Answer:
The location of the center of gravity for the entire body, relative to the floor is 1.03 m
Explanation:
To find the center of gravity of a system of particles, we use that
[tex]R_{cog} = \frac{r_{1}*m_{1}+r_{2}*m_{2}+...+r_{n}*m_{n}}{M}[/tex]
where R is the vector center of gravity of the system, formed by n particles, and n masses.
In this case, for a person standing on the floor and being their body divided in three sectors, each one with a weight and an altitud in a specific point (center of gravity of the body sector), instead of mass we have every "particle" weight in Newtons (force instead of mass), being each "particle" in the formula, a sector of the body.
On the other hand, we use only magnitude for the calculation, because the gravity force is vertical to the floor, so instead of our vector formula, we use it in the vertical direction as a magnitude formula. Thus
[tex]Y_{cog} = \frac{y_{1}*W_{1}+y_{2}*W_{2}+y_{3}*W_{3}}{Wt}[/tex]
where Y is the center of gravity, y=1, 2, 3 is every "sector point" altitude from the floor, W=1, 2, 3 is every weight of a body "sector", and Wt is the sum of the three weights.
In this way we replace in our formula with the correspondent values
[tex]Y_{cog} = \frac{1.28m*438N+0.76m*144N+0.25m*87N}{669N}[/tex]
obtaining our result
[tex]Y_{cog}=1.03 m[/tex]
A -5.45 nC point charge is on the x axis at x = 1.35 m . A second point charge Q is on the x axis at -0.595 m. What must be the sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the +x direction? What must be the sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the −x direction?
Answer:
a)Q= + 0.71 nC , For the resultant electric field at the origin to be 45.0 N/C in the +x direction
b)Q= -2.83nC ,for the resultant electric field at the origin to be 45.0 N/C in the −x direction
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻9 C
Data
k = 8.99*10⁹ N×m²/C²
q₁ =+5.45nC = 3*10⁻⁹C
d₁ =1.35 m
d₂ = 0.595m
a)Problem development : sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the +x direction
We make the algebraic sum of fields at at the origin :
[tex]E_{o} =E_{q} +E_{Q}[/tex] Equation (1)
[tex]E_{q} =\frac{k*q_{1} }{d_{1}{2} }[/tex]
Calculation of E(q)
[tex]E_{q} =\frac{8.99*10^{9} *5.45*10^{-9} }{1,35^{2} }[/tex]
[tex]E_{q} =26.88\frac{N}{C}[/tex] : in the +x direction .As the charge is negative, the field enters the charge
We replace [tex]E_{o}[/tex] and [tex]E_{q}[/tex] in the equation (1)
[tex]45=26.88+E_{Q}[/tex]
[tex]E_{Q} =45-26.88[/tex]
[tex]E_{Q} = 18.12 N/C[/tex] : in the +x direction .
Sign and magnitude of Q
Q must be positive for the field to abandon the load in the +x
[tex]E_{Q} =\frac{k*Q}{d_{2}^{2} }[/tex]
[tex]18.12=\frac{8.99*10^{9}*Q }{0.595^{2} }[/tex]
[tex]Q=\frac{18.12*0.595^{2} }{8.99*10^{9} }[/tex]
Q=0.71*10⁻⁹ C =0.71 nC
b)Sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the −x direction
We make the algebraic sum of fields at at the origin :
[tex]E_{o} =E_{q} +E_{Q}[/tex]
[tex]-45=26.88+E_{Q}[/tex]
[tex]-71.88=E_{Q}[/tex]
[tex]71.88=\frac{8.99*10^{9} *Q}{0.595^{2} }[/tex]
Q= 2.83*10⁻⁹ C
Q= -2.83nC
Q must be negative for the field to enters the charge in the −x direction
The magnitude and sign of Q is given by the required magnitude and
sign of the charge at the origin due to the sum of the charges.
Responses:
The sign and magnitude of Q when the charge is 45 N/C in the +x direction is, Q ≈ 2.83 nCThe sign and magnitude of Q when the charge is 45 N/C n the -x direction is, Q ≈ -713.4 pCHow can the charge of the two particles at the origin be found?
The charge at the origin is given as follows;
When the charge at the origin is 45.0 N/C, we have;
[tex]45 = \mathbf{\dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} + \dfrac{8.99 \times 10^{9} \times Q} {(-0.595)^2}}[/tex]
Which gives;
[tex]Q = \dfrac{\left(45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx \mathbf{2.83 \times 10^{-9}}[/tex]
When the charge at the origin is [tex]E_0[/tex] = 45 N/C, we have;
Q ≈ 2.83 × 10⁻⁹ C = 2.83 nCWhen the charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;
[tex]Q = \dfrac{\left(-45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx -7.134 \times 10^{-10}[/tex]
Therefore;
The charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;
Q ≈ -7.134 × 10⁻¹⁰ C = -713.4 pCLearn more about electric field strength here:
https://brainly.com/question/12644782
A car is able to stop with an acceleration of − 3.00 m/s^2. Justify the mathematical routine used to calculate the distance required to stop from a velocity of 100.0 km/h by choosing the correct answer below.
A.) 16.7m because the average velocity is 50 km/h, the change in velocity divided by the acceleration is the time, and the time multiplied by the average velocity is the distance.
B.) 64.4m because the average velocity is 13.9 m/s, the average velocity divided by the acceleration is the time, and the time multiplied by the average velocity is the distance.
C.) 129m because the average velocity is 13.9 m/s, the change in velocity divided by the acceleration is the time, and the time multiplied by the average velocity is the distance.
D.) 257m because the initial velocity is 27.8 m/s, the initial velocity divided by the acceleration is the time, and the time multiplied by the initial velocity is the distance.
