Answers
Free diagram of the crate:
The free body equation for the crate is given as,
[tex]F=mg\sin \theta[/tex]Here, F is the force applied, m is the mass of the crate, g is the acceleration due to gravity and θ is the angle of inclination.
Substituting all known values,
[tex]\begin{gathered} F=20\text{ kg}\times9.8\text{ m/s}^2\times\sin (15^{\circ}) \\ =50.7\text{ N} \end{gathered}[/tex]T
Related Questions
One closed organ pipe has a length of 2.24 meters. When a second pipe is played at the same time, a beat note with a frequency of 1.1 hertz is heard. By how much is the second pipe too long? Include units in your answer.
Answers
ANSWER
0.07 m
EXPLANATION
First, we have to find the frequency that the first pipe would play,
[tex]\lambda=4L=4\cdot2.24m=8.96m[/tex]The frequency is,
[tex]f=\frac{v}{\lambda}=\frac{343m/s}{8.96m}=38.28125Hz[/tex]The frequency played by the second pipe is,
[tex]f=38.25125Hz-1.1Hz=37.18125Hz[/tex]The wavelength of this note is,
[tex]\lambda=\frac{v}{f}=\frac{343m/s}{37.18125s^{-1}}\approx9.2251m[/tex]So the length of the second pipe is,
[tex]L=\frac{\lambda}{4}=\frac{9.2251m}{4}\approx2.31m[/tex]The difference between the pipes' length is,
[tex]2.31m-2.24m=0.07m[/tex]Hence, the second pipe is 0.07 meters too long
27. Write a number in scientific notation with4 sig figs, and 2 zeros
Answers
Answer:
1.600 x 10³
Explanation:
The numbers in scientific notation are multiplied by a power of 10, additionally, if there is a decimal point, all the numbers including zeros are significant figures. Therefore, a number in scientific notation with 4 significant figures and 2 zeros is:
1.600 x 10³
The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The complete set has a mass of 3,548 kg. Suppose the entire publication is placed on a cart that can move without friction. The cart is at rest, and a librarian is sitting on top of it, just having loaded the last volume. The librarian jumps off the cart with a horizontal velocity relative to the floor of 2.65 m/s to the right. The cart begins to roll to the left at a speed of 0.03 m/s. Assuming the cart’s mass is negligible, what is the librarian’s mass? Round to the hundredths.
Answers
ANSWER:
40.17 kg
STEP-BY-STEP EXPLANATION:
Given:
m1 = 3548 kg
V1 = -0.03 m/s
V2 = 2.65 m/s
We apply law of conservation of linear momentum:
[tex]\begin{gathered} P_i=P_f \\ P_i=0 \\ P_f=m_1\cdot V_1+m_2\cdot V_2 \end{gathered}[/tex]We replace and calculate the mass as follows:
[tex]\begin{gathered} m_1\cdot V_1+m_2\cdot V_2=0 \\ 3548\cdot-0.03+2.65\cdot m_2=0 \\ -106.44+2.65\cdot m_2=0 \\ m_2=\frac{106.44}{2.65} \\ m_2=40.17\text{ kg} \end{gathered}[/tex]Therefore the mass is equal to 40.17 kg
During a hockey game, two hockey players (m1 = 82kg and m2 = 76kg) collide head on in a 1 dimensional perfectly instant collision. If the first hockey player is moving to the left at a velocity 4.2 m/s and the second hockey player is moving in the opposite direction at a velocity of 3.4 m/s, how fast are they both moving after they collide, assuming they stick together after they collide? How much kinetic energy is lost as a result of the collision?
Answers
Given:
The mass of player 1 is m1 = 82 kg
The initial velocity of player 1 is
[tex]v_{i1}=\text{ 4.2 m/s}[/tex]towards left.
The mass of player is m2 = 76 kg
The initial velocity of player 2 is
[tex]v_{i2}=\text{ -3.4 m/s}[/tex]in opposite direction.
Required:
The final velocity after the collision assuming they stick together.
The loss of kinetic energy after collision.
Explanation:
According to the conservation of momentum, the final velocity will be
[tex]\begin{gathered} m1v_{i1}+m2v_{i2}=(m1+m2)v_f \\ v_f=\frac{m1v_{i1}+m2v_{i2}}{m1+m2} \\ =\text{ }\frac{(82\times4.2)-(76\times3.4)}{82+76} \\ =0.544\text{ m/s} \end{gathered}[/tex]The loss in kinetic energy will be
[tex]\begin{gathered} \Delta KE=KE_{after}-KE_{before} \\ =\frac{1}{2}(m1+m2)(v_f)^2-\frac{1}{2}m1(v_{i1})^2-\frac{1}{2}m2(v_{i2})^2 \\ =\frac{1}{2}(82+76)(0.544)^2-\frac{1}{2}\times82\times(4.2)^2-\frac{1}{2}\times76\times(3.4)^2 \\ =23.38-723.24-439.28 \\ =-1139.14\text{ J} \end{gathered}[/tex]Final Answer:
The final velocity after the collision assuming they stick together is 0.544 m/s.
