Answer:
1010 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-68^2}{2\times 250}\\\Rightarrow a=-9.248\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-68}{-9.248}\\\Rightarrow t=7.35\ s[/tex]
Time taken by the thunderbird to stop is 7.35 seconds
Time the thunderbird was at the pit is 5 seconds
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{68^2-0^2}{2\times 420}\\\Rightarrow a=5.5\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{68-0}{5.5}\\\Rightarrow t=12.36\ s[/tex]
Time taken to accelerate back to 68 m/s is 12.36 seconds
Total time to this point is 7.35+5+12.36 = 24.71 seconds
The Mercedes Benz is moving at a constant velocity hence it has no acceleration and we use the formula
Distance = Speed × Time
⇒Distance = 68 × 24.71 = 1680 m
The thunderbird has covered 250+420 = 670 m
So, the distance between them is 1680-670 = 1010 m
Final answer:
To find out how far the Thunderbird has fallen behind the Mercedes Benz, we calculate the time taken for its deceleration, pit stop, and acceleration, then determine how far the Mercedes traveled in that time. The Thunderbird falls behind by approximately 1346.4 meters.
Explanation:
The Thunderbird's driver decelerates to a stop, pauses for 5 seconds, then accelerates back to speed. The Mercedes Benz continues at a constant speed of 68.0 m/s throughout this time. We will calculate the time taken for the Thunderbird to decelerate and accelerate, and use this to determine how far the Mercedes has traveled in the same period, thus finding out how far the Thunderbird has fallen behind.
Deceleration Phase
First, we need to find the Thunderbird's deceleration. Using the formula v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (68.0 m/s), a is the acceleration, and s is the distance (250 m), we solve for a to find the deceleration rate. Rearranging the formula gives us a = (v^2 - u^2) / 2s. Plugging in the numbers, we find that the deceleration rate is approximately -9.184 m/s^2.
To find out how long this deceleration took, we use the formula v = u + at, solving for t gives us t = (v - u) / a. This calculation reveals that the deceleration phase lasted approximately 7.4 seconds.
Acceleration Phase
Assuming the Thunderbird accelerates at the same rate but in the positive direction over 420 m, we'll use the same method to find out the acceleration time. The acceleration time is also approximately 7.4 seconds.
Total time for stop and acceleration equals deceleration time + pit stop time + acceleration time, which totals 19.8 seconds.
The Mercedes Benz, traveling at 68.0 m/s, would have traveled 1346.4 meters in this time.
Therefore, the Thunderbird has fallen behind by 1346.4 meters, considering both the stopping and accelerating phases and the pit stop time.
A 0.07-kg lead bullet traveling 258 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted to heat that it alone absorbs, what is its temperature change?
Answer:
temperature change is 262.06°K
Explanation:
given data
mass = 0.07 kg
velocity = 258 m/s
to find out
what is its temperature change
solution
we know here
heat change Q is is equal to kinetic energy that is
KE = 0.5 × m× v² ...........1
here m is mass and v is velocity
KE = 0.5 × 0.07 × 258²
KE = 2329.74 J
and we know
Q = mC∆t .................2
here m is mass and ∆t is change in temperature and C is 127J/kg-K
so put here all value
2329.74 = 0.07 × 127 × ∆t
∆t = 262.06
so temperature change is 262.06°K
Final answer:
The temperature change of a 0.07-kg lead bullet that had been traveling at 258 m/s before coming to a stop is found to be approximately 260°C.
Explanation:
Calculating Temperature Change of a Lead Bullet
To calculate the temperature change of a 0.07-kg lead bullet that comes to a stop after traveling at 258 m/s, we must first determine how much kinetic energy the bullet had before stopping. The kinetic energy (KE) of the bullet can be calculated using the formula KE = 1/2 [tex]mv^2[/tex], where m is the mass of the bullet and v is its velocity. Plugging in the values gives us:
KE = 1/2 * 0.07 kg * (258 [tex]m/s)^2[/tex]
KE = 0.035 kg * 66564 [tex]m^2/s^2[/tex]
KE = 2330.74 J
This kinetic energy is converted to heat energy, which we will denote as Q. The temperature change (\ΔT) is then given by the formula Q = mc\ΔT, where m is the mass of the bullet and c is the specific heat capacity of lead.
Since c for lead is approximately 128 J/kg°C, we can rearrange the formula to solve for \Deltat:
ΔT = Q / (mc)
ΔT = 2330.74 J / (0.07 kg * 128 J/kg°C)
ΔT = 2330.74 J / 8.96 J/°C
ΔT = 259.96°C
Therefore, the temperature change of the bullet is approximately 260°C.
A race car starts at rest and speeds up to 40 m/s in a distance of 100 m. Determine the acceleration of the car.
Answer:
Acceleration of the car, [tex]a=8\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 0 (at rest)
Final speed of the car, v = 40 m/s
Distance covered, s = 100 m
We need to find the acceleration of the car. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
a is the acceleration of the car
[tex]a=\dfrac{v^2}{2s}[/tex]
[tex]a=\dfrac{(40)^2}{2\times 100}[/tex]
[tex]a=8\ m/s^2[/tex]
So, the acceleration of the car is [tex]8\ m/s^2[/tex]. Hence, this is the required solution
Consider an alcohol and a mercury thermometer that read exactly 0 oC at the ice point and 100 oC at the steam point. The distance between the two points is divided into 100 equal parts in both thermometers. Do you think these thermometers will give exactly the same reading at a temperature of , say, 60 oC? Explain.
Final answer:
Alcohol and mercury thermometers may not read the same at 60°C despite being calibrated at 0°C and 100°C because they have different rates of thermal expansion, with mercury being more linear and alcohol having a higher expansion coefficient.
Explanation:
Alcohol and mercury thermometers are calibrated at two fixed points: the freezing point of water (0°C) and the boiling point of water (100°C). However, these substances expand at different rates over the temperature range, which means the expansion is not linear. While both may be divided into 100 equal divisions between these two points, an alcohol thermometer and a mercury thermometer may not give the same reading at a temperature such as 60°C because their rates of thermal expansion differ. Mercury has a more linear expansion with temperature compared to alcohol, which has a higher coefficient of expansion, causing its physical expansion for an equal temperature increment to be larger. This inconsistency is often overlooked because it is somewhat small over the range of temperatures we normally encounter, but it becomes significant at more extreme temperatures.
An unwary football player collides head-on with a padded goalpost while running at 7.9 m/s and comes to a full stop after compressing the padding and his body by 0.32 m. Take the direction of the player's initial velocity as positive.
Answer:
Acceleration, a = [tex]97.52 m/s^{2}[/tex]
t = 0.08 s
Solution:
As per the question:
The velocity of the player, v = 7.9 m/s
distance, d = 0.32 m
Now, consider the direction of the initial velocity of the player as positive and the acceleration to be constant:
Also, the final velocity of the player, v' = 0 m/s as he finally stops
Using the third eqn of motion:
[tex]v'^{2} = v^{2} - 2ad[/tex]
[tex]0 = 7.9^{2} - 2a\times 0.32[/tex]
a = [tex]97.52 m/s^{2}[/tex]
Also, From eqn (1) of motion:
v' = v - at
0 = 7.9 - 97.52t
t = 0.08 s
The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the total force = average pressure × area)
Answer:
The hydro static force on the back of the dam is [tex]1.96\times10^{11}\ N[/tex]
Explanation:
Given that,
Width b= 1000 m
Depth d= 200 m
We need to calculate the average pressure
Using formula of average pressure
[tex]P_{avg}=\rho\times g\times d_{avg}[/tex]
Put the value into the formula
[tex]P_{avg}=1000\times9.8\times100[/tex]
[tex]P_{avg}=980000\ Pa[/tex]
We need to calculate the hydro static force on the back of the dam
Using formula of force
[tex]F = P_{avg}\times A[/tex]
Put the value into the formula
[tex]F = 980000\times1000\times200[/tex]
[tex]F=1.96\times10^{11}\ N[/tex]
Hence, The hydro static force on the back of the dam is [tex]1.96\times10^{11}\ N[/tex]
In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1there was a gentle 1.2 cm/s current flowing in the direction that the swimmers were going, while in lane 8there was a current of the same speed but directed opposite to the swimmers' direction. Suppose a swimmer could swim the 50.0 m in 25.0 s in the absence of any current. How would the time it took the swimmer to swim 50.0 m change in lane 1?
Enter negative value if the swimmer would be faster and positive value if the swimmer would be slower. Part B
How would the time it took the swimmer to swim 50.0 m change in lane 8?
Enter negative value if the swimmer would be faster and positive value if the swimmer would be slower.
A) -0.15 s
First of all, we need to velocity of the swimmer in absence of current. This is given by:
[tex]v_0 = \frac{d}{t}[/tex]
where
d = 50.0 m
t = 25.0 s
Substituting,
[tex]v_0 = \frac{50.0}{25.0}=2.0 m/s[/tex]
In lane 1, the velocity of the current is
[tex]v_c = +1.2 cm/s = +0.012 m/s[/tex]
where the + sign means it is in the same direction as the swimmer. Therefore, the net velocity of the swimmer in lane 1 will be
[tex]v=v_0+v_c = 2.0 + 0.012 = = 2.012 m/s[/tex]
And so, the time the swimmer will take to cover the 50.0 m will be:
[tex]t=\frac{d}{v}=\frac{50.0}{2.012}=24.85 s[/tex]
So, the time would change by
[tex]\Delta t = 24.85 - 25.0 = -0.15 s[/tex]
which means that the swimmer will be 0.15 s faster.
B) +0.15 s
To solve this part, we just need to consider that the current goes in the opposite direction, so its velocity actually is:
[tex]v_c = -12 cm/s = -0.012 m/s[/tex]
where the negative sign indicates the opposite direction.
So, the net velocity of the swimmer in lane 8 is
[tex]v=v_0+v_c = 2.0 - 0.012 = = 1.988 m/s[/tex]
And so, the time the swimmer will take to cover the 50.0 m will be:
[tex]t=\frac{d}{v}=\frac{50.0}{1.988}=25.15 s[/tex]
So, the time would change by
[tex]\Delta t = 25.15 - 25.0 = +0.15 s[/tex]
which means that the swimmer will be 0.15 s slower.
What resistance is needed in series with a 10-uF capacitor
at1.0kHz for a total impedance of 45 Ohm?
Answer:
The value of resistance will be 42.08 ohm
Explanation:
We have given capacitance [tex]C=10\mu F=10\times 10^{-6}F[/tex]
Frequency f = 1 kHz = 1000 Hz
Impedance Z = 45 ohm
Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\times 3.14\times 1000\times 10^{-5}}=15.92ohm[/tex]
We know that impedance is given by [tex]Z=\sqrt{R^2+X_C^2}[/tex]
[tex]45=\sqrt{R^2+15.92^2}[/tex]
Squaring both side [tex]2025=R^2+253.4464[/tex]
R = 42.08 ohm
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 18.0 m/s, and the distance between them is 58.0 m . After t1 = 5.00 secs, the motorcycle starts to accelerate at a rate of 4.00 m/s^2 . The motorcycle catches up with the car at some time t2 . A) How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1 Express the time numerically in seconds using three significant figures. B) How far does the motorcycle travel from the moment it starts to accelerate (at time t1 ) until it catches up with the car (at time t2)? Should you need to use an answer from a previous part, make sure you use the unrounded value. Answer numerically in meters using three significant figures.
The equations of motion are used to find the motion characteristics of an object with time
A) The time it takes the motorcycle to reach the car is approximately 5.39 seconds
B) The distance the motorcycle travel from the moment it starts to accelerate to the moment it catches up with the car is approximately 155 meters
The reason the above values are correct is as follows:
The given parameters are;
The initial speed of the car and the motorcycle, v₁ = 18.0 m/s
The initial distance between the motorcycle following the car = 58.0 m
The time after which the motorcycle starts to accelerate, t₁ = 5.00 secs
The acceleration of the motorcycle = 4.00 m/s²
The time at which the motorcycle catches up with the car = t₂
Required:
A) The time it takes the motorcycle to accelerate before it catches up with the car, which is to find t₂ - t₁
Method:
The distance moved by motorcycle to reach the car = The distance moved by the car in the same time + 58.0 meters
Solution;
The appropriate equation of motion to use are [tex]s = u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2[/tex] and [tex]d = u\cdot t[/tex]
Where;
s = The distance between the motorcycle and the car = 58.0 m
u = The initial velocity of the motorcycle = 18.0 m/s
t = Δt = t₂ - t₁ = The time the motorcycle accelerates
a = The acceleration of the motorcycle = 4.00 m/s²
d = The distance moved by the car
By the method, we have;
[tex]s = d + 58.0[/tex]
[tex]u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2 = u \cdot t[/tex]
Plugging in the values into the equation gives;
[tex](18.0 \times t) + \dfrac{1}{2} \times 4 \times t^2 = (18.0 \times t) + 58.0[/tex]
Cancelling the like term, (18.0 × t), on both sides of the equation gives;
[tex]\dfrac{1}{2} \times 4 \times t^2 = 58.0[/tex]
[tex]2\cdot t^2 = 58.0[/tex]
[tex]t = \Delta t = \sqrt{\dfrac{58.0}{2} } = \sqrt{29} \approx 5.39[/tex]
The time it takes the motorcycle to reach the car, t ≈ 5.39 seconds
The time it takes the motorcycle to reach the car is approximately 5.39 seconds
B) Required:
The distance the motorcycle travels from the moment it starts to accelerate, t₁ to the time it catches up with the car, t₂
Method:
The distance travelled during the time interval t₂ - t₁ should be calculated
Solution:
The distance, s, travelled during the time interval [tex]t_2 - t_1 = \Delta t = t[/tex] is given as follows;
[tex]s = u \cdot t + \dfrac{1}{2} \cdot a \cdot t^2[/tex]
From part (A), u = 18.0 m/s t = [tex]\sqrt{29}[/tex] s, and a = 4.00 m/s², therefore;
[tex]s = 18.0 \times \sqrt{29} + \dfrac{1}{2} \times 4.00 \times (\sqrt{29} )^2 \approx 154.933[/tex]
Rounded to three significant figures, the distance the motorcycle travel, s = 155 meters
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To solve the question, we use the equations of motion to equate the distances traveled by the car and motorcycle. After solving the quadratic equation, we find the time the motorcycle catches up with the car and the distance it traveled.
Explanation:The question involves a situation where a motorcycle is following a car traveling at a constant speed on a highway. Initially, both the car and the motorcycle are traveling at the same speed of 18.0 m/s, and there is a distance of 58.0 m between them. After 5.00 seconds, the motorcycle begins to accelerate at 4.00 m/s² until it catches up with the car. To find the time t2-t1, which is the duration the motorcycle catches up to the car, and the distance the motorcycle covers from the moment it starts to accelerate until it catches up with the car, we use the equations of motion.
We know that the car travels with a constant velocity, so the distance it covers in time t after the motorcycle starts accelerating can be given by:
d_car = v_car × t
For the motorcycle, which starts to accelerate at time t1, the distance it covers can be given by:
d_moto = d_initial + v_initial × (t - t1) + ½ × a × (t - t1)²
Given that the initial distance d_initial is 58.0 m, v_initial is 18.0 m/s, and a is 4.00 m/s², we can equate the distances for both the car and motorcycle when the motorcycle catches up:
18.0 × t = 58.0 + 18.0 × (t - 5.00) + ½ × 4.00 × (t - 5.00)²
By solving this quadratic equation, we can find the value of t, and then t2-t1 would simply be t - 5.00 seconds. Finally, to find out the distance the motorcycle traveled, we substitute the value of t into the distance equation for the motorcycle.
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According to a rule-of-thumb, every five seconds between a lightning flash and the following thunder gives the distance to the flash in miles. Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in m/s from this rule. What would be the rule for kilometers?
Answer:
speed of sound = 321 m/s
Rule of thumb for kilometers: 1 km every 3 seconds
Explanation:
Hi!
If a thunder originates at a place a distance D from you, and it takes time T for its sound to reach you, then:
[tex]V = \frac{D}{T}[/tex]
Whre V is the speed of sound. Time T is the time elapsed between the moment you see the flash (becuase of the assumption of tha it takes no time for the light to reach you) and the moment you hear the thunder. Then
[tex]V = \frac{1mile}{5\;s} =\frac{1609.34 \;m}{5\;s} = 321\frac{m}{s}[/tex]
To calculate the rule in kilometers:
[tex]T = \frac{1\;km}{321*10^{-3} \frac{km}{s}} \approx 3\; s[/tex]
The estimated speed of sound derived from the rule of thumb is approximately 322 m/s. For the rule in kilometers, each five seconds would represent about 1.609 kilometers.
Explanation:The rule-of-thumb that every five seconds between a lightning flash and the accompanying thunder equates to one mile of distance is based on the speed of sound. To convert this to meters per second we must know that 1 mile approximately equals 1609.34 meters. Therefore, if light travels nearly instantaneously, and one mile is covered in five seconds, the speed of sound can be estimated as about 1609.34 m/5 s or approximately 322 m/s.
For the rule in kilometers, we need to convert miles into kilometers. As 1 mile is approximately 1.609 kilometers, the rule of thumb in kilometers would be that each five seconds would represent 1.609 kilometers.
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A small 0.14 kg metal ball is tied to a very light (essentially massless) string that is 0.9 m long. The string is attached to the ceiling so as to form a pendulum. The pendulum is set into motion by releasing it from rest at an angle of 55 ∘ with the vertical.(a) What is the speed of the ball when it reaches the bottom of the arc?(b) What is the centripetal acceleration of the ball at this point?(c) What is the tension in the string at this point?
Answer:
a) v=2.743m/s
b) [tex]a_c = 8.363m/s^2[/tex]
c) T=2.543N
Explanation:
First, calculate the height of the ball at the starting point:
[tex]y' = 0.9cos(55)[/tex]
[tex]y' = 0.516[/tex]
At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:
[tex]E_p=E_k\\mgh=\frac{mv^2}{2}[/tex]
Solving for v:
[tex]v=\sqrt{2gh}[/tex]
if h is the height loss: (l-y')
v=2.743m/s
The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):
[tex]a_c=\frac{v^2}{r}[/tex]
[tex]a_c = 8.363m/s^2[/tex]
To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:
[tex]\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N[/tex]
Vector A is in the direction 44.0 degrees clockwise from the y-axis. The x-component of A Ax=-15.0 m. Part A: What is the y-component of vector A? Part B: What is the magnitude of vector A?
Answer:
a) Y component of the vector =15.54 m
b) Vector magnitude = 21.6 m
Explanation:
The given vector makes 44 degree angle with Y axis, as given. This is same as 90 -44 = 46 degrees with the horizontal or X axis.
b) X component of the given vector = [tex]A_{x}[/tex] = A cos 46 =15
⇒ A = 15/cos 16 = 21.6 m = Total vector magnitude.
a) Y component of the vector = 21.6 sin 46 = 15.54 m
b) A = 21.6 m
The y-component of vector A (Ay) is approximately 10.3 m, assuming a positive y-direction due to the angle being in the second quadrant. The magnitude of vector A, found using the Pythagorean theorem with its components, is approximately 18.0 m.
To find the y-component of vector A (Ay), we use the trigonometric relationship Ay = A sin(θ), where A is the magnitude of the vector and θ is the angle it makes relative to the x-axis. However, since the angle given is clockwise from the y-axis, we must first convert it to the counterclockwise angle from the x-axis, which would be θ = 90° + 44.0° = 134.0°.
Given that Ax = -15.0 m, and assuming θ is measured counterclockwise, we can find the magnitude of A using the Pythagorean theorem: A = √(Ax² + Ay²). To find Ay, we rearrange the equation for the y-component to Ay = A sin(θ) and solve for Ay using the magnitude we just found. Our direction angle θ is in the second quadrant, meaning Ay should be positive.
Since we only have Ax, we first find A as follows: A = √((-15.0 m)² + Ay²). To determine Ay, we need to look at signs and quadrants. Ax is negative, and because the angle is measured clockwise from the y-axis, Ay must be positive. Therefore, we can infer that Ay is approximately 10.3 m by implication of the Pythagorean theorem, knowing that the other possible value is negative, which does not fit the quadrant.
To find the magnitude of A, we use the derived components: A = √((-15.0 m)² + (10.3 m)²), yielding approximately 18.0m.
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What must be the focal length of the appropriate positive lens? If it is a bi-convex lens of refractive index 1.5, what are the values of radii of curvature?
Answer:
the radii of curvature is 30 cm.
Explanation:
given,
object is place at = 45 cm
image appears at = 90 cm
focal length = ?
refractive index = 1.5
radii of curvature = ?
[tex]\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}[/tex]
[tex]\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}[/tex]
f = 30 cm
using lens formula
[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})[/tex]
[tex]R_1 = R\ and\ R_2 = -R[/tex]
[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})[/tex]
[tex]R = (n -1)\ f [/tex]
[tex]R = 2(1.5 -1)\ 30[/tex]
R = 30 cm
hence, the radii of curvature is 30 cm.
Calculate the density, mass and weight of an object if it occupies 200 ft^3 and has a specific volume of 10 ft^3 /lb.
Answer:
a) 0.1 lb/ft³
b) 20 lb
c) 88.788 N
Explanation:
Given:
Volume of the object = 200 ft³
Specific volume = 10 ft³/ lb
Now,
Density = [tex]\frac{\textup{1}}{\textup{Specific volume}}[/tex]
or
Density = [tex]\frac{1}{10}[/tex] = 0.1 lb/ft³
also,
Specific volume = [tex]\frac{\textup{Volume}}{\textup{Mass}}[/tex]
or
10 = [tex]\frac{200}{\textup{Mass}}[/tex]
or
Mass = 20 lb
And,
Weight = Mass in kg × Acceleration due to gravity
Mass = 20 lb
and 1 lb = 0.453 kg
thus,
20 lb = 20 × 0.453 = 9.06 kg
Therefore,
weight = 9.06 × 9.8 = 88.788 N
For the object,
Density of the object is 0.1 ib per ft cubedMass of the object is 20 ib.Weight of the object is 88.788 NWhat is the specific volume of a object?
Specific volume of the object is inverse to the density of the object.
Given information-
The volume of the object is 200 ft cubed.
The specific volume of the object is 10 ft cubed.
The density of a object is the inverse of its specific volume, thus density of the given object is,
[tex]\rho=\dfrac{1}{\rm specific volume} \\\rho=\dfrac{1}{10} \\\rho=0.1 \rm ib/ft^3[/tex]
The mass of a object is the ratio of the volume to the specific volume, thus mass of the given object is,
[tex]m=\dfrac{\rm volume}{\rm specific volume} \\m=\dfrac{200}{10} \\m=20 \rm ib[/tex]
As 1 Ib is equal to the 0.453 kg.
The weight of a object is the product of the mass of the object to the acceleration due to gravity, thus weight of the given object is,
[tex]W=0.453\times20\times9.81\\W=88.788\rm N[/tex]
Thus for the object,
Density of the object is 0.1 ib per ft cubedMass of the object is 20 ib.Weight of the object is 88.788 NLearn more about the specific density here;
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A car traveling 77 km/h slows down at a constant 0.45 m/s^2 just by "letting up on the gas." --Part a : Calculate the distance the car coasts before it stops.
--Part b: Calculate the time it takes to stop.
--Part c: Calculate the distance it travels during the first second.
--Part d: Calculate the distance it travels during the fifth second. Need help with all parts A-D, please show all work and formulas used.
Answer:
(a) 508.37 m
(b) 47.53 s
(c) 21.165 m
(d) 19.365 m
Explanation:
initial velocity, u = 77 km/h = 21.39 m/s
acceleration, a = - 0.45 m/s^2
(a) final velocity, v = 0
Let the distance traveled is s.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=21.39^{2}-2 \times 0.45 \times s[/tex]
s = 508.37 m
(b) Let t be the time taken to stop.
Use first equation of motion
v = u + at
0 = 21.39 - 0.45 t
t = 47.53 s
(c) Use the formula for the distance traveled in nth second
[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]
where n be the number of second, a be the acceleration, u be the initial velocity.
put n = 1, u = 21.39 m/s , a = - 0.45m/s^2
[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 1-1 \right )}[/tex]
[tex]s_{n^{th}=21.165m[/tex]
(d) Use the formula for the distance traveled in nth second
[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]
where n be the number of second, a be the acceleration, u be the initial velocity.
put n = 5, u = 21.39 m/s , a = - 0.45m/s^2
[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 5-1 \right )}[/tex]
[tex]s_{n^{th}=19.365m[/tex]
A truck heading due west increases its speed with a constant acceleration from 12 m/s to 18 m/s in 7.5 seconds. How far does the truck travel during this time? What was the acceleration of the truck during this period?
Answer:
[tex]acceleration=0.8 \frac{m}{s^{2} } \\distance=112.5 m[/tex]
Explanation:
We can find these answers following the equations of motion.
To find the acceleration, we use the equation:[tex]a=[/tex]Δ[tex]V/t[/tex]
Where Δ[tex]V[/tex] is the difference between the final speed and the initial speed. And [tex]t[/tex] is the time spent
We replace the terms:
[tex]a=\frac{18\frac{m}{s} -12\frac{m}{s} }{7.5s}[/tex]
We solve the difference:
[tex]a=\frac{6\frac{m}{s}}{7.5s}[/tex]
We divide the terms, so we can have the answer:
[tex]a=0.8 \frac{m}{s^{2}}[/tex]
2. To find the distance traveled by the truck, we use the equation:
[tex]x=V_0t+\frac{1}{2}at^{2}[/tex]
Where [tex]x[/tex] is the distance traveled, [tex]V_0[/tex] is the initial speed, [tex]a[/tex] is the acceleration and [tex]t[/tex] is the time.
We replace the terms:
[tex]x=(12\frac{m}{s}*7.5s)+\frac{1}{2}[0.8\frac{m}{s^{2} }*(7.5)^{2} ][/tex]
We multiply and solve the exponential:
[tex]x=90m+\frac{1}{2}(0.8\frac{m}{s^{2} }*56.25s^{2} )[/tex]
Then, we multiply the terms left:
[tex]x=90m+22.5m[/tex]
And add, so we can have the answer:
[tex]x=112.5m[/tex]
An amplitude modulation transmitter radiates 20 KW. If the modulation index is 0.7, find the magnitude of the carrier power?
Answer:
carrier power is 16.1 kW
Explanation:
given data
power transmit P = 20 kW
modulation index m = 0.7
to find out
carrier power
solution
we will apply here power transmit equation that is express as
[tex]P = 1 + \frac{m^2}{2} * carrier power[/tex] ...............1
put here all value in equation 1 we get carrier power
[tex]P = 1 + \frac{m^2}{2} * carrier power[/tex]
[tex]20 = 1 + \frac{0.7^2}{2} * carrier power[/tex]
solve it we get
carrier power = 16.1
so carrier power is 16.1 kW
A train starts at rest, accelerates with constant acceleration a for 5 minutes, then travels at constant speed for another 5 minutes, and then decelerates with –a. Suppose it travels a distance of 10km in all. Find a.
Answer:
The value of acceleration equals [tex]0.0556m/s^{2}[/tex]
Explanation:
The distances covered in each of the three phases are calculated as under
1) Distance covered in accelerating phase for a period of 5 minutes or 300 seconds ( 1 minute = 60 seconds)
Using second equation of kinematics
[tex]s=ut+\frac{1}{2}at^{2}\\\\s_{1}=0\times 300+\frac{1}{2}\times a\times (300)^{2}\\\\s_{1}=0.5\times a\times (300)^{2}[/tex]
2) Distance covered in 5 minutes while travelling at a constant speed which is The speed after 5 minutes of travel is obtained by first equation of kinematics as
[tex]v=u+at\\\\v=0+a\times 300\\\\v=300a[/tex]
Thus distance traveled equals
[tex]s_{2}=300a\times 300\\\\s_{2}=(300)^{2}a[/tex]
3)
The phase when the car stops the distance it covers during this phase can be obtained using third equation of kinematics as
[tex]v^{2}=u^{2}+2as\\\\\therefore s=\frac{v^{2}-u^{2}}{2a}\\\\s_{3}=\frac{0-(300a)^{2}}{2\times -a}\\\\s_{3}=\frac{300^{2}a}{2}[/tex]
Now the sum of [tex]s_{1}+s_{2}+s_{3}[/tex] equals 10 kilometers or 10000 meters.
Thus we get
[tex]0.5\times a\times 300^{2}+300^{2}a+300^{2}\times \frac{a}{2}=10000\\\\a(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})=10000\\\\\therefore a=\frac{10000}{(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})}=0.0556m/s^{2}[/tex]
To solve for the acceleration, we can use the constant acceleration equation and find the distance traveled during each phase.
Explanation:The constant acceleration equation can be used to solve this problem. The total distance traveled by the train is equal to the sum of the distances traveled during each phase.
In the first phase, the train starts at rest and accelerates for 5 minutes. The distance traveled during this phase can be found using the equation s = 0.5at^2, where s is the distance, a is the acceleration, and t is the time. Since the initial velocity is 0, we can simplify the equation to s = 0.5at^2.
In the second phase, the train travels at a constant speed for 5 minutes. The distance traveled during this phase is equal to the product of the speed and the time.
In the third phase, the train decelerates with a negative acceleration. The distance traveled during this phase can be found using the equation s = ut + 0.5at^2, where u is the initial velocity, a is the acceleration, and t is the time. Since the final velocity is 0, we can simplify the equation to s = ut.
The total distance traveled is given as 10 km. By plugging in the values for the distances and using the given times for each phase, we can solve for the acceleration, a, and find that a = 0.014 km/s^2.
Learn more about Calculating acceleration and distance here:https://brainly.com/question/32817533
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A rookie quarterback throws a football with an initial upward velocity component of 17.0 m/s and a horizontal velocity component of 18.3 m/s . Ignore air resistance. A. How much time is required for the football to reach the highest point of the trajectory?
B. How high is this point?
C. How much time (after it is thrown) is required for the football to return to its original level?
D. How does this compare with the time calculated in part (a).
E. How far has it traveled horizontally during this time?
Answer:
(a) 1.73 s
(b) 14.75 m
(c) 3.36 s
(d) double
(e) 63.32 m
Explanation:
Vertical component of initial velocity, uy = 17 m/s
Horizontal component of initial velocity, ux = 18.3 m/s
(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.
Use first equation of motion in vertical direction
vy = uy - gt
0 = 17 - 9.8 t
t = 1.73 seconds
(B) Let the highest point is at height h.
Use III equation of motion in vertical direction
[tex]v^{2}=u^{2}-2gh[/tex]
0 = 17 x 17 - 2 x 9.8 x h
h = 14.75 m
(C) The time taken by the ball to return to original level is T.
Use second equation of motion i vertical direction.
[tex]h = ut + 0.5at^2[/tex]
h = 0 , u = 17 m/s
0 = 17 t - 0.5 x 9.8 t^2
t = 3.46 second
(D) It is the double of time calculated in part A
(E) Horizontal distance = horizontal velocity x total time
d = 18.3 x 3.46 = 63.32 m
If a marathon runner averages 9.51 mi/hr, how long does it take him to run a 26.220-mile marathon. Express your answers in hours, minutes, and seconds.
Explanation:
Speed of the marathon runner, v = 9.51 mi/hr
Distance covered by the runner, d = 26.220 mile
Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]t=\dfrac{d}{v}[/tex]
[tex]t=\dfrac{26.220\ mi}{9.51\ mi/hr}[/tex]
t = 2.75 hours
Since, 1 hour = 60 minutes
t = 165 minutes
Since, 1 minute = 60 seconds
t = 9900 seconds
Hence, this is the required solution.
A 133.7 ft3 volume of liquid hydrogen rocket fuel has a mass of 268 kg. Calculate: the weight of the fuel at standard (Earth) sea level conditions in N and Ibr, the density of the fuel in kg/m3, and the specific volume in ft3/lbm (Ibm=pound-mass, which is not the same as a a lbf or a slug)
First, let's make some convertions:
[tex]268kg*\frac{1 lbm}{0.454kg}= 590.84 lbm[/tex]
[tex]133.7 ft^3*\frac{1m^3}{35.3147ft^3} = 3.78m^3[/tex]
a) weight of the fuel:
Newtons: The weight in newtons is equal to the mass in kilograms times the gravity in m/s^2.
[tex]W = m*g = 268kg*9.81m/s^2=2629.08 N[/tex]
lbf: The weight inlbf is equal to the mass in slugs times the gravity in ft/s^2.
[tex]W= m*g = 590.84 lbm *\frac{1 slug}{32.174lbm} *32.174ft/s^2 = 590.84 lbf[/tex]
b) density:
The density is the mass in kg of the fuel divided by its volume in m^3:
[tex]d = \frac{m}{v} =\frac{268kg}{3.78m^3} =70.9 kg/m^3[/tex]
c) specific volume:
The specific volume is the volume in ft^3 of the fuel divided by its mass in lbm:
[tex]v_{sp} = \frac{v}{m} =\frac{133.7 ft^3}{590.84 lbm} = 0.226 ft^3/lbm[/tex]
A ball is thrown upward. It leaves the hand with a velocity of 14.6 m/s, having been accelerated through a distance of 0.505 m. Compute the ball's upward acceleration, assuming it to be constant.
Answer:
upward acceleration is 211.04 m/s²
Explanation:
given data
velocity = 14.6 m/s
distance = 0.505 m
to find out
upward acceleration
solution
we have given distance and velocity and ball is going upward
so acceleration will be calculated by velocity formula that is
v² - u² = 2×a×s ............1
here v is velocity and u is initial velocity and s is distance and a is acceleration
and u = o because starting velocity zero
put here all there value in equation 1
v² - u² = 2×a×s
14.6² - 0 = 2×a×0.505
solve it we get a
a = 211.04
so upward acceleration is 211.04 m/s²
A teacher sends her students on a treasure hunt. She gives the following instructions: · Walk 300 m north · Walk 400 m northwest · Walk 700 m east-southeast and the treasure is buried there. As all the other students walk off following the instructions, Jane physics student quickly adds the displacements and walks in a straight line to find the treasure. How far (in meters) does Jane need to walk?
Answer:
displacement is 481.3 m with 40.88° north east
Explanation:
given data
walk 1 = 300 m north = 300 j
walk 2 = 400 m northwest =400 cos(45) i + sin(45) j
walk 3 = 700 m east southeast = 700 cos(22.5) i- sin(22.5) j
to find out
displacement and direction
solution
we consider here direction x as east and direction y as north
so
displacement = distance 1 + distance 2 + distance 3
displacement = 300 j + 400 cos(45) i + sin(45) j + 700 cos(22.5) i- sin(22.5) j
we get here
displacement = 363.9 i + 315 j
so magnitude
displacement = [tex]\sqrt{363.9^{2}+315^{2} }[/tex]
displacement = 481.3 m
and angle will be = arctan(315/369.9)
angle = 0.713494059 rad
angle is 40.88 degree
so displacement is 481.3 m with 40.88° north east
Answer:
Magnitude of displacement = 481.24 m
Direction = 40.88 degrees north east.
Explanation:
Let east is along real axis and north is along imaginary axis. So,
First walk = d1 = J300 m
Second walk = d2 = -400Cos(45) + J400Sin(45) = (-282.84 + J282.84) m
Third walk = d3 = 700Cos(22.5) – J700Sin(22.5) = (646.71 – J267.88) m
Total displacement = d = d1 + d2 + d2 = (363.86 + J314.96)m
Magnitude = √((363.86)^2+(314.96)^2) = 481.24 m
Direction = arctan(314.96/363.86) = 40.88 degrees north east.
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.1 s and was brought jarringly back to rest in only 1.9 s. Calculate his (a) acceleration in his direction of motion and (b) acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. Report each answer rounded to one decimal point and report only positive answers (tak
Answer:
(a) 5.6g
(b) 15.1g
Explanation:
Let us assume:
u = initial speed of the rocketv = final speed of the rocketa = acceleration of the rockett = time intervalg = acceleration due to gravity = [tex]9.80\ m/s^2[/tex]Part (a):
While Dr. John Paul Stapp accelerates the rocket to its top speed, we have
[tex]u = 0 m/s\\v = 282\ m/s\\t = 5.1\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow a = \dfrac{282-0}{5.1}\\\Rightarrow a =55.29\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{55.29}{9.80}\\\Rightarrow \dfrac{a}{g}=5.6\\\Rightarrow a=5.6g[/tex]
Hence, the acceleration of the rocket in the direction of motion is 5.6g.
Part (b):
While Dr. John Paul Stapp accelerates the rocket from its top speed to rest, we have
[tex]u = 282 m/s\\v = 0\ m/s\\t = 1.9\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow a = \dfrac{0-282}{1.9}\\\Rightarrow a =-148.42\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{-148.42}{9.80}\\\Rightarrow \dfrac{a}{g}=-15.1\\\Rightarrow a=-15.1g[/tex]
Here, the negative sign represents that the motion of acceleration of the rocket is opposite to the direction of motion.
Hence, the acceleration of the rocket in the direction opposite to that of motion is 15.1g.
Final answer:
The response provides the calculation for Dr. John Paul Stapp's acceleration in his direction of motion and the acceleration opposite to his motion in terms of multiples of g.
Explanation:
Acceleration:
(a) Acceleration in his direction of motion: To find the acceleration in Stapp's direction of motion, we can use the formula a = Δv / Δt. Plugging in the values, we get a = (282 m/s) / (5.1 s) = 55.3 m/s². Converting this to multiples of g, we have 55.3 m/s² / 9.81 m/s² ≈ 5.6 g.
(b) Acceleration opposite to his direction of motion: The deceleration can be calculated using the same formula a = Δv / Δt. Substituting the values, we have a = (282 m/s) / (1.9 s) = 148.4 m/s². Expressing this in multiples of g, we get 148.4 m/s² / 9.81 m/s² ≈ 15.1 g.
For every action force, there is: A. a net force. B. an unbalanced force. C. a friction force. D. an equal and opposite force. E. all of these
Answer:
The correct option is 'D': Equal and opposite force.
Explanation:
We know from Newton's third law of motion
"For every action there exists an equal and opposite reaction".
The statement is further explained as
If there exists a force from Body 1 to Body 2 then Body 2 must also exert an equal and opposite force on the Body 1.
In simple terms we can say that the action and the reaction forces exist in pairs and are always present if any one among them is present.
The steel ball A of diameter D = 25 mm slides freely on the horizontal rod of length L = 169 mm which leads to the pole face of the electromagnet. The force of attraction obeys an inverse-square law, and the resulting acceleration of the ball is a = K/(L - x)2, where K = 100 m3/s2 is a measure of the strength of the magnetic field. If the ball is released from rest at x = 0, determine the velocity v with which it strikes the pole face.
Answer:
398 m/s
Explanation:
The acceleration is given by:
a = K/(L - x)^2
K = 100 m^3/s^2
L = 0.169 m
This acceleration will result in a force:
F = m * a
F = m * K/(L - x)^2
This force will perform a work:
W = F * L
The ball will advance only until x = L - D/2
[tex]W = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
This work will be converted to kinetic energy
W = Ek
Ek = 1/2 * m * v^2
[tex]1/2 * m * v^2 = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
[tex]1/2 * v^2 = K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
First we solve thr integral:
[tex]K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
We use the replacement
u = L - x
du = -dx
And the limits
When x = L - D/2, u = D2/2, and when x = 0, u = L
[tex]-K \int\limits^{D/2}_L {u^-2}} \, du[/tex]
K / u evaluated between L and D/2
2*K / D - K / L
Then
1/2 * v^2 = 2*K / D - K / L
1/2 * v^2 = K * (2/D - 1/L)
v^2 = 2*K*(2/D - 1/L)
[tex]v = \sqrt{2*K*(2/D - 1/L)}[/tex]
[tex]v = \sqrt{2*100*(2/0.0025 - 1/0.169)} = 398 m/s[/tex]
Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.7Ω and 15.9 W, 30.4Ω and 9.12 W, and 16.3Ω and 12.3 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?
Answer:
a) greatest voltage = 29.25 V
b) power = 16 W
Explanation:
The total resistance R of the three resistors in series is:
[tex]R = (6.7 + 30.4 + 16.3) \Omega = 53.4 \Omega[/tex]
a) The greatest current I is the one that will burn the resistor with lower power rating, which is 9.12 W:
[tex]P_{max} = I_{max}^2 R = I_{max}^2 30.4\Omega = 9.12W\\I_{max} = 0.54 A[/tex]
The voltage is:
[tex]V_{max}=IR = 0.54*53.4V= 29.25 V[/tex]
b) When the current is 0.54 A, the power is:
[tex]P = RI^2=53.4*0.3 W = 16W[/tex]
Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of 4.2 m/s from a 1.0-m-high diving board. Choosing the origin to be at the water's surface, and upward to be the positive x direction, write x-versus-t equations of motion for both Bill and Ted.
Answer:
equation of motion for Bill is
[tex]y(t) = 4.9t^2[/tex]
equation of motion for Ted is
[tex]y(t) = 2 + (-4.2)(t) + 4.9t^2[/tex]
Explanation:
Taking downward position positive and upward position negative
g = 9.8 m/s^2
equation of motion for Bill is
[tex]y(t) = y_0 +v_0 t +\frac{1}{2}gt^2[/tex]
[tex]y(t) = 0 + 0(t) +\frac{1}{2}gt^2[/tex]
[tex]y(t) = \frac{1}{2}\times (9.8t)^2[/tex]
[tex]y(t) = 4.9t^2[/tex]
equation of motion for Ted is
[tex]y_0 = 2m -1m = 2m[/tex]
[tex]y_0 = -4.2 m/s[/tex]
[tex]y(t) = y_0 +v_0 t +\frac{1}{2}gt^2[/tex]
[tex]y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2[/tex]
[tex]y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2[/tex]
[tex]y(t) = 2 + (-4.2)(t) + 4.9t^2[/tex]
Answer:
Answer:
For Bill:
[tex]x(t)=3-(4.9*t^{2})[/tex]
For Ted:
[tex]x(t)=1+(4.2*t)+(-4.9*t^{2} )[/tex]
Explanation:
For Bill:
[tex]Initial position=x_{0}=3[/tex]
[tex]Initial velocity=v_{0}=0[/tex]
Now using [tex]2^{nd}[/tex] equation of motion,we have
[tex]x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})[/tex]
[tex]x_{0} =3[/tex] ,[tex]v_{0}=0[/tex]
Thus,equation becomes
[tex]x-3=1/2*g*t^{2}[/tex]
[tex]x=3+(0.5*g*t^{2})[/tex]
Taking acceleration upward positive and downward negative.
[tex]g=-10[/tex] [tex]m/s^{2}[/tex]
[tex]x(t)=3-4.9*t^{2}[/tex] for bill
For Ted
[tex]x_{0} =1[/tex]
[tex]v_{0}=4.2[/tex] [tex]m/s[/tex]
Using the same equation
[tex]x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})[/tex]
[tex]x_{0}=1[/tex] [tex]m[/tex]
[tex]v_{0}=4.2[/tex] [tex]m/s[/tex]
Substitute values
[tex]x-1=(4.2*t)+(1/2*g*t^{2})[/tex]
[tex]g=-10[/tex] [tex]m/s^{2}[/tex]
Thus equation becomes
[tex]x(t)=1+(4.2*t)+(-4.9*t^{2})[/tex] for Ted
The bones of a saber-toothed tiger are found to have an activity per gram of carbon that is 12.9 % of what would be found in a similar live animal. How old are these bones?
Answer:
The bones are 16925 years old
Explanation:
We have to use the radioactive decay law and know that the half life of carbon-14 is [tex]t_{\frac{1}{2}}=5730 \, years[/tex]. From this information we can know the decay rate of the carbon 14,
[tex]\lambda=\frac{ln(2)}{t_{\frac{1}{2}}}=1.21\times 10^{-4} s^{-1}[/tex]
Now to know the age of the bones we must directly use the radioactive decay law:
[tex]N(t)=N_0e^{-\lambda t}=0.129N_0[/tex]
Where the rightmost part of the equation comes from the statement that the activity found is just 12.9% of the activity that would be found in a similar live animal. This means that the number of carbon-14 atoms is just 12.9% of what it was at the moment the saber-toothed tiger died.
Solving for t we have:
[tex]t=-\frac{ln(0.129)}{\lambda}=16925 \, years[/tex]
A 6cm diameter horizontal pipe gradually narrows to 4cm.
Whenwater flows throught this pipe at a certain rate, the
guagepressure in these two sections is 32.0kPa and 24kPa
respectively.What is the volume rate of flow?
Final answer:
The pressure ratio between different diameter pipes can be calculated based on the square of their diameters.
Explanation:
To find the volume rate of flow (Q), we can use the principle of continuity that states the volume flow rate must be constant throughout the pipe. Q can be calculated using the area of cross-section (A) and velocity (v) with the formula Q = A1 x v1 = A2 x v2.
The ratio in water pressure between the larger and smaller water pipes can be determined by considering the cross-sectional areas of the pipes. As pressure is inversely proportional to the cross-sectional area of the pipe, the ratio in pressure is equal to the ratio of the squares of the pipe diameters. In this case, the ratio of pressures would be 36:1.
A machine is capable of a power output of 10 W+/-3 dB. What is the possible power fluctuation?
Answer:
The possible value of power fluctuation is 3.9811 W
Solution:
As per the question:
Power output, P = 10 W
Power fluctuation is from +3 dB to -3 dB
Therefore,
Power fluctuation in dB = 6 dB
Now,
P (in dB) = [tex]10log_{10}P'[/tex]
6 dB = [tex]10log_{10}P'[/tex]
[tex]0.6 = log_{10}P'[/tex]
[tex]P' = 10^{0.6} W = 3.9811 W[/tex]