Answer:
129 m because the average velocity is 13.9 m/s, the change in velocity
divided by the acceleration is the time, and the time multiplied by the
average velocity is the distance. ⇒ answer C
Explanation:
Lets explain how to solve the problem
The given is:
The care is able to stop with an acceleration of -3 m/s²
→ The final velocity = 0 and acceleration = -3 m/s²
Calculate the distance required to stop from a velocity of 100 km/h
→ Initial velocity = 100 km/h
At first we must to change the unite of the initial velocity from km/h
to m/s because the units of the acceleration is m/s²
→ 1 km = 1000 meters and 1 hr = 3600 seconds
→ 100 km/h = (100 × 1000) ÷ 3600 = 27.78 m/s
The initial velocity is 27.78 m/s
Acceleration is the rate of change of velocity during the time,
then the time is the change of velocity divided by the acceleration
→ [tex]t=\frac{v-u}{a}[/tex]
where v is the final velocity, u is the initial velocity, t is the time and
a is the acceleration
→ v = 0 , u = 27.78 m/s , a = -3 m/s²
Substitute these values in the rule
→ [tex]t=\frac{0-27.78}{-3}=9.26[/tex] seconds
The time to required stop is 9.26 seconds
We can calculate the distance by using the rule:
→ s = ut + [tex]\frac{1}{2}[/tex] at²
→ u = 27.78 m/s , t = 9.26 s , a = -3 m/s²
Substitute these values in the rule
→ s = 27.78(9.26) + [tex]\frac{1}{2}[/tex] (-3)(9.26) = 128.6 ≅ 129 m
The distance required to stop is 129 m
Average velocity is total distance divided by total time
→ Total distance = 129 m and total time = 9.26 s
→ average velocity = 129 ÷ 9.26 = 13.9 m/s
The average velocity is 13.9 m/s
So the time multiplied by the average velocity is the distance
The answer is C
129 m because the average velocity is 13.9 m/s, the change in
velocity divided by the acceleration is the time, and the time
by the average velocity is the distance.
You've been called in to investigate a construction accidentin
which the cable broke while a crane was lifting a 5300 kg
container. The steel cable is 2.0 cm indiameter and has a safety
rating of 50,000 N. The crane is designednot to exceed speeds of
3.0 m/s or accelerations of 1.0m/s2, and your tests find
that the crane is notdefective. What is the maximum tension the
cable?
Answer:
57300 N
Explanation:
The container has a mass of 5300 kg, the weight of the container is:
f = m * a
w = m * g
w = 5300 * 9.81 = 52000 N
However this container was moving with more acceleration, so dynamic loads appear.
w' = m * (g + a)
w' = 5300 * (9.81 + 1) = 57300 N
The rating for the cable was 50000 N
The maximum load was exceeded by:
57300 / 50000 - 1 = 14.6%
Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric field when there are two fields, E1and E2, where E1 is directed vertically upward and has magnitude 100 V/m and E2 is directed 45 degrees to the left of E1 and has magnitude 150 V/m. Use a graph to show vector drawing!
The resultant electric field (Er) is approximately 231.76 V/m, directed at an angle of about 62.76° to the left of the vertical E1 direction.
Find the resultant electric field when there are two fields, E1 and E2:
Step 1: Resolve the electric fields into their x and y components.
E1:
Ex1 = 0 V/m (because E1 is directed vertically upward)
Ey1 = 100 V/m
E2:
Since E2 is directed 45 degrees to the left of E1, we can use trigonometry to find its x and y components.
Ex2 = 150 V/m * sin(45°) = 106.07 V/m (directed to the left, so negative)
Ey2 = 150 V/m * cos(45°) = 106.07 V/m
Step 2: Add the x and y components of each electric field separately.
Σ Ex = Ex1 + Ex2 = 0 V/m - 106.07 V/m = -106.07 V/m
Σ Ey = Ey1 + Ey2 = 100 V/m + 106.07 V/m = 206.07 V/m
Step 3: Find the magnitude of the resultant electric field (Er) using the Pythagorean theorem.
Er = √(Σ Ex)² + (Σ Ey)²
Er = √(-106.07 V/m)² + (206.07 V/m)²
Er ≈ 231.76 V/m
Step 4: Find the direction of the resultant electric field using arctangent.
tan(θ) = Σ Ey / Σ Ex
tan(θ) = 206.07 V/m / -106.07 V/m
θ ≈ arctan(-1.94) ≈ -117.24°
Since the arctangent function only outputs values between -90° and 90°, we need to add 180° to get the angle within the range of 0° to 180°.
θ = -117.24° + 180° ≈ 62.76°
Therefore, the resultant electric field has a magnitude of approximately 231.76 V/m and is directed approximately 62.76° to the left of E1.
Carl is eating lunch at his favorite cafe when his friend Isaac calls and says he wants to meet him. Isaac is calling from a city 175 miles away, and wants to meet Carl somewhere between the two locations. Isaac says he will start driving right away, but Carl needs 35.0 min to finish his lunch before he can begin driving. Isaac plans to drive at 65.0 mph , and Carl plans to drive at 50.0 mph . Ignore acceleration and assume the highway forms a straight line. How long will Isaac be driving before he meets Carl?
Answer:
106.52 minutes
Explanation:
Given:
Initial distance between Carl and Isaac = 175 miles
speed of Isaac = 65 mph
Speed of Carl = 50 mph
Now, the Carl starts after 35 minutes, but the Isaac has already started so the distance covered by Isaac in 35 minutes will be
= [tex]\frac{35}{60}\times65[/tex]
= 37.91 miles
Therefore,
the distance left between Isaac and Carl = 175 - 37.91 = 137.09 miles
Since Isaac and Carl are moving towards each other,
therefore the relative speed between the both = 65 + 50 = 115 mph
Hence, the time taken to meet = [tex]\frac{137.09}{115}[/tex]
or
The time taken to meet = 1.192 hours
or
The time taken = 1.192 × 60 = 71.52 minutes
Therefore the total time Isaac have been travelling = 71.52 + 35
= 106.52 minutes
A skier is gliding along at 4.2 m/s on horizontal, frictionless snow. He suddenly starts down a 10° incline. His speed at the bottom is 18 m/s . a) What is the length of the incline?
