Answers
Given data
*The given initial velocity of the puck is u = 5 m/s
*The given final velocity of the puck is v = 30 m/s
*The given distance is s = 10 m
The formula for the acceleration is given by the kinematic equation of motion as
[tex]\begin{gathered} v^2=u^2+2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=\frac{(30)^2-(5)^2}{2\times10} \\ =43.75m/s^2 \end{gathered}[/tex]Hence, the acceleration of the puck is a = 43.75 m/s^2
Related Questions
Ms. Terry travels to another planet where she drops a hamburger freely from rest. On this particular planet, g=5 m/s2. How far has the hamburger fallen after 4 s?
Answers
Answer:
40 m
Explanation:
d = 1/2 a t^2
= 1/2 * 5 m/s^2 * (4 s)^2 = 1/2 * 5 * 16 = 40 m
carts, bricks, and bands
10. Predict the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks.
a. Approximately 0.16 m/s2
b. Approximately 0.50 m/s2
c. Approximately 0.64 m/s2
d. Approximately 1.00 m/s2
Answers
D. The acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
What is acceleration?The acceleration of an object is the rate of change of velocity of the object with time.
The acceleration that would occur when four rubber bands were used to pull a cart loaded with two bricks, is determined by applying Newton's second law of motion as follows.
a = F/m
where;
a is accelerationF is the applied forcem is the massLet the mass of a brick = mass of a band = m
the mass of a cart = 2 bricks = 2m
a = (force applied by 4 rubber) / (mass of 1 cart + mass of 2 brick)
a = (4m) / (2m + 2m)
a = (4m)/(4m)
a = 1 m/s²
Thus, the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
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If the input force on the lever in problem 5 is 135 N, what is the net force on the rock?
Answers
The net force on the rock is determined as 76.2 N.
What is the net force on the rock?The net force on the rock is obtained by subtracting the weight of the rock from the input force on the lever.
F(net) = Input force - weight of the rock
The weight of the rock is calculated as follows;
W = mg
where;
m is mass of the rock = 6 kgg is acceleration due to gravity = 9.8 m/s²W = 6 x 9.8
W = 58.8 N
F(net) = 135 N - 58.8 N
F(net) = 76.2 N
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The complete question is below:
If the input force on the lever in problem 5 is 135 N, what is the net force on the rock? if the mass of the rock is 6kg.
johns mass is 92.0kg on the earth, what is his mass on mars where g= 3.72m/s^2
Answers
The mass is a fundamental property of all matter.
Mass is not the same as weight, the weight depends on the gravity while the mass does not. Therefore the John's mass is the same in mars.
A 75 kg criminal wants to escape from the 5th story window of the jail, 24 m above the ground. He has a rope, which can only support a tension force of 650 N.
a. What is the maximum acceleration he can slide down without breaking his "rope?"
Answers
Answer:
a = 1.1 m/s²
Explanation:
Given:
P = 650 N
m = 75 kg
g = 9.8 m/s²
____________
a - ?
Forse:
P = m*(g - a)
650 = 75*(10 - a)
10 - a = 650 /75
Acceleration:
a = 9.8 - 650 / 75 = 9.8 - 8.7 = 1.1 m/c²
calculate how much heat is released when 500g of platinum is cooled from 250.0K to 240.0K
Answers
m= mass
C= specific heat
AT- change in temperature (final minus initial)
Q=mCA(delta)T
Q=500g(0.133J/g*C)(-10*C)
Q= -655 J
Answer:
Q = 665 J
Explanation:
Given:
m = 500 g = 0.5 kg
T₁ = 250.0 K
T₂ = 240.0 K
c = 133 J / ( kg*⁰K) - Specific heat capacity of platinum
___________
Q - ?
