Answer:
The student needs to group variables into dimensionless quantities.
Explanation:
Large experiments take a lot of time to perform because the significant variables need to be separated from the non-significant variables. However, for large quantities of variables, it is necessary to focus on the key variables.
One technique to do that is to use the Buckingham Pi Theorem. The theorem states that the physical variables can be expressed in terms or independent fundamental physical quantities. In other words:
P = n- k
n = total number of quantities
k = independent physical quantities.
A place to start with will be to find dimensionless quantities involving the mass, length, time, and at times temperature. These units are given as M, L, T, and Θ
The grouping helps because it eliminates unwanted and unnecessary experiments.
Answer:
I just took the test, It's hypothesis then design an experiment, the last one i got wrong but its not dependent. hope that helped a little.
A student starts from the hostel and goes to his friend's house and then to his school. A is the displacement vector from the hostel to his friend's house. B is the displacement vector from his friend's house to school, and C is the direct path from the hostel to school. Which of the following vector operations is correct for the displacement from the hostel to his friend's house?
A) A - B.
B) A + B.
C) C - B.
D) C + B.
Answer:[tex]\vec{C}-\vec{B}[/tex]
Explanation:
Given
[tex]\vec{A}[/tex] is displacement of student from Hostel to house
[tex]\vec{B}[/tex] is displacement of student from house to school
And [tex]\vec{C}[/tex] is displacement of student from Hostel to school
According to the triangle law of vector addition
If two vectors are acting on a body simultaneously that are shown by magnitude and direction by two sides of a triangle taken in order then the third side is the resultant of two vectors that are taken in opposite direction.
[tex]\vec{A}+\vec{B}=\vec{C}[/tex]
[tex]\vec{A}=\vec{C}-\vec{B}[/tex]
so option c is correct
A racquetball strikes a wall with a speed of 30 m/s and rebounds with a speed of 26 m/s. The collision takes 20 ms. What is the average acceleration of the ball during collision?
a. zero
b. 1500 m/s2
c. 1300 m/s2
d. 200 m/s2
e. 2800 m/s2
Answer:
e. 2800 m/s²
Explanation:
Acceleration: This is the rate of change of velocity. The S.I unit of acceleration is m/s²
Applying Newton's equation of motion,
a = (v-u)/t ................................... Equation 1
Where a = acceleration, v = Final velocity, u = Initial velocity, t = time.
Note: Let the direction of the initial velocity be negative, and the direction of the final velocity be positive
Given: v = 26 m/s, u = -30 m/s, t = 20 ms = 0.02 s
Substitute into equation 1
a = [26-(-30)]/0.02
a = (26+30)/0.02
a = 56/0.02
a = 2800 m/s².
Hence the right option is e. 2800 m/s²
In Fig. 23-45, a small circular hole of radius R = 1.80 cm has been cut in the middle of an infinite, flat, nonconducting surface that has uniform charge density σ = 4.50 pC/m2 . A z axis, with its origin at the hole’s center, is perpendicular to the surface. In unitvector notation, what is the electric field at point P at z = 2.56 cm?
Answer:
The electric field at point a =( 0.232N/c)k
Explanation:
E = E1+E2=(6/2Eo)K+(-6/2Eo)[1-(2/(sqrtz^2+R^2)k
E= 6z/2Eo(sqrtz^2+R^2)k
E= (4.5×10^-12)×(2.56×10^-2) /2(8.85×10^-12)sqrt(2.56×10^-2)^2+(1.30×10^-2)^2k
E= 1.152×10^-13/(1.77×10^-11)×(2.8×10^-2)k
E= 1.152×10^-13/(4.956×10^-13)k
E=( 0.232N/c)K
A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=(2.80 m/s)t +(0.61 m/s^3)t^3.What is the magnitude of the force F when 4.10s ?
