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.
When a particular wave is vibrating with a frequency of 4.00 Hz, a transverse wave of wavelength 60 cm is produced. Determine the speed of waves along the wire.
Answer:
Speed of the wave, v = 2.4 m/s
Explanation:
Given that,
Frequency of vibrations, f = 4 Hz
Wavelength of the transverse wave, [tex]\lambda=60\ cm=0.6\ m[/tex]
We need to find the speed of the waves along the wire. Let it is equal to v. So, the relationship is given by :
[tex]v=f\times \lambda[/tex]
[tex]v=4\times 0.6[/tex]
v = 2.4 m/s
So, the speed of waves along the wire is 2.4 m/s. Hence, this is the required solution.
Final answer:
The speed of the transverse wave on the wire, given a frequency of 4.00 Hz and a wavelength of 60 cm, is calculated to be 2.40 meters per second (m/s) using the formula Speed = Frequency × Wavelength.
Explanation:
To determine the speed of a wave, you can use the formula:
Speed = Frequency × Wavelength
In the given scenario, the frequency is 4.00 Hz, and the wavelength is 60 cm, which is equivalent to 0.60 meters (since 1 meter = 100 cm). Thus, the speed of the wave on the wire can be calculated as follows:
Speed = 4.00 Hz × 0.60 m
Speed = 2.40 m/s
Therefore, the speed of the transverse wave along the wire is 2.40 meters per second (m/s).
How many photons are emitted per second by a He−Ne laser that emits 1.5 mW of power at a wavelength λ=632.8nm. What is the frequency of the electromagnetic waves emitted by a He−Ne laser?
Answer:
The frequency of the emitted EM wave by He-Ne laser is [tex]4.74\times 10^{14} Hz[/tex]
Given:
Power emitted by He-Ne laser, P = 1.5 mW = [tex]1.5\times 10^{- 3}[/tex]
Wavelength, [tex]\lambda = 632.8 nm = 632.8\times 10^{- 9} m[/tex]
Solution:
Now, to calculate the frequency, [tex]\vartheta[/tex] of the EM wave emitted:
Energy associated with 1 photon = Power of one photon per sec = [tex]\frac{hc}{\lambda}[/tex]
Therefore, power associated with 'N' no. of photons, P = [tex]N\frac{hc}{\lambda}[/tex]
where
[tex]h = 6.626\times 10^{-34} m^{2}kg/s[/tex] = Planck's constant
Now,
[tex]1.5\times 10^{- 3} = N\frac{6.626\times 10^{-34}\times 3\times 10^{8})}{632.8\times 10^{- 9}}[/tex]
N = [tex]4.77\times 10^{15}[/tex]
Also, we know that:
[tex]c = \vartheta \lambda[/tex]
Thus
[tex]\vartheta = \frac{c}{\lambda} = \frac{3\times 10^{8}}{632.8\times 10^{- 9}}[/tex]
[tex]\vartheta = 4.74\times 10^{14} Hz[/tex]
Find the magnitude of vector A = i - 2j + 3k O V14 10 O4
Answer:
[tex]|A|=\sqrt{1^2+(-2)^2+(3)^2}=3.74[/tex]
Explanation:
Given that,
Vector [tex]A=i-2j+3k[/tex]............(1)
We need to find the magnitude of vector A. Let us suppose vector in the form of, R = xi + yj + zk. Its magnitude is given by :
[tex]|R|=\sqrt{x^2+y^2+z^2}[/tex]
In equation (1),
x = 1
y = -2
z = 3
So, the magnitude of vector A is given by :
[tex]|A|=\sqrt{1^2+(-2)^2+(3)^2}[/tex]
|A| = 3.74
So, the magnitude of given vector is 3.74. Hence, this is the required solution.
Assume a certain adult requires 2.4 × 102 mL of pure oxygen per minute and assume inhaled air contains 20.0 percent oxygen by volume and exhaled air is 16.0 percent. (Also, assume that the volume of inhaled air is equal to that of the exhaled air.) If the person breathes 12.0 times every minute, what is the volume of air per breath?
Answer:
volume of air breath = 495.75 mL per min
Explanation:
given data
oxygen required = 2.4 × 10² mL / Min = 240 mL
inhaled air = 20 %
exhaled air = 16 %
to find out
volume of air per breath
solution
we know here that breathes 12 time every minute
and we consider here air breath = y per min
we know oxygen remain in body = 20 % - 16 % = 4%
and required 12 time air breath
so air required = 240 / 12 = 19.83 mL
so oxygen breath equation will be
y × 4% = 19.83
y = 19.83 / 0.04
y = 495.75
so volume of air breath = 495.75 mL per min
During the _____ developmental stage children are drawing their thoughts instead of merely observations.