Express your answer with the appropriate units.
b) How long does it take him to reach the bottom?
Express your answer with the appropriate units.
Answer:
a) 90m
b) 8.1s
Explanation:
The shortest way to finding the length of the incline is to apply an energetic analysis to determine the height of the incline. At the beginning, there is potential and kinetic energy that will turn into kinetic energy only:
[tex]mgh+\frac{1}{2}mv_{0}^{2} =\frac{1}{2}mv_{f}^{2}[/tex]
Before we input any information, let's solve for h:
[tex]mgh+\frac{1}{2}mv_{0}^{2} =\frac{1}{2}mv_{f}^{2}}\\\\m(gh+\frac{v_{0}^{2}}{2})=\frac{1}{2}mv_{f}^{2}}\\\\gh+\frac{v_{0}^{2}}{2}=\frac{v_{f}^{2}}{2}\\\\gh=\frac{1}{2}(v_{0}^{2}-v_{f}^{2})\\\\h=\frac{1}{2g}(v_{0}^{2}-v_{f}^{2})=\frac{1}{2*9.8\frac{m}{s^{2}}}(4.2\frac{m}{s}^{2}-18\frac{m}{s}^{2})=15.63m[/tex]
Using a sine formula we can solve for [tex]l[/tex]:
[tex]Sin(10)=\frac{h}{l}\\\\l=\frac{h}{Sin(10)}=\frac{15.63m}{0.17}=90m[/tex]
In order to find the time we will use the distance formula and final velocity formula as a system of equations to solve for t:
[tex]X=V_{0}t+\frac{at^2}{2}\\\\V_f=V_0+at[/tex]
Since the acceleration and time are both variables we will solve for acceleration in the final velocity formula and replace in the distance formula:
[tex]V_f=V_0+at\\V_f-V_0=at\\\frac{V_f-V_0}{t}=a\\\\l=V_0t+\frac{(\frac{V_f-V_0}{t})t^2}{2}\\l=V_0t+\frac{(V_f-V_0)t}{2}\\l=t(V_0+\frac{V_f-V_0}{2})\\\\t=\frac{l}{V_0+\frac{V_f-V_0}{2}}=\frac{90m}{4.2\frac{m}{s}+\frac{18\frac{m}{s}-4.2\frac{m}{s}}{2}}=8.1s[/tex]
Final answer:
In this question, we calculate the length of the incline and the time taken by a skier to reach the bottom based on given velocity and angle.
Explanation:
For part a: The length of the incline can be found using the principles of physics. Using the given information about the skier's initial and final speeds and the angle of the incline, you can calculate that the length of the incline is approximately 67.7 meters.
For part b: To determine the time it takes for the skier to reach the bottom, you can use the kinematic equations of motion along the incline. The time taken for the skier to reach the bottom is about 9.64 seconds.
A 3 inch fire hose has a water flow of 200 gallons per minute. What is the flow in liters per second? Note: Use US gallons not UK gallons. (10 points) The blower on an air conditioning unit produces 95 cubic feet per minute of air in the ductwork. What is the air flow in cubic meters per hour? m/hr (10 points) Torque (or moment) is the measure of turning force or twist on an object. If a mechanic is applying 12 Newton centimeters of torque on a bolt how many pound inches would this be?
Answer:
Case I: 12.617 L/s
Case II: 161.406 cubic meters per hour
Case III: 1.062 Pound inches
Explanation:
Given:
Speed of water flow = 200 gallons per minuteSpeed of air blow = 95 cubic feet per minuteMeasure of Torque = 12 Newton centimeterAssumptions:
1 US gallon = 3.785 L1 min = 60 s1 ft = 0.3048 m1 h = 60 min1 inch = 2.54 cm1 N = 0.2248 lbCase I:
[tex]Speed\ of\ water\ flow = 200 \dfrac{gallon}{min}\\\Rightarrow V_{water} = 200\times \dfrac{3.785\ L}{60\ s}\\\Rightarrow V_{water} = 12.617\ L/s[/tex]
Case II:
[tex]Speed\ of\ air\ blow = 95 \dfrac{ft^3}{min}\\\Rightarrow V_{air} = 95\times \dfrac{(0.3048\ m)^3}{\dfrac{1}{60}\ h}\\\Rightarrow V_{air} = 95\times (0.3048)^3\times 60\ m^3/h\\\Rightarrow V_{air} = 161.406\ m^3/h[/tex]
Case III:
[tex]Measure\ of\ torque = 12\ N cm\\\Rightarrow \tau = 12\times (0.2248\ lb)\times \dfrac{1}{2.54}\ in\\ \Rightarrow \tau = 1.062\ lb in[/tex]
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s later. You may ignore air resistance. A-If the initial speed of the first ball is v0 = 8.90 m/s what must the height h of the building be for both balls to reach the ground at the same time?
B-If v0 is greater than some value vmax, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for vmax
C-if v0 is less than some value vmin, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for vmin.