Answer: 75.05N
Explanation:
According to newton's second law,
Force = mass×acceleration
Given mass = 5.0kg
Acceleration = change in velocity/time
Velocity = change in displacement/position/time i.e dy/dt
Given position of the particle as y(t)=(2.80 m/s)t +(0.61 m/s)t³
dy/dt = 2.80 + 3(0.61)t²
V = dy/dt = 2.80 + 1.83t²
Acceleration (a) = dv/dt = 2(1.83)t
dv/dt = 3.66t
To get the force when t =4.10second
dv/dt @ t = 4.10s will be 3.66(4.10)
acceleration = 3.66×4.10 = 15.01m/s²
Magnitude of the force F = ma
F = 5.0kg × 15.01m/s²
F = 75.05N
A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20∘C?
Assume that the air bubble is always in thermal equilibrium with the surrounding water.
Answer:
0.0257259766982 m
Explanation:
[tex]P_2[/tex] = Atmospheric pressure = 101325 Pa
[tex]d_1[/tex] = Initial diameter = 1.5 cm
[tex]d_2[/tex] = Final diameter
[tex]\rho[/tex] = Density of water = 1000 kg/m³
h = Depth = 40 m
The pressure is
[tex]P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa[/tex]
From ideal gas law we have
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m[/tex]
The diameter of the bubble is 0.0257259766982 m
An athlete swims the length L of a pool in a time t1 and makes the return trip to the starting position in a time t2. If she is swimming initially in the positive x-direction, determine her average velocities symbolically for the following. (Assume that time t2 is from the other end of the pool to the starting point. Use any variable or symbol stated above as necessary. Do not substitute numerical values; use variables only. Indicate the direction with the sign of your answer.)
Answer:
[tex]v_1=\frac{L}{t_1}[/tex]
[tex]v_2=\frac{-L}{t_2}[/tex]
[tex]v_{avg}=\frac{L-L}{t_1+t_2} =0[/tex]
Explanation:
Given:
length of the pool along which the athlete swims, [tex]L[/tex]time taken to swim in the positive x-direction from the start of the pool, [tex]t_1[/tex]time taken to return from the end of the pool to the starting point, [tex]t_2[/tex]Velocity of the athlete from start to the end of the pool in positive direction:
[tex]\rm velocity=\frac{displacement}{time}[/tex]
[tex]v_1=\frac{L}{t_1}[/tex]
Velocity of the athlete from end returning to the start of the pool in negative direction:
Here we have the negative displacement.
[tex]v_2=\frac{-L}{t_2}[/tex]
Now the total average velocity:
[tex]\rm v_{avg}=\frac{total\ displacement}{total\ time}[/tex]
Here we have total displacement as zero because the athlete is finally at the initial starting point.
so,
[tex]v_{avg}=\frac{L-L}{t_1+t_2} =0[/tex]
Final answer:
The symbolic average velocities of an athlete swimming the length of a pool are positive for the trip in one direction (v1 = L / t1) and negative for the return trip (v2 = -L / t2). These reflect the direction of motion related to the positive x-direction.
Explanation:
Average Velocities of an Athlete in a Pool
The average velocity of the athlete for the trip in the positive x-direction is the length of the pool L divided by the time taken t1, which can be symbolically written as v1 = L / t1. Since the swimmer is moving in the positive x-direction, the average velocity v1 is taken as positive.
For the return trip in the negative x-direction, the average velocity is the negative of the length of the pool L divided by the time taken t2, represented as v2 = -L / t2. The average velocity v2 is negative because the swimmer is moving in the opposite direction to the initial positive x-direction.
The concept of average velocity involves both the displacement of the swimmer and the time taken. If, at the end of two trips (to the end of the pool and back), the swimmer's displacement is zero (since they return to the start), the average velocity for the entire trip is also zero.
Consider the following distinct forces:______________________________.
1.A downward force of gravity.
2.A force exerted by the channel pointing from Q to O.
3.A force in the direction of motion.
4.A force pointing from O to Q.
Which of the above forces is (are) acting on the ball when it is within the frictionless channel at position 'Q'.
n Q?
Answer:
2 and 4
Explanation:
Since it is at a position Q it is stationary ball, for a body to be at a point two opposite forces must act on it
How much kinetic energy is in a runner weighing 83 kg moving at a velocity of 16 m/s
Answer:
10,624J
Explanation:
KE = 1/2 m*v²
KE = 1/2 83*16²
KE = 41.5*256
KE = 10,624J
Calculate the magnitude of the electric field at the center of a square with sides 24.1 cm long if the corners, taken in rotation, have charges of
1.16C,
2.32μC,
3.48μC
4.64μC
(all positive).