A. scribbling
B. Schematic
C. circular
D. pre schematic
Answer:
D. pre schematic
Explanation:
During the _pre schematic____ developmental stage children are drawing their thoughts instead of merely observations.
Pre schematic stage in children is the age at which they drawing, symbols, spatial figures to express themselves. Their becomes more complex and full of lines as they age. They use single base lines, multiple base lines and fold up views.
A ball is thrown vertically downwards at speed vo from height h. Draw velocity vs. time & acceleration vs. time graphs. In terms of only the givens vo and h, derive expressions for the final speed of the ball and the elapsed time of flight.
Answer:
Explanation:
let the ball is thrown vertically downwards with velocity u.
So, initial velocity, = - u (downwards)
acceleration = - g (downwards)
let the velocity is v after time t.
use first equation of motion
v = u + at
- v = - u - gt
v = u + gt
So, it is a straight line having slope g and y intersept is u.
The graph I shows the velocity - time graph.
Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.
So, acceleration time graph is a starigh line parallel to time axis having slope zero.
the graph II shows the acceleration - time graph.
Use III equation of motion to find the final speed in terms height.
[tex]v^{2}=u^{2}+2gh[/tex]
And the time is
v = u + gt
[tex]t=\frac{v-u}{g}[/tex]
A small metal sphere of mass 3.1 g and charge 2.6 μC is fired with an initial speed of 5.6 m/s directly toward the center of a second metal sphere carrying charge 3.4 μC. This second sphere is held fixed. If the spheres are initially a large distance apart, how close do they get to each other? Treat the spheres as point charges.
Answer:
d = 1.636 m
Explanation:
Initially charged sphere are far apart so their potential energy is zero . Kinetic energy will be of small sphere
K E of small sphere
= 1/2 m v²
= .5 x 3.1 x 10⁻³ x ( 5.6)²
=48.608 x 10⁻³ J
If d be the distance of closest approach between them , potential energy of
Charges
= k q₁ x q₂ / d²
[tex]\frac{9\times10^9\times2.6\times10^{-6}\times3.4\times10^{-6}}{d^2}[/tex]
= [tex]\frac{79.56\times10^{-3}}{d^2}[/tex]
Total kinetic energy at this point will be zero.
Applying the theory of conservation of mechanical energy
Potential energy at distance d = Kinetic energy at infinity
48.608 x 10⁻³ = [tex]\frac{79.56\times10^{-3}}{d^2}[/tex]
d = 1.636 m
A student is driving to school on a road with a speed limit of 40mph (17.9 m/s). A stoplight ahead of her turns yellow, so she begins to slow down to come to a stop. On a dry day, she is able to make her car slow down with an acceleration of magnitude of 8.8 m/s^2. However, if the road is wet, she can only manage an acceleration with a magnitude 3.9 m/s^2. Part A:
On a dry day, how far (in meters) from the intersection would the student need to start braking in order to stop in time?
Part B:
If it were raining instead, how far (in meters) from the intersection would the student need to start braking?
Answer:
18.21 m
41.1 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Part A
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -8.8}\\\Rightarrow s=18.21\ m[/tex]
On a dry day she would have to start braking 18.21 m away from the intersection
Part B
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -3.9}\\\Rightarrow s=41.1\ m[/tex]
On a wet day she would have to start braking 41.1 m away from the intersection
Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at the angle 75° above horizontal. What ball will have bigger speed at the highest point of their trajectory? (section 4.3) Ball A Ball B Impossible to answer without knowing their actual initial speeds. They will have the same speeds
Answer:
Ball A
Solution:
According to the question:
Two balls A and B are kicked at angle [tex]20^{\circ}[/tex] and [tex]75^{\circ}[/tex] respectively.
The component of the velocity that provides the ball with the forward movement is the horizontal component, i.e., cosine of the angle.
Since, cos[tex]20^{\circ}[/tex] is greater than cos[tex]75^{\circ}[/tex] and thus Ball A will have more speed than ball B at the top most point of the trajectory.
Final answer:
Ball A will have a bigger speed at the highest point of its trajectory in comparison to ball B, as it was kicked at a lower angle of 20°, resulting in a larger horizontal component of the initial velocity when compared to ball B kicked at 75°.
Explanation:
The question is asking about the speeds of two balls at the highest point of their trajectories when they are kicked with the same initial velocity but at different angles. To determine the speed at the highest point, we can use the principles of projectile motion. The speed of a projectile at its highest point consists only of the horizontal component of the initial velocity because the vertical component equals zero at that instant.