The loss of kinetic energy after the collision is 1139.14 J.
i would like some help with this problem, i already tried to complete it on my own but would like to check my answer
Answers
Given:
Two circuits with multiple resistors
To find:
The net resistance of the circuits
Explanation:
For the first circuit
The equivalent series resistance of the 1 and 2 is,
[tex]\begin{gathered} R_1+R_2 \\ =20+16 \\ =36\text{ ohm} \end{gathered}[/tex]The equivalent series resistance of the 4 and 5 is,
[tex]\begin{gathered} 10+14 \\ =24\text{ ohm} \end{gathered}[/tex]The net resistance of the resistances is,
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{36}+\frac{1}{24}+\frac{1}{12} \\ \frac{1}{R}=\frac{11}{72} \\ R=\frac{72}{11}\text{ohm} \\ R=6.54\text{ ohm} \end{gathered}[/tex]Hence, the net resistance of the upper circuit is 6.54 ohms.
For the circuit below:
[tex]R_2,\text{ R}_3,\text{ R}_4\text{ are in series}[/tex]So, the equivalent resistance is,
[tex]\begin{gathered} R_2+R_3+R_4 \\ =2+2+2 \\ =6\text{ ohm} \end{gathered}[/tex]This equivalent resistance in parallel with
[tex]R_5[/tex]So, the equivalent resistance is,
[tex]\begin{gathered} \frac{6\times2}{6+2} \\ =1.5\text{ ohm} \end{gathered}[/tex]Now 1.5 ohm is in series with the rest two resistances.
So, the net resistance is,
[tex]\begin{gathered} R_1+R_6+1.5 \\ =2+2+1.5 \\ =5.5\text{ ohm} \end{gathered}[/tex]Hence, the net resistance of the circuit below is 5.5 ohms.
What will be the effect of multiplying negative scalar number with a vector.. Explain with the help of drawing.
Answers
When we multiply a vector by a scalar number, each dimension of the vector will be multiplied by the number.
For example, multiplying the vector <3, 4> by 2 would create the vector <6, 8>.
But if this scalar is negative, the dimensions of the vector will change the signal:
vector <3, 4> multiplied by -2 creates the vector <-6, -8>.
This causes the original vector to change its direction to the opposite direction, that is, it flips 180°:
Please answer the questions
Answers
Answer:
Where's the question? So that i can answer it!
Question 2
An object has a momentum of 500 kg mls. If its mass is 20 kg, its speed must be
O 25 m/sO 20 m/s50 m/sO 500
Answers
Given:
The momentum of the object is p = 500 kg m/s
The mass of the body is m = 20 kg
To find the speed of the object.
Explanation:
Speed can be calculated by the formula
[tex]\begin{gathered} p=mv \\ v=\frac{p}{m} \end{gathered}[/tex]On substituting the values, the speed of the object will be
[tex]\begin{gathered} v=\frac{500}{20} \\ =25\text{ m/s} \end{gathered}[/tex]Thus, the speed of the object is 25 m/s
Christie and Daisy measured the applied forces while pulling on opposite ends of a rope, but they forgot to fill in their table ofresults shown below. Predict the most likely values that belong in Box C and Box D.Scale 1 (Christie) Scale 2 (Daisy) Observed Effect on Motion2 kg2 kgBox A3 kgBox BRope moved toward DaisyBox CBox DRope moved toward Christie(1 point)O Box C 4 kg; Box D: 4 kgO Box C: 3 kg, Box D: 3 kgO Box C: 4 kg, Box D: 3 kgO Box C: 3 kg; Box D: 4 kg
Answers
To find:
Most likely values that belong in box C and box D.
Explanation:
From Newton's second law of motion, the net force acting on an object is equal to the product of the mass of the object and the acceleration of the object produced due to the net force.
That is,
[tex]\vec{F}=m\vec{a}[/tex]Thus the object will accelerate in the direction of the net force.
It is given that in the third trail, the box moves towards Christie. Thus the net force is directed towards the Christie. Thus the force applied by Christie must be greater than the force applied by Daisy.