Answer:
The net electric field = 1016.96 x 10 raised to power 3 N/C
Explanation:
The detailed steps and mathematical explanation is as shown in the attached file.
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb?
Explanation:
First we will convert the given mass from lb to kg as follows.
157 lb = [tex]157 lb \times \frac{1 kg}{2.2046 lb}[/tex]
= 71.215 kg
Now, mass of caffeine required for a person of that mass at the LD50 is as follows.
[tex]180 \frac{mg}{kg} \times 71.215 kg[/tex]
= 12818.7 mg
Convert the % of (w/w) into % (w/v) as follows.
0.65% (w/w) = [tex]\frac{0.65 g}{100 g}[/tex]
= [tex]\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}[/tex]
= [tex]\frac{0.65 g}{100 ml}[/tex]
Therefore, calculate the volume which contains the amount of caffeine as follows.
12818.7 mg = 12.8187 g = [tex]\frac{12.8187 g}{\frac{0.65 g}{100 ml}}[/tex]
= 1972 ml
Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.
Sarah wanted to find out if temperature has an effect on the growth of bread mold. She grew the mold in nine Petri dishes containing the same amount and type of nutrients. Three were kept at 0 C, three were kept at 90 C, and three were kept at room temperature, 27 C. The containers were examined and the growth of the bread mold was recorded each Friday for five weeks. The dependent or responding variable is:
Answer:
Growth of Bread Mold
Explanation:
The controlled variables are the type of containers used and also the different temperatures.
In Sarah's experiment about the influence of temperature on bread mold, the dependent or responding variable is the growth of bread mold. Her experiment is structured to determine if the mold's growth changes in response to different temperatures.
Explanation:In Sarah's experiment, the dependent or responding variable is the growth of the bread mold. The dependent variable is what changes or responds based on the changes in the independent variable. In this case, Sarah is checking to see if different temperatures (the independent variable) cause different rates of bread mold growth (the dependent variable). An experiment like this can demonstrate how environmental factors, like temperature, affect living organisms such as mold.
Learn more about Dependent Variable here:
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1000 kg of water initially at 6 m/s runs through a hydro-generator. If the water leaves the generator at velocity of 4 m/s, and the efficiency of the generator is 100%, how much electricity in kJ (2 significant figures) will we get?
Answer:
8.00 kJ
Explanation:
The first thing is to determine what quantities are there.
the mass of water = 1 000 kg
initial velocity, u = 6 m/s
final velocity, v = 4 m/s
the generator is operating at 100 % efficiency, so there is no energy loss.
The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.
The kinetic energy is calculated as follows:
Ek = 1/2 mv²
= 1/2×(1 000)× (4)²
= 8 000 J/s
= 8.00 kJ Ans
In which mechanical test is a specimen deformed with a gradually increasing load that is applied uniaxially along the long axis of the specimen that contracts the specimen, i.e. the specimen is squeezed?
Answer:
Compression Test
Explanation:
The Specimen is undergoing a compression test. It is similar to tensile test with the difference that the force is compressive and applied along the direction of stress. Both Tensile and compression tests are performed on Universal Testing machine. Compression test is done to determine the product's reaction when it is compressed, squashed and crushed.
The secondary line-to-line voltage of a three-phase Y-Y-connected transformer bank is 115 VAC. What would be the secondary line to neutral voltage?
A. 199.2 V
B. 66.4 V
C. 84.2 V
D. 50 V
Answer:
B. 66.4 V
Explanation:
for a Y-Y connection (star connection), the relationship between the line voltages and the phase voltages is governed by the formula:
[tex]V_{L} = \sqrt{3} (V_{ph})[/tex]
Where [tex]V_{L}[/tex] = Line Voltage = 115 V (AC there just shows that the current is alternating current )
and [tex]V_{ph}[/tex] = Phase Voltage
⇒ [tex]V_{ph}[/tex] = [tex]V_{L}[/tex] / [tex]\sqrt{3}[/tex]
[tex]V_{ph}[/tex] = 115 / [tex]\sqrt{3}[/tex]
[tex]V_{ph}[/tex] = 66.4 V
Hence the secondary Line to neutral voltage is 66.4 Volts
The secondary line to neutral voltage is Option B 66.4 V.