Since both balls A and B were kicked with the same initial speed, the horizontal component of that speed can be found using the cosine of their respective launch angles. For ball A, this component is cos(20°) multiplied by the initial speed, and for ball B, it is cos(75°) multiplied by the initial speed. Because the cosine of 20° is greater than the cosine of 75°, ball A will have a larger horizontal component of velocity and thus a higher speed at the peak of its trajectory.
Building helps children develope their _____ skills
A. number sense and operations
B. operations and geometry
C. measurement and counting
D. measurement and geometry
Answer:
D. measurement and geometry
Explanation:
Building helps children develop their _ measurement and geometry__ skills. The measurement and geometry skill of the children by observing buildings.
Geometry is a method or tool for understanding the relations among shapes and spatial properties. Children can develop spatial reasoning and can visualize shapes in different positions (orientation) when observing a building.
The molecules in metal are usually more tightly packed than the molecules in water. Most metals will sink in water. Pick your conclusion from this list. Water is more dense than metal
Metal is more dense than water
Water and Metal have the same density
Answer:
Metal is more dense than water.
Explanation:
As we know, the molecules of the metal are tightly closed as compared to that of water and the density of a material is defined as the mass of the material per unit volume.
In a certain volume of metal, there are more numbers of molecules than that of water in the same amount of volume, therefore the density of metal is greater than that of water.
Also, according to Archimedes' principle, if there is an object in a fluid, then the buoyant force on that object is equal to the weight of the fluid that it displaces.
When the density of the object is larger than that of the fluid then it overcomes that buoyant force and sinks.
Thus, an object sinks in a fluid if its density is larger than that of the fluid and floats otherwise.
Since, the metal sink in water, it means Metal is more dense than water.
What are the concepts of dynamically continuous innovation and discontinuous innovation? Can you share examples to illustrate them in action?
Answer:
Explained
Explanation:
Dynamically continuous innovation:
- Falls in between continuous and discontinuous innovation.
-Changes in customer habits are not as large as in discontinuous innovation and not as negligeble as in continuous innovation.
best example can as simple as transformation in Television. New HD TVs have flat panels, wide screens and improved performance The Added features are considered dynamically improved.
Discontinuous innovation:
- discontinuous innovation comprise of new to world product only so they are discontinuous to every customer segment.
- these product are so fundamentally different from the the product that already exist that they reshape market and competition.
For example- the mobile and the internet technology are reshaping the market through regular innovation and change.
Dynamically continuous innovation involves moderate changes to an existent product, seen, for example, in the evolution from traditional cell phones to smartphones. Discontinuous innovation involves the creation of entirely new products, directly impacting consumer behavior, such as the invention of the internet.
Explanation:Dynamically continuous innovation and discontinuous innovation are two concepts centered around improvements to existing products or introduction of radically new products. A dynamically continuous innovation signifies moderate changes to existing products. An example could be the transition from a traditional cell phone to a smartphone. The base product- phone remained the same, only the features were enhanced. This innovation still requires consumers to adapt to new functionalities, but to a lesser degree than discontinuous innovations.
On the other hand, discontinuous innovation involves creating products that did not exist before, fundamentally changing consumer behavior. An example would be the creation of the internet. Before its invention, nothing similar existed and it greatly impacted how people live and work.
Learn more about Innovation here:https://brainly.com/question/33676512
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A race car has a speed of 80 m/s when the driver releases a drag parachute. If the parachute causes a deceleration of 4m/s^2, how far will the car travel before it stops?
Answer:
The car travels 800m
Explanation:
We will need two formulas to solve this question:
[tex]V_{f}=V_{0}+at\\X=V_{0}t+\frac{at^{2}}{2}[/tex]
We can obtain the the time by using the final velocity formula and replacing with known values:
[tex]V_{f}=V_{0}+at\\\\0\frac{m}{s}=80\frac{m}{s}-4\frac{m}{s^{2}}t\\\\4\frac{m}{s^{2}}t=80\frac{m}{s}\\t=20 s[/tex]
Since we know the time, the question will be answered by inputing all the variables needed in the distance formula:
[tex]X=V_{0}t+\frac{at^{2}}{2}\\\\X=80\frac{m}{s}(20s)-\frac{4(20s)^{2}}{2}=800m[/tex]
Final answer:
The race car will travel 800 meters before coming to a stop after deploying a drag parachute causing a deceleration of 4 m/s^2 from an initial speed of 80 m/s.