Answer:
[tex]h=53.09m[/tex] (2)
[tex]v_{min}>5.05m/s[/tex]
[tex]v_{max}<10.4m/s[/tex]
Explanation:
a)Kinematics equation for the first ball:
[tex]v(t)=v_{o}-g*t[/tex]
[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]
[tex]y_{o}=h[/tex] initial position is the building height
[tex]v_{o}=8.9m/s[/tex]
The ball reaches the ground, y=0, at t=t1:
[tex]0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}[/tex]
[tex]h=1/2*g*t_{1}^{2}-v_{o}t_{1}[/tex] (1)
Kinematics equation for the second ball:
[tex]v(t)=v_{o}-g*t[/tex]
[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]
[tex]y_{o}=h[/tex] initial position is the building height
[tex]v_{o}=0[/tex] the ball is dropped
The ball reaches the ground, y=0, at t=t2:
[tex]0=h-1/2*g*t_{2}^{2}[/tex]
[tex]h=1/2*g*t_{2}^{2}[/tex] (2)
the second ball is dropped a time of 1.03s later than the first ball:
t2=t1-1.03 (3)
We solve the equations (1) (2) (3):
[tex]1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}[/tex]
[tex]g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)[/tex]
[tex]g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)[/tex]
[tex]-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)[/tex]
[tex]2.06*gt_{1}-2v_{o}t_{1}=g*1.06[/tex]
[tex]t_{1}=g*1.06/(2.06*g-2v_{o})[/tex]
vo=8.9m/s
[tex]t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s[/tex]
t2=t1-1.03 (3)
t2=3.29sg
[tex]h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m[/tex] (2)
b)[tex]t_{1}=g*1.06/(2.06*g-2v_{o})[/tex]
t1 must : t1>1.03 and t1>0
limit case: t1>1.03:
[tex]1.03>9.81*1.06/(2.06*g-2v_{o})[/tex]
[tex]1.03*(2.06*9.81-2v_{o})<9.81*1.06[/tex]
[tex]20.8-2.06v_{o}<10.4[/tex]
[tex](20.8-10.4)/2.06<v_{o}[/tex]
[tex]v_{min}>5.05m/s[/tex]
limit case: t1>0:
[tex]g*1.06/(2.06*g-2v_{o})>0[/tex]
[tex]2.06*g-2v_{o}>0[/tex]
[tex]v_{o}<1.06*9.81[/tex]
[tex]v_{max}<10.4m/s[/tex]
The brakes are applied to a car traveling on a dry, level highway. A typical value for the magnitude of the car's acceleration is 4.60 m/s2. If the car's initial speed is 31.8 m/s, how long does it take to stop and how far does it travel, starting from the moment the brakes are applied?
Final answer:
The car takes approximately 6.91 seconds to stop and travels approximately 110.7 meters before coming to a stop.
Explanation:
The car's initial speed is 31.8 m/s and it decelerates at a rate of 4.60 m/s2. To find the time it takes to stop, we can use the equation:
Final velocity = Initial velocity + (Acceleration x Time)
Since the final velocity is 0 m/s when the car stops, we can rearrange the equation to solve for time:
Time = (Final velocity - Initial velocity) / Acceleration
Plugging in the values, we get:
Time = (0 m/s - 31.8 m/s) / -4.60 m/s2
Simplifying the equation, we find that it takes approximately 6.91 seconds for the car to stop.
To find the distance traveled, we can use the equation:
Distance = Initial velocity x Time + (0.5 x Acceleration x Time2)
Plugging in the values, we get:
Distance = 31.8 m/s x 6.91 s + (0.5 x -4.60 m/s2 x (6.91 s)2)
Simplifying the equation, we find that the car travels approximately 110.7 meters before it comes to a stop.
What is the magnitude of the electric force between two point charges with Q1 = -1.5 C and Q2 = 0.8 C at a distance of 1 km?
Answer:
F = -10800 N
Explanation:
Given that,
Charge 1, [tex]q_1=-1.5\ C[/tex]
Charge 2, [tex]q_2=0.8\ C[/tex]
Distance between the charges, [tex]d=1\ km=10^3\ m[/tex]
We need to find the electric force acting between two point charges. Mathematically, it is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]F=-9\times 10^9\times \dfrac{1.5\times 0.8}{(10^3)^2}[/tex]
F = -10800 N
So, the magnitude of electric force between two point charges is 10800 N. Hence, this is the required solution.
What is the difference between average and instantaneous velocity?
Explanation:
Instantaneous velocity is specific rate of the change of the position or the displacement with respect to the time at a particular single point (x,t)
[tex]v(t)=\frac{d}{dt}x(t)[/tex]
Average velocity is average rate of the change of the position or the displacement with respect to the time over an particular interval.
[tex]{v}=\frac {\Delta x}{\Delta t}=\frac{{x}_{\text{f}}-{x}_{\text{i}}}{{t}_{\text{f}}-{t}_{\text{i}}}[/tex]
A fathom is a unit of length, usually reserved for measuring the depth of water. A fathom is exactly 6.00 ft in length. Take the distance from Earth to the Moon to be 246,000 miles, and use the given approximation to find the distance in fathoms. 1 mile = 5280 ft. Note, this one needs to be in fathoms so make sure you include "fathoms" in the units
Answer:
216480000 fathoms
Explanation:
1 fathom = 6 feet
[tex]1\ feet=\frac{1}{6}\ fathom[/tex]
Distance from Earth to the Moon = 246000 miles
Converting to feet
1 mile = 5280 feet
246000 miles = 1298880000 feet
Convert to fathom
[tex]1\ feet=\frac{1}{6}\ fathom\\\Rightarrow 1298880000\ feet=\frac{1298880000}{6}=216480000\ fathom[/tex]
So, the distance between Earth and Moon is 216480000 fathoms
At the instant the traffic light turns green, a car starts with a constant acceleration of 8.00 ft/s^2. At the same instant a truck, traveling with a constant speed of 10.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?
Answer:
they meet at distance 25 feet
Explanation:
given data
acceleration of car = 8 ft/s²
truck speed = 10 ft/s
car initial speed u = 0
truck acceleration = 0
to find out
How far from the starting point will car overtake the truck
solution
we apply here equation of motion
s = ut + 0.5 ×a×t² .............1
here s is distance and a is acceleration and t is time u is initial speed
so truck distance
s = 10t + 0.5 ×0×t²
s = 10 t ...............2
and car distance
s = 0+ 0.5 ×8×t²
s = 4×t² ..........................3
so from equation 2 and 3
10 t = 4×t²
t = 2.5 s
so both meet at distance
s = 10 (t)
s = 10 ( 2.5 ) = 25 ft
so they meet at distance 25 feet
Cosmic rays are highly energetic particles which can travel at great speeds through outer space. A typical speed for a cosmic ray is 1.29 x 10^8 m/s. What is this speed converted to km/hr (kilometers per hour)? A. 1.29 x 10^5 km/hr B. 35.8 km/hr C. 7.74 x 10^6 km/hr D. 4.64 x 10^8 km/hr
Answer:
Speed of the cosmic rays, [tex]v=4.64\times 10^8\ km/hr[/tex]
Explanation:
It is given that, Cosmic rays are highly energetic particles which can travel at great speeds through outer space.