Given that,
The secondary line-to-line voltage of a three-phase Y-Y-connected transformer bank is 115 VAC.Based on the above information, the calculation is as follows:
[tex]V_{ph} = V_L \div \sqrt{3}\\\\= 115 \div \sqrt{3}[/tex]
= 66.4 V
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If the horizontal circular path the riders follow has a 7.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 2.25 times that due to gravity?
Explanation:
Radius , r = 7 m
Centripetal acceleration, a = 2.25 g = 22.07 m/s²
We have equation for centripetal acceleration
[tex]a=\frac{v^2}{r}[/tex]
Substituting
[tex]a=\frac{v^2}{r}\\\\22.07=\frac{v^2}{7}\\\\v=12.43m/s[/tex]
We have equation for frequency
[tex]f=\frac{2\pi }{\omega }\\\\f=\frac{2\pi r}{v}\\\\f=\frac{2\pi \times 7}{12.43}=3.54Hz[/tex]
Revolutions per minute = 3.54 x 60 = 212.4
Revolutions per minute of rider is 212.4
Four charges are the corners of a square with B and C on opposite corners. Charges A and D on the other two corners have equal charge while both B and C have a charge of +1.0 C. What is the charge on A so that the force B is zero?
Final answer:
The charge on A should be -2.0 C in order for the force on B to be zero.
Explanation:
In order for the force on charge B to be zero, the charge on A must be equal in magnitude but opposite in sign to the combined charge of charges C and D at the other corners of the square. Since both B and C have a charge of +1.0 C, A must have a charge of -2.0 C. This creates a symmetric distribution of charges, resulting in a net force of zero on charge B.
A tractor drives down highway 71 at a steady 12 km/hr. A Subaru WRX passes it at a steady 96 km/hr (possibly with a police car following it). Which vehicle has the smallest net force on it? (assume that the road at this point is both level and straight)
Answer:
neither one has a net force on it
Explanation:
Both vehicles are moving at steady velocity hence the acceleration zero.
since acceleration is zero , the net force on it is zero. therefore neither one has a net force on it
If a positive test charge is placed in an electric field, what is the direction of the force on the test charge?
Explanation:
If a positive test charge is placed in an electric field, it will exert the force in the test charge in the direction of electric field vector. We know that the direction of electric field is given by electric field lines. The field lines for a positive charge is outwards. The electric force acting on the charge is given by :
F = q E
Hence, this is the required solution.
Final answer:
A positive test charge placed in an electric field will experience a force in the direction of the field lines, being repelled by positive charges and attracted to negative charges.
Explanation:
When a positive test charge is placed in an electric field, it experiences a force. This force is in the direction of the electric field lines. So, a positive test charge will be repelled by a positive charge and attracted to a negative charge, consistent with Coulomb's law, which states that like charges repel and unlike charges attract. Additionally, if a positive test charge is placed near two negative charges, it will be attracted towards both, as the electric field lines point towards negative charges.
The electric field vectors point in the direction that a positive test charge would move, which is away from positive charges and towards negative charges. By studying various positions of a test charge and the forces experienced, we can map out these vectors to visualize the electric field.
At what angle with the ground should a projectile be thrown to reach the maximum distance? a.) 0° with the ground b.) 45° with the ground c.) 62° with the ground d.) 82° with the ground
Answer:
B
Explanation:
With projectiles, the maximum range is always achieved at a 45 degree launch angle.
the measure of each exterior angle of a regular pentagon is ___ the measure of each exterior angle of a regular nonagon
Answer:
(a) 72°
(b) 40°
Explanation:
PENTAGON
First, we calculate the total angles in a Pentagon using:
180(n - 2)
Where n = number of sides of the polygon, in this case, 5.