Explanation:
To calculate how far the race car will travel before it stops after deploying a drag parachute, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled. Considering that the deceleration is constant, the kinematic equation we can use is v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Since the car comes to a stop, v = 0 m/s. The initial speed is u = 80 m/s and the deceleration provided by the drag parachute is a = -4 m/s2 (negative because it's deceleration). Plugging these values into the equation gives us:
0 = (80 m/s)2 + 2(-4 m/s2)s
Solving for s, we find:
s = (80 m/s)2 / (2 * 4 m/s2)
s = 6400 m2/s2 / 8 m/s2
s = 800 meters
Therefore, the race car will travel 800 meters before coming to a stop after the drag parachute is deployed.
Calculate the potential energy of a +1.0μC point charge sitting 0.1m from a -5.0μC point charge.
Answer:
P.E. = -0.449 J
Explanation:
Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.
Now Potential energy is defined by this formula,
P.E. = k q₁ q₂/ r
where P.E. is the potential energy.
k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)
q₁ = charge of one particle = +1.0μC
q₂ = charge of another particle = -5.0μC
r = distance = 0.1 m
Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m
P.E. = -0.449 J
A vertical electric field of magnitude 1.85 x 10^4 N/C exists above the Earth's surface on a day when a thunderstorm is brewing. A car with a rectangular size of 6.00 m by 3.00 m is traveling along a dry gravel roadway sloping downward at 19.0°. Determine the electric flux through the bottom of the car.
Answer:
The electric flux through the car's bottom is [tex]1.75\times 10^{4} Wb[/tex]
Solution:
As per the question:
Magnitude of vertical Electric field, [tex]E_{v} = 1.85\times 10^{4} N/C[/tex]
The area of the rectangular surface of the car, [tex]A_{bottom} = 6\times 3 = 18 m^{2}[/tex]
Downward slope at an angle, [tex]\angle\theta = 19.0^{\circ}[/tex]
Now, the electric flux, [tex]\phi_{E}[/tex] is given by:
[tex]\phi_{E} = \vec{E_{v}}.\vec{A_{bottom}} = E_{v}A_{bottom}cos\theta[/tex]
Now, substituting the appropriate values in the above formula:
[tex]\phi_{E} = 1.85\times 10^{4}\times 18cos19.0^{\circ}[/tex]
[tex]\phi_{E} = 1.75\times 10^{4} Wb[/tex]
An opera singer in a convertible sings a note at 550 Hz while cruising down the highway at 92 km/h . The speed of sound in the air is 343 m/s. A) What is the frequency heard by a person standing beside the road in front of the car?
B) What is the frequency heard by a person standing beside the road behind the car?
Answer:
A) 594 Hz
B) 512 Hz
Explanation:
Howdy!
This is a typical Doppler effect problem, whose formula is:
[tex]f = \frac{v+v_{r} }{v+v_{s}} f_{0}[/tex]
Where :
[tex]v_{r}[/tex]
is the velocity of the receiver (in our case is equal to zero)
[tex]v_{s}[/tex]
is the velocity of the source (in our case is 92 km/h) which is positive when the source is moving away (second case) and negative otherwise.
[tex]v_{s}[/tex]
Is the velocity of the wave in the medium.
Before we start calculating we need to have the velocity of the source in the same units as the velocity of the waves:
92km/h = 92*(1000)/3600 m/s = 25.5 m/s
A)
When the source is moving towards the receiver the sign of the velocity is negative, so:
[tex]f = \frac{343}{343-25.5} 550 Hz[/tex]
f = 594.1 Hz
B) Now the velocity of the source must change sign:
[tex]f = \frac{343}{343+25.5} 550 Hz[/tex]
f = 511.9 Hz
As a sanity check we know that when the source is moving towards the source the frequency is higher and when the source is moving away from the source the frequency is lower.
A) The frequency heard by a person standing beside the road in front of the car is 595.3 Hz. B) The frequency heard by a person standing beside the road behind the car is 514.5 Hz.
To solve this problem, we will use the Doppler effect formula, which relates the observed frequency to the actual frequency of the source, the speed of sound, and the relative velocity between the source and the observer. The formula for the observed frequency [tex]\( f' \)[/tex] when there is relative motion between the source and the observer is given by:
[tex]\[ f' = \left( \frac{v + v_o}{v + v_s} \right) f \][/tex]
where:
f is the frequency of the source,
v is the speed of sound in the medium,
[tex]v_o[/tex] is the speed of the observer relative to the medium (positive if moving towards the source, negative if moving away),
[tex]v_s[/tex] is the speed of the source relative to the medium (positive if moving away from the observer, negative if moving towards).
A) For the person standing beside the road in front of the car:
The observer is stationary, so [tex]\( v_o = 0 \)[/tex].
The car is moving towards the observer, so [tex]\( v_s = -92 \text{ km/h} \)[/tex].