Heat is released:
Q = c*m*( T₁ - T₂) = 133*0.5*(250.0 - 240.0) =
= 133*0.5*10 ≈ 665 J
A cat chases a mouse across a 0.66 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor taylor (jdt3899) – Homework 3, 2d motion 22-23 – tejeda – (LermaHPHY1 1) 3 2.4 m from the edge of the table. The acceleration of gravity is 9.81 m/s 2 . What was the cat’s speed when it slid off the table?
Answers
The cat’s speed when it slid off the table will be 6.552 m/s
The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.
a = -g = 9.8 m[tex]/s^{2}[/tex]
using equation of motion
x = u(horizontal )*t + 1/2 * a (horizontal) * [tex]t^{2}[/tex]
since , a (horizontal) = 0
x = u(horizontal )*t
u = x / t equation 1
similarly
y = u(vertical)*t + 1/2 * a (vertical) * [tex]t^{2}[/tex]
u(vertical) = 0
t = [tex]\sqrt{2y / a}[/tex] equation 2
substituting the value of equation 2 in equation 1
u = x / [tex]\sqrt{2y / a}[/tex]
= [tex]\sqrt{\frac{-9.81}{2*-0.66} } * 2.4[/tex]
= 6.552 m/s
The cat’s speed when it slid off the table will be 6.552 m/s
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What is the index of refraction for a medium where light has a velocity of 2.75 x10³ m/s?0.9176.751.090.250
Answers
n = c/v, where:
n: index of refraction
c: speed of light in a vacuum (3*10^8 m/s)
v: speed of light in medium
n = (3*10^8)/(2.75*10^8)
n = 3/2.75
n = 1.09
A football player runs from his own goal line to the opposing team's goal line, returning to his twenty-yard line, all in 27.0 s. Calculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.)HINTApply the definitions of average speed and average velocity.Click the hint button again to remove this hint.(a) Calculate his average speed. ____yards/s(b) Calculate the magnitude of his average velocity. ____yards/s
Answers
time = 27 s
d = 20 yard
a) speed = distance / time
d1 = 100 yard ( own goal to opposing team's goal line)
d2 = 80 yard ( returning to 20 yard line)
d= d1+d2
d= 100 + 80 = 180 yards
Speed = 180 y / 27s = 6.667 y/s
b) velocity = displacement / time
d = 100 - 80 = 20 y
Velocity = 20 / 27 = 0.74 y/s
How many moles of a gas sample are in a 5.0 L container at 251 K and 370 kPa?(The gas constant is8.31L kPamol K)Round your answer to one decimal place and enter the number only with no units.
Answers
Given:
The volume of the gas, V=5.0 L
The temperature of the gas, T=251 K
The pressure of the gas, P=370 kPa
The gas constant, R=8.31 L kPa/(mol K)
To find:
The moles of the gas sample.
Explanation:
From the ideal gas equation,
[tex]PV=\text{nRT}[/tex]Where n is the moles of the gas present.
On substituting the known values,
[tex]\begin{gathered} 370\times5.0=n8.31\times251 \\ \Rightarrow n=\frac{370\times5.0}{8.31\times251} \\ =0.9\text{ mol} \end{gathered}[/tex]Final answer:
The moles of the gas present in the sample is 0.9 mol
Sheena can row a boat at 2.90 mi/h in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at 2.00 mi/h. Not having her calculator ready, she guesses that to go straight across, she should head upstream at an angle of 25.0° from the direction straight across the river.What is her speed with respect to the starting point on the bank?How long does it take her to cross the river?How far upstream or downstream from her starting point will she reach the opposite bank? If upstream, enter a positive value and if downstream, enter a negative value.In order to go straight across, what angle upstream should she have headed?