Hence, the total angle in a polygon is
180(5 - 2) = 180 * 3 = 540°
Therefore, each angle will be:
540°/5 = 108°
Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular pentagon will be:
180 - 108 = 72°
The exterior angle of a regular Pentagon is 72°
NONAGON
First, we calculate the total angles in a Nonagon using:
180(n - 2)
Where n = number of sides of the polygon, in this case, 9.
Hence, the total angle in a polygon is
180(9 - 2) = 180 * 7 = 1260°
Therefore, each angle will be:
1260°/9 = 140°
Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular nonagon will be:
180 - 140 = 40°
The exterior angle of a regular Nonagon is 40°
Final answer:
Each exterior angle of a regular pentagon, which measures 72 degrees, is greater than each exterior angle of a regular nonagon, which measures 40 degrees, because the sum of the exterior angles of any polygon is always 360 degrees.
Explanation:
The question asks how the measure of each exterior angle of a regular pentagon compares to the measure of each exterior angle of a regular nonagon. To answer this, first, we need to understand that the sum of the exterior angles of any polygon is always 360 degrees, no matter the number of sides. Therefore, to find the measure of each exterior angle of a polygon, we simply divide 360 degrees by the number of sides.
For a pentagon, which has five sides, each exterior angle has a measure of 360 ÷ 5 = 72 degrees. For a nonagon, which has nine sides, each exterior angle has a measure of 360 ÷ 9 = 40 degrees. Therefore, the measure of each exterior angle of a regular pentagon is greater than the measure of each exterior angle of a regular nonagon.
You take a bucket outside and hold it as you normally would to hold water. As the rain comes down, the flux of water out of the bucket is Incorrect: Your answer is incorrect. . If you were to then cut out the bottom of the bucket and again hold it in the rain, the flux of water out of the bucket would be (postive, negative, zero) ?
Answer:
the bucket-surface area vector and the direction of the rain is 180° which gives a negative flux in the outward direction.If we cut the base of the bucket then hold it in the rain then the there is no area intercepted by the rainfall and hence we have a zero flux.Explanation:
We know that flux is defined as the density of any energy field passing through a given area parallel to the area vector.
Mathematically:
[tex]\phi=\bar E.A\cos\theta[/tex]
where:
[tex]\phi=[/tex] flux through the area
[tex]A=[/tex] area of concern
[tex]\bar E=[/tex] density of energy field per unit area
[tex]\theta=[/tex] angle between the energy field lines and the normal to area.
When we hold a bucket in the rain then assuming that the base of the bucket is normal to the direction of rainfall then the angle between the area vector and the rain drop is zero when observed from inside.
For the flux outside the bucket we are concerned with the outer surface of the bucket, now the angle formed between the bucket-surface area vector and the direction of the rain is 180° which gives a negative flux in the outward direction.If we cut the base of the bucket then hold it in the rain then the there is no area intercepted by the rainfall and hence we have a zero flux.The angle between the raindrops and the surface area of the bucket is 180 degrees. Hence the flux will be negative and in the outward direction of the bucket.
The value of flux will be zero when the base of the bucket has been cut and then hold in rain.
Electric fluxThe electric flux can be defined as the property of an electric field that shows the number of electric field lines intersecting a given area.
The electric flux can be written in an equation form as given below.
[tex]\phi = EA cos\theta[/tex]
Where [tex]\phi[/tex] is the electric flux, E is the electric field, A is the area of surface and [tex]\theta[/tex] is the angle between the electric field lines and the normal (perpendicular) to A.
Let us consider that the base of the bucket is normal to the direction of the rainfall when we hold the bucket in rain. In this case, the angle between the raindrops and the normal to the surface will be zero inside the bucket.
From the outer surface, the angle formed between the surface area of the bucket and the direction of the rain is 180 degrees. When we put the value of the angle in the formula of flux, we will get a negative flux for the angle of 180 degrees in the outward direction of the bucket.
The flux depends on the surface area. If we cut the base of the bucket and hold it in rain then there will be no surface area that is intersecting by the rainfall. Hence the value of flux will be zero for the no surface area of the bucket.
To know more about the flux, follow the link given below.