The speed of the cosmic rays, [tex]v=1.29\times 10^8\ m/s[/tex]
We need to convert the speed of cosmic rays to kilometers per hour. We know that :
1 kilometers = 1000 meters
1 hour = 3600 seconds
[tex]v=1.29\times 10^8\ m/s=\dfrac{1.29\times 10^8\times (1/1000\ km)}{(1/3600\ h)}[/tex]
On solving the above expression,
[tex]v=464400000\ km/h[/tex]
or
[tex]v=4.64\times 10^8\ km/hr[/tex]
So, the speed of the cosmic rays is [tex]4.64\times 10^8\ km/hr[/tex]. Hence, this is the required solution.
The skateboarder in the drawing starts down the left side of the ramp with an initial speed of 7.03 m/s. Neglect nonconservative forces, such as friction and air resistance, and find the height h of the highest point reached by the skateboarder on the right side of the ramp. (g = 9.80 m/s2)
Answer:
h=5.04m
Explanation:
At the bigining he has only kinetic energy because he is at the lowest point. When he is at the highest point, he is no longer moving because he will start moving downwards: all his kinetic energy transformed into potential gravitational energy:
[tex]E_k=E_p[/tex]
[tex]\frac{mv^2}{2}=mgh\\ h= \frac{v^2}{g}[/tex]
h=5.04m
Answer:
The height of the highest point reached by the skateboarder on the right side of the ramp is [tex]h=2.52m[/tex]
Explanation:
In this problem, we use the Energy Conservation Principle, as there are not nonconservative forces, the energy is constant and we will call it E. Then, we choose two different moments that will be advantageous for our analysis.
The first moment is the moment where the skateboarder is situated at the height 0 meters (meaning that the potential energy is zero), i.e, the energy is totally kinetic. We will call it K.
The second moment is the moment where the skateboarder is situated at the highest point reached, this means that all energy will be potential, because for an instant, the velocity of the skateboarder is zero (and in the following instant he will go back the way he came due to the pull of gravity). We will call it V. This is the point with the height that we want to calculate.
Therefore, we can equalize both energies, as they are constant and equal to E. We write
[tex]E=K=V=\frac{1}{2}mv^2 =mgh\Leftrightarrow \frac{1}{2}v^2=gh\Leftrightarrow h=\frac{v^2}{2g}[/tex]
Now, we can replace the given data, to obtain
[tex]h=\frac{7.03^2}{2*9.80}m=2.52m[/tex]
which is the outcome.
A 6.0 kg ball is dropped from a 10-m height. Its kinetic energy, just before it hits the ground is : 588 J
392 J
280 J
140 J
882 J
Answer:
588 J
Explanation:
mass of ball, m = 6 kg
Height from it dropped, h = 10 m
initial velocity, u = 0
acceleration due to gravity, g = 9.8 m/s^2
Let it hits the ground with velocity v.
Use third equation of motion
[tex]v^{2} = u^{2}+2as[/tex]
[tex]v^{2}=0^{2}+2\times 9.8\times 10[/tex]
v = 14 m/s
The formula for the kinetic energy is given by
[tex]K=\frac{1}{2}mv^{2}[/tex]
where, m is the mass of the ball and v be the velocity of the ball as it hits the ground
K = 0.5 x 6 x 14 x 14
K = 588 J
Two point charges, A and B, are separated by a distance of 16.0cm. The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 43.0 N on the other, find the magnitudes of the charges. Charge A: ____ in C
Charge B: _____ in C
Answer:
Charge on A is [tex]q=0.7820\times 10^{-5}C[/tex]
Charge on B is [tex]2q=2\times 0.7820\times 10^{-5}C=1.5640\times 10^{-5}C[/tex]
Explanation:
We have given one charge is twice of other charge
Let [tex]q_1=q[/tex], then [tex]q_2=2q[/tex]
Distance between two charges = 16 cm = 0.16 m
Force F = 43 N
According to coulombs law force between tow charges is given by
[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex], here K is constant which value is [tex]9\times 10^9[/tex]
So [tex]43=\frac{9\times 10^92q^2}{0.16^2}[/tex]
[tex]q^2=0.0611\times 10^{-9}[/tex]
[tex]q^2=0.611\times 10^{-10}[/tex]
[tex]q=0.7820\times 10^{-5}C[/tex] so charge on A is [tex]q=0.7820\times 10^{-5}C[/tex]
And charge on B is [tex]2q=2\times 0.7820\times 10^{-5}C=1.5640\times 10^{-5}C[/tex]
The magnitudes of the charges are:
Charge A: [tex]\underline {1.10 \times 10^{-5} \text{C}}[/tex].