Answers
Given
v = 2.90 mi/h
x = 1.20 mi
vs = 2.0 mi/h
θ = 25°
Procedure
Part a)
Velocity of the boat with respect to water stream is given as
[tex]\begin{gathered} v_b=v-v_s \\ v_b=(2.90\cdot\sin 25-2.00)\hat{j}+2.90\cdot\cos 25\hat{i} \\ v_b=-0.77\hat{j}+2.62\hat{i} \end{gathered}[/tex]magnitude of the speed is given as
[tex]\begin{gathered} v_b=\sqrt[]{0.77^2+2.62^2} \\ v_b=2.73mi/h \end{gathered}[/tex]Part b)
Time to cross the river is given as:
[tex]\begin{gathered} t=\frac{x}{v_x} \\ t=\frac{1.2mi}{2.90\cdot\cos25} \\ t=0.46h \end{gathered}[/tex]Part c)
Distance moved by boat in downstream is given as
[tex]\begin{gathered} x=v_yt \\ x=-0.77\cdot0.46 \\ x=-0.354mi \end{gathered}[/tex]Part d)
In order to go straight, we must net speed along the stream must be zero
so we will have
[tex]\begin{gathered} v\sin \theta=v_s \\ 2.90\sin \theta=2.00_{} \\ \sin \theta=\frac{2.00}{2.90} \\ \theta=43.60^{\circ} \end{gathered}[/tex]Object a attracts objects be the gravitational force of 10 N from a given distance the distance between the two objects is doubled what is the new force of attraction between
Answers
We are given that two objects are being attracted by a gravitational force between each other of 10N. The gravitational force between two masses is given by the following equation:
[tex]F_g=G\frac{m_Am_B}{r^2}[/tex]Where:
[tex]\begin{gathered} F_g=\text{ Gravitational force} \\ m=\text{mass} \\ G=\text{ Gravitational constant} \\ r=\text{ distance between the masses} \end{gathered}[/tex]Replacing the given values for the 10N force:
[tex]10=G\frac{m_Am_B}{r^2_1}[/tex]Where:
[tex]r_1=\text{ initial distance}[/tex]Now we will solve for the product of the masses and the gravitational constant by multiplying both sides by the distance squared:
[tex]10r^2_1=Gm_Am_B[/tex]Now, the product of the masses and the gravitational constant won't change if we double the distance, therefore, if we apply the equation for the gravitational force for the new distance we get:
[tex]F_{g2}=G\frac{m_Am_B}{r^2_2}[/tex]We can replace the value we determined earlier:
[tex]F_{g2}=\frac{10r^2_1}{r^2_2}[/tex]Since the distance is double, we have:
[tex]r_2=2r_1[/tex]Replacing in the previous equation:
[tex]F_{g2}=\frac{10r^2_1}{(2r_1)^2}[/tex]Solving the square:
[tex]F_{g2}=\frac{10r^2_1}{4r^2_1}[/tex]Now we can cancel out the distances squared:
[tex]F_{g2}=\frac{10}{4}[/tex]Solving the operation:
[tex]F_{g2}=2.5[/tex]Therefore, doubling the distance the new gravitational force is 2.5N.
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Which of the following best represents R= A - B ?
Please help, it’s due soon!
Answers
Option C is the best representation for the resultant vector representing the resultant R=A+B because the result is represented by the closing side of the triangle, so it is the correct answer.
What exactly is a vector quantity?A measure with both magnitude and direction It is typically represented by an arrow with the same direction as the quantity and a length proportional to the magnitude of the quantity. A vector, while including magnitude and demand, does not have a position. Vector quantities are physical quantities that are distinguished by the presence of both magnitude and direction. Displacement, force, torque, momentum, acceleration, velocity, and so on. Vector quantities are physical quantities that have distinct magnitude and direction definitions. For example, suppose a boy is riding a bike at 30 km/hr in the north-east direction. In nature, vectors have velocity, acceleration, force, electromagnetic fields, and importance. (Weight is the force produced by gravity's acceleration acting on a mass.) A Scalar is a quantity or phenomenon that has only magnitude and no specific direction.To learn more about vector quantity, refer to:
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8) If the volume of the liquid in graduated cylinder B is 90 mL, then whatis the volume of the rock?AYour answer8060B100180
Answers
Answer:
20 mL
Explanation:
The volume of the rock is equal to the difference of volume of A and B. So, it is equal to
90 mL - 70 mL = 20 mL
Because 90 mL is the volue in cylinder B and 70 mL is the volume in cylinder A.