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Which of the following describes how net work is determined?
A. The difference in the initial and final forces applied
B. The net force on an object applied for a brief second
C. The initial force applied to an object over a distance
D. The net force on an object applied over a distance
Answer:
D. The net force on an object applied over a distance
Explanation:
Work can be defined as the product of force multiplied by a distance
Where
F = force [Newtons]
d = distance = [meters]
Therefore:
Work = F * d
Work in units of Joules [J]
If the half-life of an element is about 5,000 years, then a fossil that has one-eighth the normal proportion of the parent isotope to the daughter isotope should be about how many years old?
Answer : The time passed in years is [tex]1.49\times 10^4\text{ years}[/tex]
Explanation :
Half-life = 5000 years
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{5000\text{ years}}[/tex]
[tex]k=1.39\times 10^{-4}\text{ years}^{-1}[/tex]
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]1.39\times 10^{-4}\text{ years}^{-1}[/tex]
t = time passed by the sample = ?
a = let initial amount of the reactant = X
a - x = amount left after decay process = [tex]\frac{X}{8}[/tex]
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{1.39\times 10^{-4}}\log\frac{X}{(\frac{X}{8})}[/tex]
[tex]t=14962.706\text{ years}=1.49\times 10^4\text{ years}[/tex]
Therefore, the time passed in years is [tex]1.49\times 10^4\text{ years}[/tex]
0 mph and half the distance at 69.0 mph . On her return trip, she drives half the time at 31.0 mph and half the time at 69.0 mph .
Part AWhat is Julie's average speed on the way to Grandmother's house?
Express your answer with the appropriate units.
Part BWhat is her average speed on the return trip?
Express your answer with the appropriate units.
Answer:
a) v = 34.5 mph, b) v = 50 mph
Explanation:
The average speed is defined.
v = (x₂ -x₁) / (t₂ -t₁)
If the movement is with uniform speed we can also obtain the average speed, averaging the speeds with a time weight factor, if the time is equal for each speed
v = (v₁ + v₂) / 2
a) For the outward journey
v = (69.0 +0) / 2
v = 34.5 mph
b) For the return trip
v = (69.0 + 31.0) / 2
v = 50 mph
Which style of jazz music released performers from traditional rules and placed importance on each performer contributing his or her own feelings for the music in the moment? Question 9 options: hard bop jazz fusion free jazz emo jazz
Answer:
The correct answer is: Free Jazz.
Explanation:
Free jazz is a music genre developed around the year 1964 and is considered a subgenre of Jazz where musicians are asked to break the traditional conventions related to the interpretation of jazz music.
In free jazz, the most important technique is improvisation, where musicians simply perform and play their instruments freely.
The key to performing free jazz is to try to express spontaneous feelings through the instrument, while not being aware of certain conventions traditional to jazz music.
In this particular case, to the question Which style of jazz music released performers from traditional rules and placed importance on each performer contributing his or her own feelings for the music in the moment, the correct answer is: Free Jazz.
An airplane is flying at 95.0 m/s when the engine is turned off. As it glides, it experiences a force of drag so that is slows down at -0.700 m/s2. How far in meters will it travel in 1.00 minute ?
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 95 m/s
Acceleration, a = -0.700 m/s²
Time, t = 1 minute = 60 seconds
Substituting
s = ut + 0.5 at²
s = 95 x 60 + 0.5 x -0.700 x 60²
s = 4440 m
Distance traveled is 4440 m
The airplane travels 4440 m in 1 minute
To find the distance the airplane will travel in 1.00 minute, use the equation d = [tex]vi * t + (1/2) * a * t^2[/tex]e, vi is the initial velocity, t is the time, and a is the acceleration. Substitute the given values into the equation and solve to find the distance.
Explanation:To find the distance the airplane will travel in 1.00 minute, we need to use the equation:
d = vi * t +[tex](1/2) * a * t^2[/tex]
Where:
d = distance traveled
vi = initial velocity
t = time
a = acceleration
Using the given values:
vi = 95.0 m/s (initial velocity)
a = -[tex]0.700 m/s^2[/tex] (deceleration)
t = 1.00 minute = 60 seconds
Now we can substitute the values into the equation:
d = (95.0 * 60) + (1/2) * [tex](-0.700) * (60^2)[/tex]
Solving this equation will give us the distance traveled by the airplane in 1.00 minute.