We need to convert the speed of the car from km/h to m/s to match the units of the speed of sound:
[tex]\[ 92 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 25.56 \text{ m/s} \][/tex]
Now we can plug the values into the Doppler effect formula:
[tex]\[ f' = \left( \frac{343 \text{ m/s} + 0}{343 \text{ m/s} - 25.56 \text{ m/s}} \right) \times 550 \text{ Hz} \][/tex]
[tex]\[ f' = \left( \frac{343}{317.44} \right) \times 550 \][/tex]
[tex]\[ f' \approx 595.3 \text{ Hz} \][/tex]
B) For the person standing beside the road behind the car:
The observer is again stationary, so [tex]\( v_o = 0 \)[/tex].
The car is moving away from the observer, so [tex]\( v_s = 92 \text{ km/h} \)[/tex] or [tex]\( 25.56 \text{ m/s} \)[/tex].
Using the Doppler effect formula:
[tex]\[ f' = \left( \frac{343 \text{ m/s} + 0}{343 \text{ m/s} + 25.56 \text{ m/s}} \right) \times 550 \text{ Hz} \][/tex]
[tex]\[ f' = \left( \frac{343}{368.56} \right) \times 550 \][/tex]
[tex]\[ f' \approx 514.5 \text{ Hz} \][/tex]
A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wall at a point having coordinates (3.1, 0.5), where the units are meters, what is the distance of the fly from the corner of the room? Answer needs to be in appropriate significant figures.
Answer:
The distance is 3.1 m
Explanation:
The position vector of the fly relative to the corner of the wall is
r = (3.1, 0.5).
The distance of the fly from the corner will be calculated as the magnitude of the vector "r"
magnitude of vector [tex] r = \sqrt{(3.1 m)^{2} + (0.5 m)^{2}} = 3.1 m[/tex]
Since the numbers to be added have only one decimal place 3.1 and 0.5, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.
A particle (q = 3.0 mC, m = 20 g) has a speed of 20 m/s when it enters a region where the electric field has a constant magnitude of 80 N/C and a direction which is the same as the velocity of the particle. What is the speed of the particle 3.0 s after it enters this region?
Answer:
56 m/s
Explanation:
The electirc force applied on the particle by the field will be
F = q * E
F = 3*10^-3 * 80 = 0.24 N
This force will cause an acceleration:
F = m * a
a = F/m
a = 0.24 / 0.02 = 12 m/s^2
The equation for speed under constant acceleration is:
V(t) = V0 + a*t
V(3) = 20 + 12 * t = 56 m/s
The final speed will be 56 m/s.
A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?
Answer:
The time for final 15 cm of the jump equals 0.1423 seconds.
Explanation:
The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as
[tex]v^2=u^2+2as[/tex]
where
'v' is the final velocity of the player
'u' is the initial velocity of the player
'a' is acceleration due to gravity
's' is the height the player jumps
Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get
[tex]0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s[/tex]
Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as
[tex]v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s[/tex]
Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as
[tex]v=u+at[/tex]
where symbols have the usual meaning
Applying the given values we get
[tex]t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds[/tex]
A 1-m3 tank containing air @ 25 oC & 500 kPa is connected to another tank containing 5 kg of air at 35 oC & 200 kPa through a valve. The valve is opened and the whole system is brought to thermal equilibrium with the surrounding of 20 oC. Determine the volume of the second tank and the final equilibrium pressure of air. Take air gas constant (R) = 0.287 kJ/(kg.oK)
Answer:
Volume of Tank 2, V' = [tex]2.17 m^{3}[/tex]
Equilibrium Pressure, [tex]P_{eq} = 278.82 kPa[/tex]
Given:
Volume of Tank 1, V = 1 [tex]m^{3}[/tex]
Temperature of Tank 1, T = [tex]25^{\circ}C[/tex] = 298 K
Pressure of Tank 1, P = 500 kPa
Mass of air in Tank 2, m = 5 kg
Temperature of tank 2, T' = [tex]35^{\circ}C[/tex] = 303 K
Pressure of Tank 2, P' = 200 kPa
Equilibrium temperature, [tex]20^{\circ}C[/tex] = 293 K
Solution:
For Tank 1, mass of air in tank can be calculated by:
PV = m'RT
[tex]m' = \frac{PV}{RT}[/tex]
[tex]m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg[/tex]
Also, from the eqn:
PV' = mRT
V' = volume of Tank 2
Thus
V' = [tex]\frac{mRT}{P}[/tex]
V' = [tex]\frac{5\times 0.287\times 303}{200} = 2.17 m^{3}[/tex]
Now,
Total Volume, V'' = V + V' = 1 + 2.17 = 3.17[tex]m^{3}[/tex]
Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg
Final equilibrium pressure, P'' is given by:
[tex]P_{eq}V'' = m''RT_{eq}[/tex]
[tex]P_{eq} = \frac{m''RT_{eq}}{V''}[/tex]
[tex]P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa[/tex]
[tex]P_{eq} = 287.82 kPa[/tex]
The separation between two 1 kg masses is (a) decreased by 2/3, and (b) increased by a factor of 3. How is the mutual gravitational force affected in each case?