Therefore, the volume of the rock is 20 mL
A type of wave that is a combination of a transverse and a longitudinal wave is called aQuestion 17 options:Slinky waveSurface waveSound waveLight wave
Answers
The type of wave that is a combination of longitudinal and Transverse wave is called Surface wave
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.
Answers
Considering the given position function, it is found that:
a) The velocity function is: v(t) = 5 m/s.
b) The functions are graphed at the end of the answer.
Position and velocity functionThe position function in this problem, after t seconds, is defined according to the following rule:
s(t) = 5t.
The velocity function is the derivative of the position function, hence it is calculated as follows:
v(t) = s'(t) = [5t]' = 5 m/s. (position in meters, hence velocity in meters per second).
The derivative rule applied was the power rule, [5t]' = 5[t'] = 5 x 1 x t^(1 - 1) = 5t^0 = 5.
These two functions are graphed at the end of the answer, considering a domain of t ≥ 0, as time cannot assume negative values.
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Isaac Newton in his 1670's lectures on optics and later in his 1704 publicationOpticks, he described light as composed of in his corpuscular hypothesisarguing that the perfectly straight lines of reflection demonstrated this nature.linesparticlesphotonswaves
Answers
The perfectly straight lines of reflection demonstrated in the nature is due to particle nature of the light. Therefore, Newton described that the light consists of 'particles' which means second option is correct.
A bunny and a tortoise start a race from rest. The bunny accelerates at a rate "a" for a time "t" until it reaches itsmaximum speed (vb) after traveling a distance d_b. The tortoise accelerates at a rate one fourth as great as the bunny andtakes three times as long to reach its maximum speed of vt after traveling a distance d_t.a) What is the maximum speed of the tortoise? Answer in terms of vbb)How far did the tortoise travel? Answer in terms of db
Answers
We are given that a Bunny starts a race from rest, this means that its initial speed is zero:
[tex]v_{b0}=0,(1)[/tex]We are also given that it accelerates at a rate "a" for a time "t" and reaches its maximum speed. We can use the following equation of motion to relate the final speed with its acceleration:
[tex]v_b=at,(2)[/tex]We are also given that it travels a distance "d". We can use the following equation of motion to relate the distance travel with the velocity and the acceleration:
[tex]2ad_b=v^2_b,(3)[/tex]Now. We turn our attention to the turtle. We are told that its acceleration is one-fourth of the acceleration of the bunny, therefore, we have:
[tex]a_t=\frac{1}{4}a[/tex]And it also says that it takes 3 times as long to reach its maximum speed, therefore the time of the turtle is:
[tex]t_t=3t[/tex]Now we use the equation of motion that relates velocity with time and acceleration for turtle:
[tex]v_t=a_tt_t[/tex]Since we are required to express this velocity in terms of the final velocity of the bunny we replace the values of acceleration and time of the turtle that we determined previously:
[tex]v_t=(\frac{1}{4}a)(3t)[/tex]Rearranging terms:
[tex]v_t=\frac{3}{4}at[/tex]From the equation of motion from the bunny we know that the product of acceleration and time of the bunny equals its final velocity , therefore:
[tex]v_t=\frac{3}{4}v_b[/tex]And thus we have determined the final velocity of the turtle in terms of the final velocity of the bunny.