The function s(t) represents the position of an object at time t moving along a line. Suppose s (2 )equals 146and s (6 )equals 254.Find the average velocity of the object over the interval of time [2 comma 6 ].
Answer:
v(t) = 27 units
Explanation:
The function s(t) represents the position of an object at time t moving along a line such that,
[tex]s(2)=146[/tex]
and
[tex]s(6)=254[/tex]
We need to find the average velocity of the object over the interval of time [2,6]. The velocity of the object is equal to the total distance divided by time. It is given by :
[tex]v(t)=\dfrac{s(6)-s(2)}{6-2}[/tex]
[tex]v(t)=\dfrac{254-146}{6-2}[/tex]
v(t) = 27 units
So, the average velocity of the object is 27 units. Hence, this is the required solution.
Vector has a magnitude of 4.40 m and is directed east. Vector has a magnitude of 3.40 m and is directed 39.0° west of north. What are (a) the magnitude and (b) the direction (counterclockwise from east) of ? What are (c) the magnitude and (d) the direction (counterclockwise from east) of ?
Answer:
Magnitude = 4.056 m
Direction = 42.3⁰
Explanation:
The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.
The vector 4.40 is directed East. This automatically becomes a horizontal component.
But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.
Resolving the vectors should yield the horizontal and vertical components:
Horizontal components
The first component is 4.40 m
The second one is derived by resolving 3.40 to the horizontal like this 3.40 × - cos 61⁰ = -1.648 m
Adding the horizontal component gives 4.40 m + ( -1.648 m) = 2.752 m
Vertical components
Resolve 3.40 with the angle 61⁰ like this: vertical comp = 3.41 × sin 61
= 2.98 m
The magnitude is given by √[(2.98)²+ (2.752)²] = 4.056 m Ans
The direction us given by tan⁻¹ (2.98/2.752) = 42.3⁰ Ans
An 8.6 m, 267 kg uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is 0.53 , and the ladder makes a 32.7 ◦ angle with the ground. How far up the ladder can a 1068 kg person climb before the ladder begins to slip? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Answer:
[tex]r_{P}=2.58 m[/tex]
Explanation:
If we want to find how far up the ladder can a person climb before the ladder begins to slip, the total force in the y-direction and the total torque must be zero, it is an equilibrium condition.
So, let's start whit y-direction forces.
[tex]\Sigma F_{y}=0[/tex]
[tex]N_{L}-W_{L}-W_{P}=0[/tex]
NL is the normal forceWL is the weight of the ladderWP is the weight of the personSo, we can find the normal force:
[tex]N_{L}=m_{L}g+m_{P}g=g(m_{L}+m_{P})=9.8(267+1068)=13083 N[/tex]
We know that the friction force is the product of the coefficient of static friction times the normal force. Let's find it.
[tex]F_{f}=\mu N_{L}=0.53*13083=6934 N[/tex]
Referent to the x-direction forces, we can say that the total sum must be equal to zero too.
[tex]\Sigma F_{x}=0[/tex]
[tex] -N_{wall}+F_{f}=0[/tex]
[tex] N_{wall}=F_{f}=6934 N[/tex]
Where N(wall) is the normal force due to the wall. (negative direction)Now, as we said before, the total torque must be zero, so we have:
[tex]\Sum \tau=\tau_{wall}+\tau_{L}+\tau_{P}=0[/tex] (1)
We will choose the pivot point at the bottom of the ladder, then we have:
Each force acting on the ladder must be perpendicular to the ladder, so we need to use the components.
[tex]\tau_{wall}=r_{wall}*N_{wall}sin(32.7)=8.6*6934*sin(32.7)=32215.8Nm[/tex] (positive due to the counterclockwise rotation)
[tex]\tau_{L}=r_{L}*W_{L}sin(57.3)=(8.6/2)*2616.6*sin(57.3)=-9468.2 Nm[/tex] (negative due to the clockwise rotation)
[tex]\tau_{P}=r_{P}*W_{P}sin(57.3)=r_{P}*10466.4*sin(57.3)=-r_{P}8807.6 Nm[/tex] (negative due to the clockwise rotation)
We use 57.3° because is the angle between the ladder weight vector and the ladder surface, we use the same angle to the person.
