallel" is true? Question 7 options: The voltage across each resistor is the same. The current through each resistor is the same. The total current through the resistors is the sum of the current through each resistor. The power dissipated by each resistor is the same.
Answer: The correct statement is:
--> The voltage across each resistor is the same.
Explanation:
RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:
--> Resistors in parallel and
--> Resistors in series
RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.
A cat's displacement is 15 meters to the right in 7.0 seconds. If, at the start of the 7.0 seconds, the cat was moving at a velocity of 2.0 m/s left what was its final velocity?
Answer:
6.3 m/s
Explanation:
From the given information:
The displacement (x) = 15 m
time (t) = 7.0 s
initial velocity = -2.0 m/s (since it is moving in the opposite direction)
We need to determine the acceleration then find the final velocity.
By applying the kinematics equation:
[tex]x = ut + \dfrac{1}{2}at^2[/tex]
[tex]15 = (-2.0)(7.0) + \dfrac{1}{2}a(7.0)^2\\ \\ 15 = -14.0 + \dfrac{49}{2}a \\ \\ 29= 24.5a \\ \\ a= \dfrac{29}{24.5} \\ \\ a = 1.184 \ m/s^2[/tex]
Now, to determine the final velocity by using the equation:
v = u + at
v = -2 + 1.184(7.0)
v = 6.288 m/s
v ≅ 6.3 m/s