Now we are told to determine the final velocity in terms of the distance traveled by the bunny. First, we will use the equation of motion that relates the distance, the acceleration, and the time with the final velocity:
[tex]2a_td_t=v^2_t[/tex]Now we replace the value of the acceleration:
[tex]2(\frac{1}{4}a)d_t=v^2_t[/tex]Simplifying:
[tex]\frac{1}{2}ad_t=v^2_t[/tex]Now we use the relationship between velocities we determined in part A:
[tex]\frac{1}{2}ad_t=(\frac{3}{4}v_b)^2[/tex]Simplifying:
[tex]\frac{1}{2}ad_t=\frac{9}{16}v^2_b[/tex]From equation (3) we can replace the value of the velocity of the bunny:
[tex]\frac{1}{2}ad_t=\frac{9}{16}\times2ad_b[/tex]We can cancel out the acceleration:
[tex]\frac{1}{2}d_t=\frac{9}{16}\times2d_b[/tex]Now we multiply both sides by 2:
[tex]d_t=2\times\frac{9}{16}\times2d_b[/tex]Simplifying:
[tex]d_t=\frac{9}{4}d_b[/tex]And thus we have found the distance traveled by the turtle in terms of the distance traveld by the bunny.
Physics
Hi dears how could we solve this question basically how to plug it in calculator
Answers
The new length of the steel bar is 11.65 cm.
What is linear expansion?
Linear expansion can be defined as the increase in length of a material due to increase in the temperature of the material.
Mathematically, the linear expansion of a material is given as;
ΔL = L₀αΔT
where;
ΔL is the change in length or increase in lengthL₀ is the initial length of the steelα is the coefficient of linear expansion of steel = 11 x 10⁻⁶/⁰CΔT is change in temperatureThe change in the length of the steel is calculated as follows;
ΔL = (11.5 cm) x (11 x 10⁻⁶/⁰C) x (1221 ⁰C - 22 ⁰C)
ΔL = 0.152 cm
The new length of the steel bar is calculated as follows;
L = ΔL + L₀
L = (0.152 cm) + (11.5 cm)
L = 11.65 cm
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A) A recipe calls for 5.0 qt of milk. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.B) A gas can holds 2.0 gal of gasoline. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.
Answers
A.
[tex]\begin{gathered} 1qt=946.353cm^3 \\ 5qt=\text{?} \\ \text{quantity}=946.353\times5 \\ \text{quantity}=4731.765cm^3 \end{gathered}[/tex]B.
[tex]\begin{gathered} 1\text{ gal=}3785.41\text{ }cm^3 \\ 2\text{ gal}=\text{?} \\ \text{quantity}=2\times3785.41 \\ \text{quantity}=7570.82cm^3 \end{gathered}[/tex]A star of mass 3.0e30 kg is moving in a circle of radius 1.0e12 metres, with a period of 100 years. This is due to the gravity of a second unseen object. what is the mass of the unseen object in kg
Answers
The mass of the unseen object is 1.21 x 10²⁴ kg.
What is the centripetal acceleration of the star?
The centripetal acceleration of the star is calculated as follows;
a = v²/r
where;
v is the linear speed of the starr is the radiusv = 2πr/T
where;
T is period of the motion = 100 years = 3.154 x 10⁹ sv = (2π x 1 x 10¹²) / (3.154 x 10⁹)
v = 1,992.1 m/s
a = (1992.1)² / ( 1 x 10¹²)
a = 3.97 x 10⁻⁶ m/s²
The centripetal force of the star is calculated as follows;
F = ma
F = (3 x 10³⁰) x ( 3.97 x 10⁻⁶)
F = 1.19 x 10²⁵ N
The mass of the unseen object is calculated as follows;
F = mg
m = F/g
where;
g is the acceleration due to gravitym = (1.19 x 10²⁵) / (9.8)
m = 1.21 x 10²⁴ kg
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If you have 100.g of a radioactive isotope with a half-life of 10. years, how much of the isotopewill you have left after 20. years?
Answers
Use the radioactive decay formula using t_0 as the half life:
[tex]A=A_0\cdot2^{-t/t_0}[/tex]Substitute the initial amount of radioactive element A_0=100g, the half life t_0=10y and the time period t=20y to find the remaining amount after 20 years:
[tex]\begin{gathered} A=100g\times2^{-20y/10y} \\ =100g\times2^{-2} \\ =100g\times\frac{1}{4} \\ =25g \end{gathered}[/tex]Therefore, after 20 years, there are 25 grams left.