Now we can use (1) to find the distance
[tex]\Sum \tau=\tau_{wall}+\tau_{L}+\tau_{P}=32215.8-9468.2-r_{P}8807.6=0[/tex]
Therefore [tex]r_{P}=2.58 m[/tex]
I hope it helps you!
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.95 m/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 39.9 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 6.50 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.
Part A. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
Express your answer using three significant figures.
Part B. Where is Henrietta when she catches the bagels?
Express your answer using three significant figures.
Answer:
Part A 12.9 m/s
Part B 36.9 m
Explanation:
Henrietta ran at 3.95 m/s. She was 6.5 seconds away from from the window when Bruce throws the bagel. This equates to a distance of d
[tex]d= 3.95\times6.5=25.675[/tex] (distance = speed x time)
If Bruce throws the bagel horizontally, the vertical component of the motion is purely under gravity with initial speed of 0 m/s. Using one of the equations of motion for the vertical motion,
[tex]h=ut+0.5at^2[/tex]
h = 39.9 m (height of the building)
a = 9.8 (acceleration of gravity)
[tex]t=\sqrt{\dfrac{2\times39.9}{9.8}}= 2.85[/tex]
During this time, Henrietta will have travelled another distance e
[tex]e =3.95\times2.85=11.2575[/tex]Therefore, Henrietta would have jogged a total distance of 25.675 + 11.2575 m = 36.9325 m.
The horizontal speed with which Bruce must throw the bagel is the total distance travelled by Henrietta divided by the time it takes to land vertically i.e.
[tex]v=\dfrac{36.9325}{2.85}=12.96[/tex]
Therefore, the speed of throw is 13.0 m/s.
Henrietta was 36.9 m from the window when she caught the bagel.
Part A. The speed at which the bagel was thrown horizontally is 12.9 m/s.
Part B. The total distance traveled by Henrietta when she caught the bagel from the window is 36.9325 m.
How do you calculate the speed of the bagel?Given that, the speed of Henrietta is 3.95 m/s. When Bruce throws the bagel, the time difference between both of them is 6.5 seconds. Hence the distance can be calculated as given below.
Distance D = Speed [tex]\times [/tex] Time
[tex]D = 3.95 \times 6.50[/tex]
[tex]D = 25.675 \;\rm m[/tex]
Part A
Given that Bruce throws the bagel horizontally, then during the motion, the initial speed will be zero and the vertical component will have gravitational acceleration. In this case, the time required in motion can be calculated by the equation given below.
[tex]s = ut + \dfrac {1}{2} at^2[/tex]
Where, s is the distance, a is the acceleration, t is time and u is the initial speed.
For the horizontal motion, s = 39.9 m, g = 9.8 m/s^2 and u = 0 m/s. The time will be,
[tex]39.9 = 0 + \dfrac {1}{2}\times 9.8\times t^2[/tex]
[tex]t = 2.85 \;\rm s[/tex]
The time required to reach the bagels is 2.85 seconds. But in this time, the distance traveled by Henrietta is given below.
[tex]D' = 2.85 \times 3.95[/tex]
[tex]D' = 11.2527 \;\rm m[/tex]
Hence the total distance traveled by Henrietta is given below.
Distance = 25.675 + 11.2527
Distance = 36.9325 m.
The speed of the bagel is calculated as given below.
Speed = Total Distance / Time
[tex]v = \dfrac {36.9325 }{2.85}[/tex]
[tex]v = 12.9 \;\rm m/s[/tex]
Hence we can conclude that the speed at which the bagel was thrown horizontally is 12.9 m/s.
Part B
The total distance traveled by Henrietta is 36.9325 m. At this distance, Henrietta catches the bagel from the window.
To know more about the speed, follow the link given below.
https://brainly.com/question/7359669.