Final answer:
Thus the correct answer is option C.
a student who weighs 556 Newtons climbs the stairway (vertical height of 4.0 m) in 25 s (a) how much work is done (b) what is the power output of the student
Answers
Given data:
Weight of the student;
[tex]\begin{gathered} F=mg \\ =556\text{ N} \end{gathered}[/tex]Height;
[tex]h=4.0\text{ m}[/tex]Time;
[tex]t=25\text{ s}[/tex]Part (a)
The work done by the student is given as;
[tex]W=Fh[/tex]Substituting all known values,
[tex]\begin{gathered} W=(556\text{ N})\times(4.0\text{ m}) \\ =2224\text{ J} \end{gathered}[/tex]Therefore, the work done by the student is 2224 J.
Part (b)
The power is defined as the rate of doing work. Mathematically,
[tex]P=\frac{W}{t}[/tex]Substituting alll known values,
[tex]\begin{gathered} P=\frac{2224\text{ J}}{25\text{ s}} \\ =88.96\text{ W} \end{gathered}[/tex]Therefore, the power output of the student is 88.96 W.
a clay ball (0.65kg) is thrown at a wall and sticks. The speed of the ball before hitting the wall was 15 m/s. It took 45 miliseconds for the clay to come to a stop and make contact with the wall. What was the average force the wall exerted on the wall during the collision?
Answers
In order to determine the average force the wall exerted on the ball, use the following formula:
[tex]F=\frac{\Delta p}{\Delta t}[/tex]where
Δp: change in momentum
Δt: change in time = 45 ms = 0.045 s
F: force = ?
calculate the change in momentum as follow:
[tex]\Delta p=m\Delta v=(0.65kg)(15\frac{m}{s}-0\frac{m}{s})=9.75\text{ kg}\cdot\text{m/s}[/tex]next, replace Δp and Δt into the formula for F:
[tex]F=\frac{9.75\text{ kgm/s}}{0.045s}=216.67N[/tex]Hence, the average force exterded by the wall on the ball was approximately 216.67N
Which of the following is the type of force that pulls objects towards the center of motion?A. Inertia B. MomentumC. AccelerationD. Centripetal
Answers
When an object is in a circular motion, the acceleration responsible for changing the direction of the movement is known as the centripetal acceleration:
This acceleration points towards the center of the circular motion.
Using the second law of newton, we can find the centripetal force Fc:
[tex]F_c=m*a_c[/tex]Therefore the force that pulls objects towards the center of motion is the centripetal force.
Correct option: D.
Which element is a good material for use as a semiconductor?copperoxygensiliconsodium
Answers
To find:
Which of the given elements is a good material for semiconductor?
Explanation:
The silicon is widely used for the manufacturing of semiconductors. The stable structure of the silicon makes it a suitable candidate for the manufacture of semiconductors. The silicon is widely abundant in nature. This makes it economical.
Which of the following is NOT a characteristic of average speed?elloa. Average speed includes both the rate of motion and its direction.b. Average speed is the rate of motion.C. Average speed is always a positive value.d. The greater the speed of an object the faster it moves.
Answers
The average speed is defined as the total distance traveled by an object divided into the time that it takes for the object to travel that distance.
Average speed is a scalar quantity, which means that it can be represented using a number with no need of specifying a direction. Since the total distance is a positive value, then average speed is always also a positive value.
From the options, the statement "aveage speed includes both rate of motion and its direction" is false.
Therefore, the answer is: Option A.
What is the escape speed (in km/s) from an Earth-like planet with mass 4.9e+24 kg and radius 70.0 × 105 m? Use the gravitational constant G = 6.67 × 10-11 m3 kg-1 s-2.
Answers
The escape velocity = 9.66 km/s
Explanation:The mass of the planet is represented by m
[tex]m=4.9\times10^{24}\operatorname{kg}[/tex]The radius is represented by r
[tex]r=70.0\times10^5m[/tex]The gravitational constant is represented by G
[tex]G=6.67\times10^{-11}m^3kg^{-1}s^{-2}[/tex]The escape velocity (v) is given by the formula:
[tex]\begin{gathered} v=\sqrt[]{\frac{2GM}{r}} \\ v=\sqrt[]{\frac{2\times6.67\times10^{-11}\times4.9\times10^{24}}{70\times10^5}} \\ v=\sqrt[]{\frac{65.366\times10^{13}}{70\times10^5}} \\ v=\sqrt[]{0.9338\times10^8} \\ v=\sqrt[]{93380000} \\ v=9663.33\text{ m/s} \\ v=9.66\text{ km/s} \end{gathered}[/tex]The escape velocity = 9.66 km/s
Consider a container filled with coconut oil having a density of 0.903 g/ml, What is the gauge pressure (Pa) 6.35 cm below the surface? 1ml = 1cm^3, 1000g = 1kg
Answers
The gauge pressure is 562.5 Pa.
Given data:
The density of oil is ρ=0.903 g/ml.
The distance below the surface is h=6.35 cm.
The density in kg/cm³ will be,
[tex]\begin{gathered} \rho=0.903\text{ g/ml }\times\frac{1\text{kg}}{1000\text{ g}}\times\frac{1\text{ml}}{1\text{ }cm^3}\times\frac{10^6\text{ }cm^3}{1m^3} \\ \rho=\frac{903kg}{m^3} \end{gathered}[/tex]The gauge pressure can be calculated as,
[tex]\begin{gathered} p=\rho gh \\ p=(\frac{903kg}{m^3})(\frac{9.81m}{s^2})(6.35cm\times\frac{1\text{ m}}{100cm}) \\ p=\frac{562.5kg}{ms^2}\times\frac{1\text{ Pa}}{\frac{kg}{ms^2}} \\ p=562.5\text{ Pa} \end{gathered}[/tex]Thus, the gauge pressure is 562.5 Pa.
The spectral classification for spica is B1V. This means it is A. A red giant star since B1 means it is very large B. A whit dwarf star since it is B1,hence it lies along the left edge of the HR for blue C.a blue giant star since it is B1, whee B stands for blue D. A main sequence star as indicated by V
Answers
The given spectral classification for spica is B1V.
B1 means it is B type star which have very high temperature and high mass. Due to its high temperature it will emit the light of lower wavelength which corresponds to blue color.
Thus, B1V is a white dwarf star since it is B1, hence it lies along the left edge of HR for blue which means option (B) is correct.
At the bottom of the first hill, the 300 kg car is traveling 39.6 m/s. What is the kinetic energy of the car at this point? (ignore friction forces)
Answers
ANSWER:
1st option: 235,000 J
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 300 kg
Speed (v) = 39.6 m/s
We can calculate the kinetic energy using the following formula:
[tex]\begin{gathered} E_K=\frac{1}{2}mv^2 \\ \text{ We replacing:} \\ E_K=\frac{1}{2}\cdot300\cdot39.6^2 \\ E_K=235224\cong235000\text{ J} \end{gathered}[/tex]The kinetic energy is 235000 J.
The example of a book falling off of a table shows a(n) _____.1) contact force2) scalar quantity3) absence of acceleration4) field force
Answers
anotherGravity is the force of attraction between two objects. A gravitational force field is modeled as space around a massive body in which other bodies experiences force.
In Newtonian gravity, a particle of mass M creates a gravitational force field around itself given as,
[tex]g=\frac{GM}{r^2}[/tex]Here, G is the universal gravitational constant, and r is the separation between the bodies.
Hence, the example of a book falling off of a table shows a(n) field force. Therefore, option (4) is the correct choice.
The zombies are closing in on him and Mr. Mangan sees a ramp to his left. Hedrives at it and hits the ramp with a velocity of 30 m/s. If it took him 7 s toget to the ramp, what was his acceleration?
Answers
Given data
*The given velocity is v = 30 m/s
*The given time is t = 7 s
The formula for the acceleration is given as
[tex]a=\frac{v}{t}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} a=\frac{30}{7} \\ =4.28m/s^2 \end{gathered}[/tex]Hence, the acceleration is a = 4.28 m/s^2
an ocean moves at 22.9 m/s and has a wavelength of 27.0. with what frequency do the waves pass you?
Answers
Anyone available for teaching me simple pendulum in physics
Answers
Simple pendulum is a device that has a periodic motion.
In periodic motion, the object repeats its path after an interval of time.
The simple pendulum can be drawn as
It has a simple bob connected to a fixed end through a massless string.
Resistances of 2.0 Ω, 4.0 Ω, and 6.0 Ω and a 24-V emf device are all in parallel. The current in
the 2.0-Ω resistor is
Answers
Answer:
The current in the 2.0 Ω resistor is 12 A
Explanation:
Given:
R₁ = 2.0 Ω
R₂ = 4.0 Ω
R₃ = 6.0 Ω
ξ = 24 V
______________
I₁ - ?
With parallel connection:
U₁ = U₂ = U₃ = ξ
Ohm's law:
I₁ = U₁ / R₁ = 24 / 2,0 = 12 A
Resistances of 2.0 Ω, 4.0 Ω, and 6.0 Ω and a 24-V emf device are all in parallel, the current in the 2.0-Ω resistor is 6 A.
The voltage across each resistor in a parallel circuit is the same, while the current divides among the different branches.
The sum of the currents passing through each resistor equals the total current flowing into the parallel circuit.
1/RTotal = 1/R1 + 1/R2 + 1/R3
1/RTotal = 1/2.0 Ω + 1/4.0 Ω + 1/6.0 Ω
1/RTotal = 3/12 Ω
1/RTotal = 1/4 Ω
RTotal = 4 Ω
Now,
I = V / RTotal
I = 24 V / 4 Ω
I = 6 A
Thus, the current in the 2.0-Ω resistor is 6 A.
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Our galaxy, the Milky Way, contains approximately 4.0 x 1011 stars with anaverage mass of 2.0 X 1030 kg each. How far away is the Milky Way from ournearest neighbor, the Andromeda Galaxy, if Andromeda contains roughly thesame number of stars and attracts the Milky Way with a gravitational force of2.4 x 1030 N?
Answers
We will have the following:
First, we remember:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Then, from the problem we will have that:
[tex]2.4\ast10^{30}N=\frac{G(2.0\ast10^{30}kg)(2.0\ast10^{30}kg)}{r^2}[/tex][tex]\Rightarrow r^2=\frac{G(2.0\ast10^{30}kg)^2}{2.4\ast10^{30}N}\Rightarrow r\approx1.1116666667\ast10^{20}m[/tex]So, the Andromeda galaxy is approximately 1.1*10^20 meters from the milky way.
The acceleration of the Andromeda galaxy towards the milky way is:
[tex]2.4\ast10^{30}N=(2.0\ast10^{30}kg)a\Rightarrow a=1.2m/s^2[/tex]So, the acceleration towards the milky ways is 1.2m/s^2.
A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the force of friction is not significant.
Answers
The given problem can be exemplified in the following diagram:
Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:
[tex]\Sigma F=ma[/tex]Replacing the values:
[tex]mg\sin 40=ma[/tex]We may cancel out the mass:
[tex]g\sin 40=a[/tex]Using the gravity constant as 9.8 meters per square second:
[tex]9.8\frac{m}{s^2}\sin 40=a[/tex]Solving the operations:
[tex]6.3\frac{m}{s^2}=a[/tex]Therefore, the acceleration is 6.3 meters per square second.
37.Why is hot water used to loosen a sealed jar lid?Select one:a. The jar contracts when heated.b. The lid expands when heated.c. The lid contracts when heated.d. The jar and the lid both contract.
Answers
Answer:
Explanation:
Heat causes expansion of the lid at a faster rate th
A 5.0n force is applied to a 3.0kg ball to change its velocity from +9.”m/s to +3.0 m/s. The impulse on the ball is _I think it’s -18
Answers
Given,
The force is F=5 N.
The mass is m=3.0 kg
The velocities are 9 m/s to 3 m/s
The impulse is:
[tex]\begin{gathered} I=m(v_f-v_i) \\ \Rightarrow I=3(9-3) \\ \Rightarrow I=\frac{18kgm}{s} \end{gathered}[/tex]The answer is 18 kgm/s
a ray of light hits a flat mirror with an incident angle of 39.20 degrees. What angle does it reflect at? (do not put in units)
Answers
Given
A ray of light hits a flat mirror with an incident angle of 39.20 degrees.
To find
What is the angle of reflection?
Explanation
According to the law of reflection,
The angle of incidence is equal to the angle of reflection.
Here the angle of incidence is 39.20 deg
So the angle of reflection is 39.20 deg
Conclusion
The angle of reflection is 39.20 deg
MULTIPLE CHOICE QUESTION Which word is synonymous to tension? Push Pull
Answers
Pull
Explanation:Tension is a pulling force that is transmitted in a string, rope or cable.
Therefore, tension is synonymous to pull
The diagram below shows an example of an object on which tension is transmitted.
It is a pull because it is tranmitted towards the source rather than away from the source
Find the work W W done by the 30-newton force.
Answers
The work done by the 30 newton force is 4156.8 J
How do I determine the work done?We know that work done is defined according to the following formula:
Work done (Wd) = force (F) × distance (d)
Wd = fd
Where angle projection is involved, the work done is defined as follow::
Wd = FdCosθ
Haven know the formula for calculating work done, we shall thus, determine the work done by the 30 newton force as follow:
Force (F) = 30 newtonAngle (θ) = 30 degreesDistance (d) = 160 mWorkdone (Wd) =?Wd = FdCosθ
Wd = 30 × 160 × Cos 30
Wd = 4800 × 0.8660
Wd = 4156.8 J
Thus, from the calculation made above, we can conclude that the work done is 4156.8 J
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