Final answer:
When the separation between two 1 kg masses is decreased by 2/3, the mutual gravitational force between them will increase. Conversely, when the separation is increased by a factor of 3, the mutual gravitational force will decrease.
Explanation:
When the separation between two 1 kg masses is decreased by 2/3, the mutual gravitational force between them will increase. As the separation decreases, the gravitational force between the masses becomes stronger. On the other hand, when the separation is increased by a factor of 3, the mutual gravitational force will decrease. As the separation increases, the gravitational force between the masses becomes weaker.
An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard by the observer if the source emits a 2550-Hz sound?
Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz
[tex]f = f_0(\dfrac{v-v_0}{v-v_s})[/tex]
[tex]f = 2550\times (\dfrac{343+17}{343-0})[/tex]
velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
an alternator consists of a coil of area A with N turns that rotates in a uniform field B around a diameter perpendicular to the field with a rotation frequency f.
a)find the fem in the coil
b)what is the amplitude of the alternating voltage if N= 100 turns, A = 10 ^ -2 m ^ 2, B = 0.1T, f = 2000 rev/min
Answer:
(a): emf = [tex]\rm 2\pi f NBA\sin(2\pi ft).[/tex]
(b): Amplitude of alternating voltage = 20.942 Volts.
Explanation:
Given:
Area of the coil = A.Number of turns of coil = N.Magnetic field = BRotation frequency = f.(a):
The magnetic flux through the coil is given by
[tex]\phi = \vec B \cdot \vec A=BA\cos\theta[/tex]
where,
[tex]\vec A[/tex] = area vector of the coil directed along the normal to the plane of the coil.
[tex]\theta[/tex] = angle between [tex]\vec B[/tex] and [tex]\vec A[/tex].
Assuming, the direction of magnetic field is along the normal to the plane of the coil initially.
At any time t, the angle which magnetic field makes with the normal to the plane of the coil is [tex]2\pi ft[/tex]
Therefore, the magnetic flux linked with the coil at any time t is given by
[tex]\phi(t) = NBA\cos(2\pi ft)[/tex]
According to Faraday's law of electromagnetic induction, the emf induced in the coil is given by
[tex]e=-\dfrac{d\phi}{dt}\\=-\dfrac{d(NBA\cos(2\pi ft))}{dt}\\=-NBA(-2\pi f\sin(2\pi ft))\\=2\pi f NBA\sin(2\pi ft).[/tex]
(b):
The amplitude of the alternating voltage is the maximum value of the emf and emf is maximum when [tex]\sin(2\pi ft)=1.[/tex]
Therefore, the amplitude of the alternating voltage is given by
[tex]e_o = 2\pi ft NBA.[/tex]
We have,
[tex]N=100\\A=10^{-2}\ m^2\\B=0.1\ T\\f=2000\ rev/ min = 2000\times \dfrac{1}{60}\ rev/sec=33.33\ rev/sec.[/tex]
Putting all these values,
[tex]e_o = 2\pi \times 33.33\times 100\times 0.1\times 10^{-2}=20.942\ Volts.[/tex]
Two blocks of weight 3.0N and 7.0N are connected by a massless string and slide down a 30 degree inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.13; that between the heavier block and the plane is 0.31. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string
Answer:
a. a = [tex]6.41 m/s^2[/tex]
b. T = -0.81 N
Explanation:
Given,
weight of the lighter block = [tex]w_1\ =\ 3.0\ N[/tex]weight of the heavier block = [tex]w_2\ =\ 7.0\ N[/tex]inclination angle = [tex]\theta\ =\ 30^o[/tex]coefficient of kinetic friction between the lighter block and the surface = [tex]\mu_1\ =\ 0.13[/tex]coefficient of kinetic friction between the heavier block and the surface = [tex]\mu_2\ =\ 0.31[/tex]friction force on the lighter block = [tex]f_1\ =\ \mu_1N_1\ =\ \mu_1 w_1cos\theta[/tex]friction force on the heavier block = [tex]f_2\ =\ \mu_2N_2\ =\ \mu_2w_2cos\theta[/tex]Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.