Do you think we can see the planets without telescope? explain your answer
Answers
Answer:
No, how do you tell that?
Telescope is an example of technology that use on seeing the object on far place try to do that, you acn see a planet like small pieceof like an tungaw
Accomplishment of SPTA Officers and all parents this 1st Quarter.. Enjoy watching..Mabbalo
Explanation:
HOPE IT HELP TO YOUA marble was thrown vertically up the air and came back down to the samelevel in 1.6 s. (using g = 9.81 ms?).(a) What is it initial velocity?(b) What would be the highest height it can reach?(c) How long would it take to reach the highest point?
Answers
ANSWER:
(a) 15.7 m/s
(b) 12.56 m
(c) 0.8 sec
STEP-BY-STEP EXPLANATION:
We can calculate the initial velocity using the following formula:
(a)
[tex]\begin{gathered} v=u+a\cdot t \\ v=0 \\ \text{therefore:} \\ u=-a\cdot t \\ \text{ replacing} \\ u=-(-9.81\cdot1.6) \\ u=15.7\text{ m/s} \end{gathered}[/tex](b)
We calculate the height in this way:
[tex]\begin{gathered} h=\frac{(-u)^2}{2g} \\ \text{ replacing} \\ h=\frac{(-15.7)^2}{2\cdot9.81} \\ h=12.56\text{ m} \end{gathered}[/tex](c)The time to reach this height would be half the time it takes for the entire journey, that is, 0.8 seconds
The wiring in a house must be thick enough so it does not become so hot as to start a fire.part aWhat diameter must a copper wire be if it is to carry a maximum current of 34 A and produce no more than 1.6 W of heat per meter of length?
Answers
Given:
The maximum current in the circuit is,
[tex]i=34\text{ A}[/tex]The power per length is,
[tex]\frac{P}{l}=1.6\text{ W/m}[/tex]To find:
The diameter of the copper wire
Explanation:
The power (P) produced by current i, through a copper wire of resistance R and length l is given by,
[tex]\begin{gathered} Pl=i^2R \\ \frac{R}{l}=\frac{P}{i^2} \\ \frac{R}{l}=\frac{1.6}{34\times34} \end{gathered}[/tex]Now,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ R=\frac{\rho l}{\pi r^2} \end{gathered}[/tex]The resistivity of copper is,
[tex]\rho=1.72\times10^{-8}\text{ ohm.m}[/tex]So, we can write,
[tex]\begin{gathered} \frac{R}{l}=\frac{\rho}{\pi r^2} \\ \frac{1.6}{34\times34}=\frac{1.72\times10^{-8}}{\pi r^2} \\ r^2=\frac{1.72\times10^{-8}\times34\times34}{1.6} \\ r=3.5\times10^{-3}\text{ m} \\ diamer\text{ is,} \\ 2r=7.0\times10^{-3}\text{ m} \end{gathered}[/tex]Hence, the diameter is,
[tex]7.0\times10^{-3}\text{ m}[/tex]Bart, mass 32.4 kilograms, and Milhouse, mass 27.6 kilograms, play on the schoolyard seesaw. If Bart and Milhouse want to sit 4.0 meters apart, how far from the center of the seesaw should Bart sit? Include units in your answer. Answer must be in 3 significant digits.
Answers
Given data:
* The mass of the Bart is m_1 = 32.4 kg.
* The mass of the Milhouse is m_2 = 27.6 kg.
* The distance between the Millhouse and Bart is d = 4 m.
Solution:
To balance the seesaw, the net moment about the center should be zero.
The diagrammatic representation of the given system is,
The distance between the Bart and Milhouse can be written as,
[tex]\begin{gathered} d=d_1+d_2 \\ 4=d_1+d_2 \\ d_2=4-d_1 \end{gathered}[/tex]where d_2 is the distance of Milhouse from the center and d_1 is the distance of Bart from the center,
Consider the moment as positive if it is in an anticlockwise direction and negative if it is a clockwise direction.