Charge B: [tex]\underline {5.52 \times 10^{-6} \text{C}}[/tex]
Use Coulomb's Law, which states:
[tex]F = k \frac{ |q_1 \, q_2| }{ r^2 }[/tex]
where:
[tex]F[/tex] is the magnitude of the force between the charges,[tex]k[/tex] is Coulomb's constant [tex](8.99 \times 10^9 \; \text{Nm}^2/\text{C}^2)[/tex],[tex]q_1[/tex] and [tex]q_2[/tex] are the magnitudes of the charges,[tex]r[/tex] is the distance between the charges.Given:
The distance [tex]r = 0.16 \; \text{m}[/tex] (converted from 16.0 cm)The force [tex]F = 43.0 \; \text{N}[/tex]The magnitude of charge on A is twice that of charge on B [tex](q_1 = 2q_2)[/tex]Let's substitute these values into Coulomb's Law:
[tex]43.0 \; \text{N} = (8.99 \times 10^9 \; \text{Nm}^2/\text{C}^2) \frac{ |2q \cdot q| }{ (0.16 \; \text{m})^2 }[/tex]
Simplify the equation:
[tex]43.0 \; \text{N} = (8.99 \times 10^9) \frac{ 2q^2 }{ 0.0256 } \text{Nm}^2/\text{C}^2[/tex]
[tex]43.0 \; \text{N} = ( 7.03 \times 10^{11} ) 2q^2[/tex]
Solving for [tex]q^2[/tex]:
[tex]43.0 = 1.406 \times 10^{12} q^2[/tex]
[tex]q^2 = \frac{ 43.0 }{ 1.406 \times 10^{12} }[/tex]
[tex]q = \sqrt{ \frac{ 43.0 }{ 1.406 \times 10^{12} } }[/tex]
[tex]q \approx 5.52 \times 10^{-6} \text{C}[/tex]
This gives us the magnitude of charge B [tex]q_2[/tex].
Since [tex]q_1 = 2q_2[/tex]:
Charge A: [tex]q_1 = 2 \times 5.52 \times 10^{-6} \text{C} = 1.10 \times 10^{-5} \text{C}[/tex]
Charge B: [tex]q_2 = 5.52 \times 10^{-6} \text{C}[/tex]
What is the magnitude (in N/C) and direction of an electric field that exerts a 4.00 x 10^−5 N upward force on a −2.00 µC charge? magnitude N/C
direction: upward, downward, to the left, to the right
Answer:
The magnitude of electric field is [tex]1.25\times10^{14}\ N/C[/tex] in downward.
Explanation:
Given that,
Force [tex]F= 4.00\times10^{-5}\ N[/tex]
Charge q= -2.00 μC
We know that,
Charge is negative, then the electric field in the opposite direction of the exerted force.
We need to calculate the magnitude of electric field
Using formula of electric force
[tex]F = qE[/tex]
[tex]E = \dfrac{F}{q}[/tex]
[tex]E=\dfrac{4.00\times10^{-5}}{-2.00\times1.6\times10^{-19}}[/tex]
[tex]E=-1.25\times10^{14}\ N/C[/tex]
Negative sign shows the opposite direction of electric force.
Hence, The magnitude of electric field is [tex]1.25\times10^{14}\ N/C[/tex] in downward.
A solid cylinder of cortical bone has a length of 500mm, diameter of 2cm and a Young’s Modulus of 17.4GPa. Determine the spring constant ‘k’. Please explain.
Answer:
The spring constant k is[tex]1.115\times 10^{9} N/m[/tex]
Solution:
As per the question:
Length of the solid cylinder, L = 500 mm = [tex]500\times 10^{- 3} = 0.5 m[/tex]
Diameter pf the cylinder, D = 2 cm = 0.02 m
As the radius is half the diameter,
Radius, R = 1 cm = 0.01 m
Young's Modulus, E = 17.4 GPa = [tex]17.4\times 10^{9} Pa[/tex]
Now,
The relation between spring constant, k and Young's modulus:
[tex]kL = EA[/tex]
where
A = Area
Area of solid cylinder, A = [tex]2\piR(L + R)[/tex]
[tex]0.5k = 17.4\times 10^{9}\times 2\piR(L + R)[/tex]
[tex]k = \frac{17.4\times 10^{9}\times 2\pi\times 0.01(0.01 + 0.5)}{0.5}[/tex]
k = [tex]1.115\times 10^{9} N/m[/tex]
Young's modulus, E is the ratio of stress and strain
And
Stress = [tex]\frac{Force or thrust}{Area}[/tex]
Strain = [tex]\frac{length, L}{elongated or change in length, \Delta L}[/tex]
Also
Force on a spring is - kL
Therefore, we utilized these relations in calculating the spring constant.
If a body travels half its total path in the last 1.10 s of its fall from rest, find the total time of its fall (in seconds).
Answer:3.75 s
Explanation:
Given Body travels half of its motion in last 1.1 sec
Let h be the height and t be the total time taken
here initial velocity is zero
[tex]h=ut+\frac{gt^2}{2}[/tex]
[tex]h=0+\frac{gt^2}{2}[/tex]
[tex]h=\frac{gt^2}{2}------1[/tex]
Now half of the distance traveled will be in t-1.1 s and half distance traveled is in last 1.1 s
[tex]\frac{h}{2}=\frac{g\left ( t-1.1\right )^2}{2}-----2[/tex]
from 1 & 2 we get
[tex]gt^2=2g\left ( t-1.1\right )^2[/tex]
[tex]t^2-4.4t+2.42=0[/tex]
[tex]t=\frac{4.4\pm \sqrt{4.4^2-4\left ( 1\right )\left ( 2.42\right )}}{2}[/tex]
[tex]t=\frac{4.4\pm 3.11}{2}[/tex]
Therefore two value of t is satisfying the equation but only one value is possible
therefore t=3.75 s
Suppose a man's scalp hair grows at a rate of 0.49 mm per day. What is this growth rate in feet per century?
Answer:
58.703 ft/centuri
Explanation:
Length of the hair grow, l = 0.49 mm
Time to grow, t = 1 day
Convert mm into feet.
We know that, 1 mm = 0.00328 feet
So, 0.49 mm = 0.49 x 0.00328 = 0.0016072 ft
Convert day into century.
We know that, 1 century = 36525 days
So, 1 day = 1 / 36525 centuri
growth rate of hair = [tex]\frac{0.0016072}{\frac{1}{36525}}[/tex]
= 58.703 ft/centuri
There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric field that occurs if a q charge is placed between the plates at a distance d/2.
Answer:
The total electric potential at mid way due to 'q' is [tex]\frac{q}{4\pi\epsilon_{o}d}[/tex]
The net Electric field at midway due to 'q' is 0.
Solution:
According to the question, the separation between two parallel plates, plate A and plate B (say) = d
The electric potential at a distance d due to 'Q' is:
[tex]V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}[/tex]
Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':
For plate A,
[tex]V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}[/tex]
Similar is the case with plate B:
[tex]V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}[/tex]
Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:
[tex]V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}[/tex]
[tex]V_{total} = \frac{q}{4\pi\epsilon_{o}d}[/tex]
Now,
The Electric field due to charge Q at a distance is given by:
[tex]\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}[/tex]
Now, if the charge q is mid way between the field, then distance is [tex]\frac{d}{2}[/tex].
Electric Field at plate A, [tex]\vec{E_{A}}[/tex] at midway due to charge q:
[tex]\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}[/tex]
Similarly, for plate B:
[tex]\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}[/tex]
Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.
A mass is attached to a spring, which is attached to a wall. The distance from the mass and the equilibrium distance, x0 = 0, is given by x. The spring constant is 5N/m. The equilibrium distance is 1m from the wall. i. What is the force exerted on the mass at x = 3m?
ii. What is the force exerted on the mass when the mass is touching the wall?
iii. What work must be done on the mass to move it from the wall to x = 3m?
Answer:
Explanation:
Spring constant k = 5N/m
I ) x = 3 m means , spring is stretched by 3 m
Restoring force by spring = kx = 5 x 3 = 15 N
II ) When mass is touching the wall, extension in spring = 1 m
Force by spring on the body
= 1 x 5 = 5 N .
iii ) . It is touching the wall , x = 1 m
Stored energy in the spring = 1/2 k x² = .5 x 5 x 1 x 1
= 2.5 J
When x = 3 , energy stored in it
Potential energy stored in it = 1/2 k x²
= .5 x 5 x 3 x 3
= 22.5 J
Increase in stored energy = 22.5 - 2.5
20 J
This must be the work done to stretch it from 1 m to 3 m .
When an object is thrown vertically upward from the surface of the Earth: What is the instantaneous velocity in the point of maximum height?What is the acceleration in the point of maximum height?
Answer:
zero, acceleration due to gravity = 9.8 m/s^2
Explanation:
When an object throws vertically upwards, the acceleration acting on the object is acceleration due to gravity which is acting in vertically downwards direction.
As the object moves upwards, its velocity goes on decreasing because the direction of velocity and the acceleration both are opposite to each other. At maximum height, the velocity of object becomes zero and then object starts moving in downwards direction.
Thus, the value of instantaneous velocity at maximum height is zero but the vale of acceleration is acceleration due to gravity which is acting vertically downward direction.
Two points are given in polar coordinates by : (r, θ) = (2.60 m, 50.0°)
and
(r, θ) = (3.60 m, −46.0°)
, respectively. What is the distance between them?
Two points are given in polar coordinates, the distance between the two points is approximately 3.12m.
You can use the polar-to-cartesian conversion formula to convert each point to Cartesian coordinates (x, y), then apply the distance formula in
Cartesian coordinates to determine the separation between two points supplied in polar coordinates.
The polar-to-cartesian conversion formulas are:
[tex]\[ x = r \cdot \cos(\theta) \][/tex]
[tex]\[ y = r \cdot \sin(\theta) \][/tex]
Given the points:
[tex]\( (r_1, \theta_1) = (2.60 \, \text{m}, 50.0^\circ) \)[/tex]
[tex]\( (r_2, \theta_2) = (3.60 \, \text{m}, -46.0^\circ) \)[/tex]
Converting these points to Cartesian coordinates:
For the first point:
[tex]\[ x_1 = 2.60 \, \text{m} \cdot \cos(50.0^\circ) \approx 1.66 \, \text{m} \][/tex]
[tex]\[ y_1 = 2.60 \, \text{m} \cdot \sin(50.0^\circ)\\\\ \approx 1.98 \, \text{m} \][/tex]
For the second point:
[tex]\[ x_2 = 3.60 \, \text{m} \cdot \cos(-46.0^\circ)\\\\ \approx 2.53 \, \text{m} \][/tex]
[tex]\[ y_2 = 3.60 \, \text{m} \cdot \sin(-46.0^\circ)\\\\ \approx -2.57 \, \text{m} \][/tex]
Now, you can use the distance formula in Cartesian coordinates to find the distance between the two points:
[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Plug in the values and calculate:
[tex]\[ \text{distance} = \sqrt{(2.53 \, \text{m} - 1.66 \, \text{m})^2 + (-2.57 \, \text{m} - 1.98 \, \text{m})^2}\\\\ \approx 3.12 \, \text{m} \][/tex]
Thus, the distance between the two points is approximately 3.12 m.
For more details regarding Cartesian coordinates, visit:
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A concert loudspeaker suspended high off the ground emits 31 W of sound power. A small microphone with a 1.0 cm^2 area is 42 m from the speaker. What is the sound intensity at the position of the microphone? (include units)
What is the sound intensity level at the position of the microphone? (in dB)
Answer:
intensity of sound at level of microphone is 0.00139 W / m 2
sound intensity level at position of micro phone is 91.456 dB
EXPLANATION:
Given data:
power of sound P = 31 W
distance betwen microphone & speaker is 42 m
a) intensity of sound at microphone is calculated as
[tex]I = \frac{P}{A}[/tex]
[tex]= \frac{34}{4 \pi ( 44m )^ 2}[/tex]
= 0.00139 W / m 2
b) sound intensity level at position of micro phone is
[tex]\beta = 10 log \frac{I}{I_o}[/tex]
where I_o id reference sound intensity and taken as
[tex]= 1 * 10^{-12} W / m 2[/tex]
[tex]\beta = 10 log\frac{0.00139}{10^[-12}}[/tex]
= 91.456 dB
A test specimen in a tensile test has a gage length of 2.0 in and an area = 0.5in^2. During the test the specimen yields under a load of 32,000 lb. The corresponding gage length = 2.0083 in. This is the 0.2% yield point. The maximum load of 60,000 lb is reached at a gage length = 2.6 in. Determine a) yield strength, b) madulus of elasticity, and c) ten
Answer:
yield strength is 64000 lb/ in²
modulus of elasticity is 29.76 ×[tex]10^{6}[/tex] lb/in²
Explanation:
given data
length L = 2 in
area A = 0.5 in²
load = 32000 lb
gage length L1 = 2.0083 in
yield point = 0.2%
maximum load = 60000 lb
to find out
yield strength and modulus of elasticity
solution
we apply here yield strength that is express as
yield strength = [tex]\frac{load}{area}[/tex] ...........1
yield strength = [tex]\frac{32000}{0.5}[/tex]
yield strength = 64000 lb/ in²
and
modulus of elasticity is calculated as
modulus of elasticity = [tex]\frac{yield strength}{strain}[/tex] ..........2
here strain = [tex]\frac{L1 - L}{L}[/tex]
strain = [tex]\frac{2.0083 - 2}{2}[/tex]
strain = 0.00415
so new strain after offset is here 0.00415 - 0.002
new strain = 0.00215
so from equation 2
modulus of elasticity = [tex]\frac{yield strength}{strain}[/tex]
modulus of elasticity = [tex]\frac{64000}{0.00215}[/tex]
modulus of elasticity is 29.76 ×[tex]10^{6}[/tex] lb/in²
Final answer:
The yield strength is 64,000 psi, modulus of elasticity is approximately 15.422 x 106 psi, and the tensile strength is 120,000 psi based on the data given from the tensile test of a specimen.
Explanation:
Yield Strength, Modulus of Elasticity, and Tensile Strength
To address the given tensile test problem, we first need to determine the yield strength, which is found by dividing the load at yield by the original area. Hence, the yield strength (\(\sigma_y\)) is 32,000 lb divided by 0.5 in2, which equals 64,000 psi.
The modulus of elasticity (E) can be calculated using Hooke's Law, which is the stress over the strain. In this scenario, the initial stress (\(\sigma_i\)) is the yield load (32,000 lb) over the area (0.5 in2) and the initial strain (\(\epsilon_i\)) is the change in length (0.0083 in) over the original gage length (2.0 in). Therefore, E is 64,000 psi divided by 0.00415, which equals approximately 15,422,000 psi or 15.422 x 106 psi.
As for the tensile strength, it is the maximum stress that the material can withstand while being stretched or pulled before necking, which is the maximum load (60,000 lb) divided by the original cross-sectional area (0.5 in2), which gives us 120,000 psi.
An airplane can fly because the air with the ____________ velocity will apply a greater force to that side of the wing. O greatest O least
Answer:
Least velocity.
Explanation:
According to the Bernauli's equation
[tex]p^{2}+\frac{1}{2}\rho v^{2}+\rho gh= constant[/tex]
Here, v is the velocity, m is the mass, h is the height, P is the pressure, [tex]\rho[/tex] is the density
Now according to question.
[tex]P_{1}^{2}+\frac{1}{2}\rho v_{1} ^{2}+\rho gh_{1} =P_{2}^{2}+\frac{1}{2}\rho v_{2} ^{2}+\rho gh_{2}[/tex]
Here airplane height is same means [tex]h_{1}=h_{2}[/tex] then the required equation will become.
[tex]P_{1}^{2}+\frac{1}{2}\rho v_{1} ^{2}=P_{2}^{2}+\frac{1}{2}\rho v_{2} ^{2}[/tex]
Therefore,
[tex]P_{1}-P_{2}=\frac{1}{2}\rho (v_{2} ^{2}-v_{1} ^{2})[/tex]
Therefore according to the situation [tex]P_{1}>P_{2}[/tex]
This will give the velocity relation [tex]v_{2} >v_{1}[/tex]
Therefore, airplane can fly with least velocity.
A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.
Answer:
Image is virtual and formed on the same side as the object, 19.29 cm from the lens.
The height of the image is 0.40509 cm
Image is upright as the magnification is positive and smaller than the object.
Explanation:
u = Object distance = 100 cm
v = Image distance
f = Focal length = -23.9 cm (concave lens)
[tex]h_u[/tex]= Object height = 2.1 cm
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-23.9}-\frac{1}{100}\\\Rightarrow \frac{1}{v}=\frac{-1239}{23900} \\\Rightarrow v=\frac{-23900}{1239}=-19.29\ cm[/tex]
Image is virtual and formed on the same side as the object, 19.29 cm from the lens.
Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-19.29}{100}\\\Rightarrow m=0.1929[/tex]
[tex]m=\frac{h_v}{h_u}\\\Rightarrow 0.1929=\frac{h_v}{2.1}\\\Rightarrow h_v=0.1929\times 2.1=0.40509\ cm[/tex]
The height of the image is 0.40509 cm
Image is upright as the magnification is positive and smaller than the object.
A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt "coasted" across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, (a) calculate his maximum speed and (b) his acceleration
Answer:
Vmax = 12.21m/s
[tex]a = 4.07m/s^{2}[/tex]
Explanation:
For the first 3 seconds:
[tex]V_{3}=V_{o}+a*t=0+a*3=3a[/tex] This is the maximum speed. But we need acceleration.
[tex]X_{3}=V_{o}*t+\frac{a*t^{2}}{2} =\frac{9*a}{2}[/tex]
For the other 6.69s with constant speed:
[tex]X_{t}=100=X_{3}+V_{3}*t[/tex]
[tex]100=\frac{9*a}{2} +3*a*6.69[/tex] Solving for a:
[tex]a = 4.07m/s^{2}[/tex] Now we replace this value on [tex]V_{3}=3a[/tex]:
[tex]V_{3}=V_{max}=12.21m/s[/tex]