From the f.b.d. of the lighter block,
[tex]w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]
From the f.b.d. of the heavier block,
[tex]w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)[/tex]
From eqn (1) and (2), we get,
[tex]w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\[/tex]
[tex]\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\[/tex]
[tex]\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.[/tex]
part (b)
From the eqn (2), we get,[tex]T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N[/tex]
Answer:
(a) 2.793 m/s^2
(b) 0.335 N
Explanation:
Let a be the acceleration in the system and T be the tension in the string.
W1 = 3 N
W2 = 7 N
m1 = 3 /1 0 = 0.3 kg
m2 = 7 / 10 = 0.7 kg
θ = 30°
μ1 = 0.13, μ2 = 0.31
Let f1 be the friction force acting on block 1 and f2 be the friction force acting on block 2.
By the laws of friction
f1 = μ1 x N1
Where, N1 be the normal reaction acting on block 1.
So, f1 = 0.13 x W1 Cosθ = 0.13 x 3 x cos 30 = 0.337 N
By the laws of friction
f2 = μ2 x N2
Where, N2 be the normal reaction acting on block 2.
So, f2 = 0.31 x W2 Cosθ = 0.31 x 7 x cos 30 = 1.88 N
Apply Newton's second law for both the blocks
W1 Sin30 - T - f1 = m1 a
3 Sin 30 - T - 0.337 = 0.3 x a
1.163 - T = 0.3 a ..... (1)
W2 Sin30 + T - f2 = m2 a
7 Sin30 + T - 1.88 = 0.7 x a
1.62 + T = 0.7 a ..... (2)
By solving equation (1) and (2) we get
a = 2.793 m/s^2
(b) Put the value of a in equation (2), we get
1.62 + T = 0.7 x 2.793
T = 0.335 N
Does percent difference give indication of accuracy or precision? Discuss
Answer:
Discussed
Explanation:
Percentage error is the measure of how far is your measured value from the actual value. It give the magnitude of error in form of percentage.
it is calculated as ratio of difference of calculated and the actual value to the actual value multiplied by 100.
If you are repeating the experiment the percent difference in your results will give the measure of how much variation there is in your own techniques, and therefore will give the precision of your experimental procedure.
A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and the corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load of 87,000 N is reached at a gage length = 64.2 mm. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 67.3 mm, determine the percent elongation. (e) If the specimen necked to an area = 53 mm2, determine the percent reduction in area.
Answer:
a) yield strength
[tex]\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa[/tex]
b) modulus of elasticity
strain calculation
[tex]\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046[/tex]
strain for offset yield point
[tex]\varepsilon_{new} = \varepsilon_0 -0.002[/tex]
=0.0046-0.002 = 0.0026
now, modulus of elasticity
[tex]E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}[/tex]
= 184615.28 MPa = 184.615 GPa
c) tensile strength
[tex]\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa[/tex]
d) percentage elongation
[tex]\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%[/tex]
e) percentage of area reduction
[tex]\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%[/tex]
The mechanical properties of the tensile test specimen, such as yield strength, modulus of elasticity, tensile strength, percent elongation, and percent reduction in area, are determined by performing calculations based on the given dimensions, yield force, maximum load, and deformations observed during the test.
Explanation:To determine the mechanical properties of a tensile test specimen, the following calculations are performed:
Yield strength (a) is the stress at which a material begins to deform plastically. Here, yield strength = Yield force / Area = 48,000 N / 100 mm² = 480 MPa (since 1 N/mm² = 1 MPa).Modulus of elasticity (b) (also known as Young's Modulus) which is a measure of the stiffness of a material, can be calculated using the formula Stress / Strain. The stress before yielding is 480 MPa, and strain is the change in length over the original length, which is (50.23 mm - 50 mm) / 50 mm = 0.0046. Thus, Modulus of elasticity = 480 MPa / 0.0046 = 104,347.83 MPa.Tensile strength (c) is the maximum stress that a material can withstand while being stretched or pulled before necking. Tensile strength = Maximum load / Original cross-sectional area = 87,000 N / 100 mm² = 870 MPa.To determine the percent elongation (d) at fracture, use the formula ((Final gage length - Initial gage length) / Initial gage length) * 100. Percent elongation = ((67.3 mm - 50 mm) / 50 mm) * 100 = 34.6%.The percent reduction in area (e) is calculated by ((Original area - Final area) / Original area) * 100. Percent reduction in area = ((100 mm² - 53 mm²) / 100 mm²) * 100 = 47%.A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
[tex]a=\dfrac{v-u}{t}[/tex].............(1)
Now, the second equation of motion is :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put the value of a in above equation as :
[tex]s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2[/tex]
[tex]s=\dfrac{t(u+v)}{2}[/tex]
[tex]u=\dfrac{2s}{t}-v[/tex]
[tex]u=\dfrac{2\times 47}{8.6}-2.3[/tex]
u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.
A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car? O a. 0.72 kN O b.0.68 kN O c. 0.64 kN O d.0.76 kN O e. 0.52 kN
Answer:
c)[tex]F_{net} = 0.640 kN[/tex]
Explanation:
As we know that resultant force is the net force that is acting on the system
As per Newton's II law we know that net force is product of mass and acceleration
so we will have
[tex]F_{net} = ma[/tex]
here we know
m = 80 kg
for circular motion acceleration is given as
[tex]a_c = \frac{v^2}{R}[/tex]
[tex]a_c = \frac{40^2}{200} = 8 m/s^2[/tex]
now we have
[tex]F_{net} = 80 \times 8[/tex]
[tex]F_{net} = 640 N[/tex]
[tex]F_{net} = 0.640 kN[/tex]
Cheetahs can accelerate to a speed of 21.8 m/s in 2.55 s and can continue to accelerate to reach a top speed of 28.1 m/s . Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x direction point in the direction the cheetah runs. Express the cheetah's top speed vtop in miles per hour (mi/h) .
Starting from a crouched position, how much time totall does it take a cheetah to reach its top speed and what distance d does it travel in that time?
If a cheetah sees a rabbit 122.0 m away, how much time ttotal will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?
Answer:
Explanation:
Given
Cheetah speed=21.8 m/s in 2.55 sec
i.e. its [tex]a=\frac{21.8}{2.55}=8.55 m/s^2[/tex]
Cheetah top speed=28.1 m/s
And 1 m/s is equal to 2.23694 mph
therefore 28.1 m/s is [tex]2.236\times 28.1=62.858 mph[/tex]
Time taken to reach top speed
v=u+at
[tex]28.1=0+8.55\times t[/tex]
[tex]t=\frac{28.1}{8.55}=3.286 s [/tex]
Distance traveled during this time
[tex]v^2-u^2=2as[/tex]
[tex]28.1^2=2\times 8.55\times s[/tex]
[tex]s=\frac{789.61}{2\times 8.55}=46.176 m[/tex]
If cheetah sees a rabbit 122 m away
time taken to reach rabbit
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]122=0+\frac{1}{2}\times 8.55\times t^2[/tex]
[tex]t^2=\frac{244}{8.55}[/tex]
[tex]t=\sqrt{28.53}=5.34 s[/tex]
A 11400 kg lunar landing craft is about to touchdown on the
surface of the moon. acceleration due to gravity onmoon is 1.0
m/s2. At an altitude of 165 m the craft's downwardvelocity is 18.0
ms . To slow down the craft, a retrorocket isfired to provide an
upward thrust. Assuming the descent isvertical, find the magniturde
of the trhust needded to reduce thevelocity to zero at the instant
when the craft touches the lunarsurface.
Answer:
F = 22572N
Explanation:
We will first calculate the acceleration needed to achieve landing speed equals zero:
[tex]V_{f}^{2}=V_{o}^{2}+2*a*\Delta Y[/tex]
[tex]0=18^{2}+2*a*(-165)[/tex] Solving for a:
[tex]a=0.98 m/s^{2}[/tex]
Now that we have the required acceleration, we need to check all forces acting on the craft:
[tex]F_{thrust} - W = m*a[/tex] where W = 11400N; m = 11400Kg; a = 0.98m/s2
Solving the equation:
[tex]F_{thrust}=22572N[/tex]
The magniture of thrust required to reduce the lunar craft's velocity to zero can be calculated using the principles of kinematics and forces. The change in kinetic energy is calculated and equated with the work done by the retro-rocket's thrust, in order to find the required thrust.
Explanation:The physics involved in this problem is related to kinematics and forces. To calculate the amount of thrust needed, we first need to find the change in kinetic energy. The initial kinetic energy is 0.5 * 11400 kg * (18 m/s)^2 and the final kinetic energy is zero since the craft is at rest. Therefore, the change in kinetic energy is -0.5 * 11400 kg * (18 m/s)^2.
By the work-energy theorem, the work done by the retro-rocket's thrust is equal to the change in kinetic energy. Since work done is also equal to force (thrust) multiplied by distance, we can equate these two expressions to find the thrust: Thrust * 165 m = -0.5 * 11400 kg * (18 m/s)^2. Solve for thrust to get the magnitude of thrust needed.
Learn more about Physics of Lunar Landing here:https://brainly.com/question/5527401
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