Thus, the net moment about the center is,
[tex]M=m_1d_1-m_2d_2[/tex]Substituting the known values,
[tex]\begin{gathered} 0=32.4\times d_1-27.6\times(4-d_1) \\ 0=32.4\times d_1-110.4+27.6d_1 \\ 0=60d_1-110.4 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} 60d_1=110.4 \\ d_1=\frac{110.4}{60} \\ d_1=1.84\text{ m} \end{gathered}[/tex]Thus, the distance of the Bart from the center is 1.84 meters.
Hunter pushed a couch across the room. He did 800 J of work in 20 seconds.The couch weighed 500 N. How much power did he have?A. 16,000 WB. 1.6WC. 800 WD. 40 W
Answers
Power = 40 W
Option D
Explanation:The workdone by the Hunter = 800 Joules
The weight of the couch = 500 N
Time, t = 20 seconds
Power = Workdone/Time
Power = 800/20
Power = 40 W
A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 219 m, and the car completes the turn in 32.0 s.
(a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors î and ĵ.
(b) Determine the car's average speed.
(c) Determine its average acceleration during the 32.0-s interval.
Answers
a.) the acceleration when the car is at B located at an angle of 35.0° is 2.689i -0.42818j
b.) the car's average speed is v= 6.84375 m/s
c.) average acceleration during the 32.0-s interval is (−0.181 i+0.181 j)m/s²
What is acceleration?Acceleration is described as the rate of change of the velocity of an object with respect to time.
(a) The car’s speed around the curve is found from
v= 219/32.0
v= 6.84375 m/s
This is the answer to part (b) of this problem. We calculate the radius of the curve from
(1/4) X 2πr = 219 m
which gives r = 139.4 m
The car’s acceleration at point B is then
ar = (V²/r ) towards the center
= ( 6.84375)² / 139.4 at 35.0° north of west
= (2.9761 m/s²)(cos 35.0)(-i) + (sin 35.0j)
= -2.689i -0.42818j
(b) From part (a), v= 6.84375 m/s
(c) We find the average acceleration from
A avg = (Vf - Vi)/ change in t
A avg = ( 6.84375 j - 6.84375 i ) / 32.0 s
A avg = (−0.181 i+0.181 j)m/s²
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I need help with this question There is 4 answersa)b)c)d)
Answers
Given,
The velocity of the joggers, v=-3.50 m/s
The mass of Jim, M=100 kg
The mass of Tom, m=59 kg
(a) The kinetic energy of the system is given by,
[tex]\begin{gathered} E_a=\frac{1}{2}mv^2+\frac{1}{2}Mv^2 \\ =\frac{1}{2}v^2(m+M) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_a=\frac{1}{2}\times3.50^2(59.0+100) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the kinetic energy of the system is 973.88 J
(b)
The total momentum of the system is given by,
[tex]\begin{gathered} p_b=mv+Mv \\ =(m+M)v \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_b=(59+100)\times3.50 \\ =556.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system is 556.5 kg m/s
(c)
Given that the velocity of Tom is -v
The total kinetic energy of the system is given by,
[tex]\begin{gathered} E_c=\frac{1}{2}Mv^2+\frac{1}{2}m(-v)^2 \\ =\frac{1}{2}(Mv^2+m(-v)^2) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_c=\frac{1}{2}(100\times3.50^2+59\times(-3.50)^2) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the total kinetic energy of the system, in this case, is 973.88 J
(d)
The total momentum of the system is given by,
[tex]\begin{gathered} p_d=Mv+m(-v) \\ =v(M-m) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_d=3.50(100-59) \\ =143.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system, in this case, is 143.5 kg m/s
Forces that are equal in magnitude
but opposite in direction will:
Answers
Answer:
The two forces equal in magnitude but acting opposite in direction on a body are called balanced forces
Explanation: