Answers
Explanation
Step 1
Diagram
A)Friction is force that resists the sliding or rolling of one solid object over another, so the direction of the friction force must be the opposite of the applied force
Frictional force always acts in a direction opposite to the direction of applied force.
so,if the direction of the force is to the right, the direction of the friction force must be to the left
Step 2
Free body diagram:
A free body diagram is a simplified representation in a problem of an object , and the force vectors acting on it,is a graphical illustration used to visualize the applied forces , so
I hope this helps you
Related Questions
Pls answer this is really important!!!
Answers
Answer:
Explanation:
a)
we need to find the horizontal range of the projectile
the formula for horizontal range is x = (u² sin2α)/g
= (255² × sin(2 × 73.7))/9.81
= 3571.2 m
Distance between ship and enemy ship = 2460 + 601 = 3061 m
So the projectile lands (3571.2 - 3061) = 510.2 m away from the enemy ship
b)
The peak is situated at 2460 m from the projectile.
The time required to reach the peak needs to be found out so that we can calculate the vertical displacement in that time.
To reach 2460 m time needed is x = u × cosα × t
⇒ 2460 = 255 × cos(73.7)° × t
⇒ t = 34.37s
Now, vertical displacement after 34.37s is
y = (u × sinα × t) - (1/2 × g × t²)
= (255 × sin(73.7) × 34.37) - (1/2 × 9.81 × (34.37)²)
=2617.81 m
The peak is 1760m high
So the projectile comes (2617.81 - 1760) = 857.8m away from the peak
If you need notes on this, I could send you my pdf on this one cause I'm not sure if I explained why I used which formula
Calculate the speed of sound in air for the following temperatures:A. 0°C B. 25°CC. 30°C D. -15°C
Answers
The speed of sound in air at a given temperature is given by:
[tex]v=331\sqrt[]{1+\frac{T}{273}}[/tex]Then we need to plug the value of the temperature to determine each case.
A.
[tex]v=331\sqrt[]{1+\frac{0}{273}}=331[/tex]B.
[tex]v=331\sqrt[]{1+\frac{25}{273}}=345.82[/tex]C.
[tex]v=331\sqrt[]{1+\frac{30}{273}}=348.71[/tex]D.
[tex]v=331\sqrt[]{1+\frac{(-15)}{273}}=321.78[/tex]Therefore the speeds for the given temperatures are:
A. 331 m/s
B. 345.82 m/s
C. 348.71 m/s
D. 321.78 m/s
hello, i need help with my zearn.. Sebastian must have at least 5 feet of rope to jump rope.Which description best represents the length of rope Sebastian needs?Any value less than or equal to 5Any value equal to 5Any value greater than 5Any value greater than or equal to 5
Answers
Sebastian must have at least 5 feet of rope to jump rope.
At least means greater or equal.
Correct option:
Any value greater than or equal to 5
If an object is placed at twice the focal length of a convex lens, what is the position of the image?1) between the object and the lens and inverted2) between the object and the lens and upright3) behind the lens and inverted4) behind the lens and upright
Answers
When the object is placed at twice the focal length of a convex lens (2F) then the position of the image is behind the lens and inverted.
Graphically,
As you can see from the above diagram, the image is formed behind the lens and inverted.
Therefore, the correct answer is the 3rd option.
"behind the lens and inverted"
Standing wavesIn material that is open on one end and closed on the other, only the [ ]a)evenb)odd
Answers
In the material that is open on one end and closed on the other, only the odd harmonic exists.
This is because the even harmonic does not produce the correct end behaviors to match the fundamental standing wave.
A train travels 110 miles in the same time that a plane covers 528 miles. If the speed of the plane is 10 miles per hour less than 5 times thespeed of the train, find both speeds.
Answers
We will have the following:
First, we will have that the expressions of distance for the train and plane are respectively:
[We remember that velocity times time equals distance]
[tex]\begin{cases}t\colon v\cdot t=110mi \\ \\ p\colon(5v-10)\cdot t=528mi\end{cases}[/tex]From this we solve for time "t" in each expression:
[tex]\begin{cases}t=\frac{110mi}{v} \\ \\ t=\frac{528mi}{5v-10}\end{cases}[/tex]Now, since the time is the same, we equal both expressions, that is:
[tex]\frac{110mi}{v}=\frac{528mi}{5v-10}[/tex]Now, we solve for "v"
[tex]\Rightarrow110(5v-10)=528(v)\Rightarrow550v-1100=528v[/tex][tex]\Rightarrow22v=1100\Rightarrow v=50[/tex]Now, we determine each vehicle's velocity:
[tex]\begin{cases}t\colon50mi/h \\ \\ p\colon240mi/h\end{cases}[/tex]So, the train is moving at 50 miles per hour and the plane is moving at 240 miles per hour.
We corroborate by determining the time it takes for each to reach the distances stated, that is:
*Train:
[tex](50mi/h)\cdot t=110mi\Rightarrow t=\frac{11}{5}h\Rightarrow t=2.2h[/tex]*Plane:
[tex](240mi/h)\cdot t=528mi\Rightarrow t=\frac{11}{5}h\Rightarrow t=2.2h[/tex]Thus proving that the velocities are the ones calculated.
What are the types of precipitation? Select four correct answers. fog sleet dew snow wind rain hail
Answers
The four main types of precipitation are Rain, sleet drizzle, and snow.
Precipitation is any product of the condensation of atmospheric water vapours.
Thus the correct answer is Rain, sleet, drizzle(dew), and snow.
The image depicts two bar magnets immersed inof that contains small needle-In particles ofIron.What is the BEST Interpretation of the patterncreated by the iron particles?A The pattern of iron particles is created by thenorth pole of one magnet facing the south poleof the other magnetB. The pattern of kroparticles is created by thesouth pols of one magnet facing the south poleof the other magnet because south poles attracteach otherC. The pattern of iron particles is created by therepulsion of like magnetic poles, we cannotdetermine if the poles are north or southpolarizationD. The pattern of fron particles is created by thenorth pole of one magnet facing the north poleof the other magnet because north poles attracteach other.
Answers
option A is correct
The pattern of iron particles is created by the
north pole of one magnet facing the south pole
of the other magnet
as the lines of force emerges from north and ends at south
28. A 40.0-kg pack is carried up a 2500-m-high mountain in 10.0 h. How much work is done?
Answers
ANSWER
[tex]980,000J[/tex]EXPLANATION
To find the work done in carrying the bag pack, we can apply the formula:
[tex]W=F\cdot d[/tex]where F = force ; d = distance moved
The force here refers to the weight of the bag pack, given by:
[tex]\begin{gathered} m\cdot g \\ 40\cdot9.8 \\ \Rightarrow392N \end{gathered}[/tex]where m = mass; g = acceleration due to gravity
Hence, the work done is:
[tex]\begin{gathered} W=392\cdot2500 \\ W=980,000J \end{gathered}[/tex]That is the answer.
Mass of Car (kg)0.036Displacement in (m)1.05Trial Time (t) in (s)1 0.732 0.723 0.73Average Time0.73a. calculate gravitational potential energyb. calculate average speedc. calculate final speed d. calculate kinetic energye. calculate efficiency
Answers
We will have the following:
a. We will determine the gravitational potential energy as follows:
[tex]\begin{gathered} U_g=(0.036kg)(9.8m/s^2)(1.05m)\Rightarrow U_g=\frac{1323}{2500}J \\ \\ \Rightarrow U_g=0.5292J \end{gathered}[/tex]So, the gravitational potential energy is 0.5292 Joules.
b. Average speed:
[tex]\begin{gathered} v_a=\frac{1.05m}{0.73s}\Rightarrow v_a=\frac{105}{73}m/s \\ \\ \Rightarrow v_a\approx1.44m/s \end{gathered}[/tex]So, the average speed was 105/73 m/s, that is approximately 1.44 m/s.
c. The final speed will be determined as follows:
[tex]v_f=(9.8m/s^2)(0.73s)\Rightarrow v_f=7.154m/s[/tex]So, the final velocity will be 7.154 m/s.
d. The kinetic energy will be:
[tex]\begin{gathered} k=\frac{1}{2}(0.036kg)(7.154m/s)^2\Rightarrow k=0.921234888...J \\ \\ \Rightarrow k\approx0.92J \end{gathered}[/tex]So, the kinetic energy is approximately 0.92 Joules.
e. We will determine the efficiency as follows:
[tex]\begin{gathered} E=\frac{0.92J}{0.5292}\ast100\Rightarrow E=173.8373167... \\ \\ \Rightarrow E\approx173.84 \end{gathered}[/tex]So, the efficiency is approximately 173.84%.
A hunter aims directly at a target (on the same level as the arrow being shot) 38.0 m away. A) If the arrow leaves the bow at a speed of 23.1 m/s, by how much will it miss the target? B) At what angle should the bow be aimed in order to hit the target?
Answers
a) 13.25 m
b)22.3 degrees
Explanation
Step 1
[tex]\text{distance = velocity }\cdot time[/tex]Let
[tex]\begin{gathered} \text{distance = 38} \\ \text{time}=\text{ t} \\ \text{speed}=\text{ }23.1\frac{m}{s} \end{gathered}[/tex]so
[tex]\begin{gathered} \text{distance = velocity }\cdot time \\ 38\text{ m=23.1}\cdot t \\ t=\frac{38\text{ m}}{23.1}=1.645\text{ s} \\ \text{time}=\text{ 1.645} \end{gathered}[/tex]now , to find the disntace to the targe let's use the formula
[tex]\begin{gathered} y=V_0\sin _{}\text{ (}\propto\text{)}+\frac{1}{2}gt^2 \\ as\text{ the angle is 0 } \\ y=V_0\sin _{}\text{ (0)}+\frac{1}{2}gt^2 \\ y=\frac{1}{2}gt^2 \end{gathered}[/tex]hence
[tex]\begin{gathered} y_f=-\frac{1}{2}g\cdot t^2 \\ \text{replace} \\ y_f=-\frac{1}{2}(9.8)\cdot(1.645^2) \\ y_f=-\frac{1}{2}(9.8)\cdot(2.7) \\ y_f=-13.25 \\ \end{gathered}[/tex]so,target missed by 13.25 meters.
Step 2
b)At what angle should the bow be aimed in order to hit the target?
to solve this we can use the expression
[tex]\begin{gathered} t=\frac{dis\tan ce(x)}{velocity\text{ (x)}} \\ so \\ \text{distance}=\text{ 38 m} \\ V_x=V\text{ cos (}\propto) \\ V_x=23.1\text{ cos (}\propto) \\ \end{gathered}[/tex]now, replace and solve for the angle
[tex]\begin{gathered} t=\frac{dis\tan ce(x)}{velocity\text{ (x)}} \\ =\frac{38}{23.1\cos \propto} \\ \\ \frac{38}{23.1\cos\propto}=\frac{23.1\sin \propto}{4.9} \\ 38\cdot4.9=23.1\sin \propto\cdot23.1\cos \propto \\ \frac{38\cdot4.9}{23.1^2}=\sin c\propto\cos \propto \\ so \\ \propto=22.3 \end{gathered}[/tex]so, the angle should be
22.3 degrees
A weight of 5.00 x 10^3 N is suspended in equilibrium by two cables. Cable 1 applies a horizontal force to the right of the object and has a tension, FT1. Cable 2 applies a force upward and to the left at an angle of 37.0° to the negative x-axis and has a tension, FT2. What is FT2?
Answers
8310 N
Explanation
Step 1
Free Body Diagram
Step 1
Newton's first law says that if the net force on an object is zero, then that object will have zero acceleration
so
[tex]\begin{gathered} \sum ^{\square}_{\square}x=F_1-F_2\cos 37=0 \\ \text{replace} \\ F_1-F_2\cos 37=0\rightarrow equation\text{ (1)} \end{gathered}[/tex]and
[tex]\begin{gathered} \sum ^{\square}_{\square}y=F_2\sin 37-w=0 \\ \text{replace} \\ F_2\sin 37-w=0 \\ F_2\sin 37-5.0\cdot10^3=0 \\ so \\ F_2=\frac{5\cdot10^3}{\sin \text{ 37}} \\ F_2=8308.2\text{ N} \\ \text{rounded} \\ F_2=8310\text{ N} \end{gathered}[/tex]therefore, the answer
8310 N
I hope this helps you
A car battery with a potential difference of 11.4 volts is connected to a starter motor with a resistance of 0.242 ohms. If the startup time is 1.54 seconds, how many electrons flow out of the battery during this time? Answer must be in 3 significant digits.
Answers
Given
The potential difference is V=11.4 V
Resistance is R=0.242 ohm
Time taken, t=1.54 sec
To find
The number of electron flow.
Explanation
Let the number of electron be n
We know current is charge per unit time
Thus,
[tex]I=\frac{q}{t}=\frac{ne}{t}[/tex]By Ohm's law,
[tex]\begin{gathered} V=RI \\ \Rightarrow V=\frac{Rne}{t} \\ \Rightarrow11.4=\frac{0.242\times n\times1.6\times10^{-19}}{1.54} \\ \Rightarrow n=4.53\times10^{20} \end{gathered}[/tex]Conclusion
The number of electron flow is
[tex]4.53\times10^{20}[/tex]A .71-kg billiard ball moving at 2.5 m/s in the x-direction strikes a stationary ball of the same mass. After the collision, the first ball moves at 2.17 m/s, at an angle of 30.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), answer the following questions to find the struck ball's momentum after the collision.Using the conservation of momentum in the y-direction, find the struck ball's y-component of momentum.
Answers
After the collision, the momentum didn't change, so the total momentum in x and y are the same as the initial.
The x component was calculated by subtracting the initial momentum (total) minus the momentum of the first ball after the collision
In the y component, as at the beginning, the total momentum was 0 in this axis, the sum of both the first and struck ball has to be the same in opposite directions. In other words, both have the same magnitude but in opposite directions
[tex]\begin{gathered} Py=0.71kg\cdot2.17\cdot sin(30) \\ Py=0.77kg\cdot m/s \end{gathered}[/tex]This is for both balls after the collision, but one goes in a positive and the other in a negative direction.
Two thin rods of length L are rotating with the same angular speed, W (in rad/s) about axes that pass perpendicularly through one end. Rod A is massless but has particle of mass 0.63 kg attached to its free end. Rod B has mass of 0.63 kg, which is distributed uniformly along its length. The length of each rod is 0.82 m, and the angular speed is 4.2 rad/s. Find the kinetic energies of rod A with its attached particle and of rod B.
Answers
ANSWER:
Rod A: 3.74 J
Rod B: 3.74 J
STEP-BY-STEP EXPLANATION:
Rod A:
Length of the Rod = l = 0.82 m
Mass of particle attached = m = 0.63 kg
Moment of inertia of the system about given axis:
[tex]\begin{gathered} I=m\cdot l^2 \\ \text{ replacing} \\ I=0.63\cdot(0.82)^2 \\ I=0.424\text{ kg}\cdot m^2 \end{gathered}[/tex]Angular speed of the rod = 4.2 rad/s
Kinetic energy of the rod is:
[tex]\begin{gathered} KE=\frac{1}{2}\cdot I\cdot w^2 \\ \text{ replacing:} \\ KE=\frac{1}{2}\cdot0.424\cdot4.2^2 \\ KE=3.74\text{ J} \end{gathered}[/tex]Rod B:
Length of the Rod = l = 0.82 m
Mass of particle attached = m = 0.63 kg
Moment of inertia of the system about given axis:
[tex]\begin{gathered} I=m\cdot l^2 \\ \text{ replacing} \\ I=0.63\cdot(0.82)^2 \\ I=0.424\text{ kg}\cdot m^2 \end{gathered}[/tex]Angular speed of the rod = 4.2 rad/s
Kinetic energy of the rod is:
[tex]\begin{gathered} KE=\frac{1}{2}\cdot I\cdot w^2 \\ \text{ replacing:} \\ KE=\frac{1}{2}\cdot0.424\cdot4.2^2 \\ KE=3.74\text{ J} \end{gathered}[/tex]What is the current through each resistor in the parallel portion?
Answers
Given:
Two resistors each of resistance R = 60 ohms are connected in parallel.
A resistance R' = 30 ohms is connected in series with the parallel combination.
The voltage across the circuit is V = 120 V.
To find the current through each resistor in the parallel portion.
Explanation:
The equivalent resistance of the resistors connected in parallel will be
[tex]\begin{gathered} \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R} \\ =\frac{1}{60}+\frac{1}{60} \\ R_p=\text{ }\frac{60}{2} \\ =30\text{ }\Omega \end{gathered}[/tex]In a series combination, the voltage will be the sum of voltages across each resistor while the current remains the same.
In parallel combination, the voltage remains the same while the current is the sum of currents through each resistor.
Voltage in series combination will be
[tex]\begin{gathered} V=V_{Rp}+V^{\prime} \\ V=V_{30}+V_{30} \\ V_{30}=\text{ }\frac{V}{2} \\ =\frac{120}{2} \\ =60\text{ V} \end{gathered}[/tex]The voltage across each resistor in the parallel combination will be 60 ohms.
The current in the series combination will be the same.
According to the Ohm's law, the current will be
[tex]\begin{gathered} V^{\prime}=I^{\prime}R^{\prime} \\ I^{\prime}=\frac{V^{\prime}}{R^{\prime}} \\ =\frac{60}{30} \\ =2\text{ A} \end{gathered}[/tex]The current in the parallel combination will be the sum of individual current
[tex]I^{\prime}=I_1+I_2[/tex]Since the voltage across the parallel combination is the same and the resistance is also the same, so
[tex]\begin{gathered} I_1=I_2 \\ =I_p \end{gathered}[/tex]Thus, the current through each resistor in parallel combination will be
[tex]\begin{gathered} I^{\prime}=I_p+I_p \\ I_p=\frac{I^{\prime}}{2} \\ =\frac{2}{2} \\ =\text{ 1 A} \end{gathered}[/tex]Hence, the current through each resistor in the parallel portion is 1 A
The distance traveled by a race car is 7km in 100sec. What is it's speed?
Answers
Answer:
[tex]\text{ Speed=}0.07\text{ m/s}[/tex]Step-by-step explanation:
Average speed is the measure of the rate of movement of a body expressed as the distance traveled divided by the time taken:
[tex]\begin{gathered} \text{ Speed=}\frac{dis\tan ce}{\text{time}} \\ \end{gathered}[/tex]Therefore if the car traveled 7 km in 100 sec:
*Remember to convert kilometers to meters:
[tex]\begin{gathered} \text{ Speed=}\frac{7}{100} \\ \text{ Speed=}0.07\text{ m/s} \end{gathered}[/tex]According to Coulomb's law, how does distance affect electric force?A.The electric force is the same no matter how far apart the objects are.B.Objects must be in contact with one another for the electric force to have an effect.C.Only sometimes can the electric force act without objects touching.D.Electric force can act at a distance but is stronger when objects are closer.
Answers
According to Coulomb's law,
[tex]Force\text{ }\propto\frac{1}{(distance)^2}[/tex]Thus, electric force can act at a distance but is stronger when objects are closer.
Hence, option D is correct.
Hello I could really use some help answering this question thank you!
Answers
In order to find the false statement about radiation, let's analyze each one:
A.
This statement is true, all objects emit radiation.
B.
This statement is also true.
C.
This statement is false, because radiation can transfer energy in vacuum (no matter) as well.
D.
This statement is true.
Therefore the correct answer is C.
Which compound is a molecule?Question 10 options:NaClMgBr2C6H12O6AgCl3
Answers
It's important to know the difference between a molecular compound and an ionic compound.
Ionic compounds are formed by the interaction between a metal and a non-metal, the first ones have the property of easily losing electrons, while non-metals have the property of easily gaining electrons.
In this case, C6H12O6 is a molecular compound because it's not using ionic interaction.
Therefore, the third option is correct.
The parallel circuit at the right depicts two resistors connected to a voltage source. The voltage source (ΔVtot) is a 12-V source and the resistor values are 7.4 Ω (R1) and 3.9 Ω (R2).a. Determine the equivalent resistance of the circuit. b. Determine the current in each branch resistor:Current in R1 = Current in R2 = c. Determine the total current in the circuit.
Answers
Given:
Two resistors are connected in parallel.
The resistance of resistor 1 is
[tex]R1=\text{ 7.4}\Omega[/tex]The resistance of resistor 2 is
[tex]R2\text{ = 3.9 }\Omega[/tex]The value of the voltage source is
[tex]\Delta V_{tot}=\text{ 12 V}[/tex]Required:
(a)The equivalent resistance of the circuit.
(b) The current in each branch of the resistor.
(c) The total current in the circuit.
Explanation:
(a) In a parallel circuit, the equivalent resistance can be calculated as
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R1}+\frac{1}{R2} \\ =\frac{1}{7.4}+\frac{1}{3.9} \\ R_{eq}=\text{ 2.55 }\Omega \end{gathered}[/tex](b) In a parallel circuit, the voltage across each resistor is the same.
The current through the resistor R1 can be calculated according to Ohm's law
[tex]\begin{gathered} I_{R1}=\frac{V}{R1} \\ =\frac{12}{7.4} \\ =1.62\text{ A} \end{gathered}[/tex]The current through the resistor R2 can be calculated as
[tex]\begin{gathered} I_{R2}=\frac{V}{R2} \\ =\frac{12}{3.9} \\ =3.08\text{ A} \end{gathered}[/tex](c) The total current in the circuit is the sum of the current through each branch.
Thus, the total current can be calculated as
[tex]\begin{gathered} I=I_{R1}+I_{R2} \\ =\frac{12}{7.4}+\frac{12}{3.9} \\ =\text{ 4.7 A} \end{gathered}[/tex]Final Answer:
(
Suppose that the 1.3 km span of the Golden Gate Bridge had no expansion joints. Calculate how much longer the bride would be if the temperature of the bridge increases by 25°C.
Answers
Given:
Δt = change in temperature = 25°c
L = Lenght = 1.3 km= 1300 m
Suppose the bride is made out of steel.
coefficient of Linear expansion of steel = 11 x 10^-6 /°C
Apply:
ΔL = L x coeficcient x Δt = 1300 ( 11x10^-6) 25 = 0.3575 m
I need to figure out this type of 2-d momentum question, I attached the one I'm stuck on as an image.
Answers
Given:
The mass of the pucks, m=0.300 kg
The initial velocity of the 1st puck, u₁=3.33 m/s
The initial velocity of the 2nd puck, u₂=1.87 m/s
The final velocity of the 1st puck, v₁=2.10 m/s
To find:
The y-component of the final velocity of the 2nd puck.
Explanation:
Let us assume that the north is the positive y-axis and the east is the positive x-axis.
As the 1st puck is going northwest, the angle made by the velocity of the 1st puck with ponegative-axis is 45 °
The velocities can be represented in the vector form as,
[tex]\begin{gathered} \vec{u_1}=3.33cos45\degree(-\hat{i})+3.33sin45\degree\hat{j} \\ =-2.35\hat{i}+2.35\hat{j} \end{gathered}[/tex]And
[tex]\vec{u_2}=1.87(-\hat{j})=-1.87\hat{j}[/tex]And
[tex]\vec{v_1}=-2.10\hat{i}[/tex]Where i-cap and j-cap are the unit vectors along the positive x and y axes respectively.
From the law of conservation of momentum, the total momentum of a system remains the same at all times. The momentum is conserved simultaneously and independently along both axes.
Considering the y-axis,
[tex]\begin{gathered} mu_1sin45\degree\hat{j}+mu_2(-\hat{j})=0+mv_{2y}\hat{j} \\ \Rightarrow u_1sin45\degree-u_2=v_{2y} \end{gathered}[/tex]Where v_2y is the y-component of the final velocity of the second puck.
On substituting the known values,
[tex]\begin{gathered} 3.33sin45\degree-1.87=v_{2y} \\ \Rightarrow v_{2y}=0.48\text{ m/s} \end{gathered}[/tex]Final answer:
The y-component of the final velocity of the second puck is 0.48 m/s
Two people are initially at rest and touching each other. Theypush off each other and they move away in opposite directions.Person 1 has a mass of 87.5 kg and a velocity of -10 meters persecond. Person 2 has a mass of 39.2 kg. What is the velocity ofPerson 2?V =unitHow far apart will the two people be after 3 seconds? (Hint,both are moving away. Calculate the distance each will traveland add those distances, or get the net velocity and calculatethe distance)d =unit
Answers
Answer:
[tex]\begin{gathered} v_2=22.321ms^{-1} \\ d=96.964m\Rightarrow\text{ ( Distance apart )} \end{gathered}[/tex]Explanation: This problem can be solved using the conservation of momentum concept, the equations used are as follows:
[tex]\begin{gathered} p_i=p_f \\ p_i=p_1+p_2=0\Rightarrow p_i=(m_1+m_2)v_i=0\Rightarrow(1) \\ p_f=p_1+p_2=0\Rightarrow p_f=m_1v_1+m_2v_2=0\Rightarrow(2) \end{gathered}[/tex]The knowns and unknowns in equation (1) and (2) are as follows:
[tex]\begin{gathered} m_1=87.5\operatorname{kg} \\ v_1=-10ms^{-1} \\ m_2=39.2\operatorname{kg} \\ v_2=\text{?} \\ v_i=0 \end{gathered}[/tex]Plugging these values in (1) and (2) gives us the following result:
[tex]\begin{gathered} (1)=(2) \\ \therefore\Rightarrow \\ (m_1+m_2)(0)=m_1v_1+m_2v_2=0 \\ (87.5\operatorname{kg})(-10ms^{-1})+(39.2\operatorname{kg})v_2=0 \\ v_2=\frac{(87.5\operatorname{kg})(10ms^{-1})}{(39.2\operatorname{kg})}=22.321ms^{-1} \\ v_2=22.321ms^{-1} \end{gathered}[/tex]To calculate the distance travelled in 3 seconds, we simply do as follows:
[tex]\begin{gathered} s=vt \\ \therefore\Rightarrow \\ (1)\rightarrow s_1=(-10ms^{-1})\cdot(3s)=-30m \\ (2)\rightarrow s_2=(22.321ms^{-1})(3s)=66.964m \\ \therefore\Rightarrow \\ d=|s_1|+|s_2|=|-30m|+|66.964m|=96.964m \\ d=96.964m \end{gathered}[/tex]Find the gravitational force between Saturn (5.68 x 10^26kg ) and the sun (1.99 x 10^30 kg). Saturn orbits at a distance of 1,404,219,991,220 m.
Answers
ANSWER
[tex]3.826\cdot10^{22}N[/tex]EXPLANATION
To find the gravitational force between them, apply the formula for gravitational force:
[tex]F=\frac{GmM}{r^2}[/tex]where G = gravitational constant = 6.6743 * 10^(-11) Nm²/kg²
m = mass of Saturn
M = mass of the sun
r = distance between them (radius of Saturn's orbit)
Therefore, the gravitational force between them is:
[tex]\begin{gathered} F=\frac{6.6743\cdot10^{-11}\cdot5.68\cdot10^{26}\cdot1.99\cdot10^{30}}{(1,404,219,991,220)^2} \\ F=\frac{75.4409\cdot10^{-11+26+30}}{1.9718\cdot10^{24}}=\frac{7.54409\cdot10^{46}}{1.9718\cdot10^{24}} \\ F=3.826\cdot10^{22}N \end{gathered}[/tex]Calculate the recoil velocity of a 5 kg rifle that fires a 24 g bullet that travel 1.5 m in 0.0023 s
Answers
The mass of rifle is M = 5kg . It fires the bullet of mass m =24 g = 0.024 kg
It travels a distance, d = 1.5 m and the time taken is t = 0.0023 s
The speed of the bullet will be
[tex]\begin{gathered} v=\frac{d}{t} \\ =\frac{1.5}{0.0023} \\ =652.17\text{ m/s} \end{gathered}[/tex]Let the recoil velocity be V. The initial velocity will be zero as rifle and bullet are at rest.
According to the conservation of momentum
[tex]\begin{gathered} \text{Initial momentum = final momentum} \\ (M+m)\times0=MV-mv\text{ } \end{gathered}[/tex]Here, rifle and bullet travel in opposite directions.
Substituting the values, the recoil velocity will be
[tex]\begin{gathered} V=\frac{mv}{M} \\ =\frac{0.024\times652.17}{5} \\ =3.13\text{ m/s} \end{gathered}[/tex]One of the tallest radio towers is in Fargo, North Dakota. The tower is 629m tall, or about 44 percent taller than the Sears Tower in Chicago. if a bird lands on top of the tower, so that the gravitational potential energy associated with the bird is 2033 J, what is the mass?
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The graph below represents the variation of the velocity versus time of a moving particle moving along the x- axis. The acceleration of this particle between 2s and 4s is equal to:
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ANSWER:
2nd option: 0 m/s^2
STEP-BY-STEP EXPLANATION:
From second 2 to second 4, we can see on the graph that the velocity remains constant, therefore the acceleration is 0.
The correct answer is 2nd option 0 m/s^2
Emily, with a mass of 54.1 kilograms, is riding a wagon of mass 15.0 kilograms with a velocity of +4.12 meters per second. She then jumps off the wagon and lands on the ground with a horizontal velocity of +5.59 meters per second. What is the velocity of the wagon after Emily jumps off? Answer must be in 3 significant digits.
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The initial momentum of the system can be expressed as,
[tex]p_i=(m_1+m_2)u[/tex]The final momentum of the system can be expressed as,
[tex]p_f=m_1v_1+m_2v_2[/tex]According to conservation of momentum,
[tex]p_i=p_f[/tex]Plug in the known expressions,
[tex]\begin{gathered} (m_1+m_2)u=m_1v_1+m_2v_2 \\ m_2v_2=(m_1+m_2)u-m_1v_1 \\ v_2=\frac{(m_1+m_2)u-m_1v_1}{m_2} \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} v_2=\frac{(54.1\text{ kg+15.0 kg)(4.12 m/s)-(54.1 kg)(5.59 m/s)}}{15.0\text{ kg}} \\ =\frac{284.692\text{ kgm/s-}302.419\text{ kgm/s}}{15.0\text{ kg}} \\ =-\frac{17.727\text{ kgm/s}}{15.0\text{ kg}} \\ =-1.18\text{ m/s} \end{gathered}[/tex]Thus, the final velocity of wagon is -1.18 m/s where negative indicates that the wagon moves in the opposite direction.
What is the length contraction of an automobile 3.133 m long when it is traveling at 51.35km/h? (Hint:for x << 1,(1-v2/c2)1/2 ~ 1 - x2/2) Compare this to the diameter of a hydrogen atom by expressing your answer in femto meters.
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The length L of a moving object whose rest length is L_0 is:
[tex]L=\sqrt{1-\frac{v^2}{c^2}}L_0[/tex]The length contraction can be calculated as the difference L_0-L:
[tex]L_0-L=\left(1-\sqrt{1-\frac{v^2}{c^2}}\right)L_0[/tex]For v<, we have:
[tex]\begin{gathered} L_0-L\approx\left(1-\left[1-\frac{1}{2}\left(\frac{v}{c}\right)^2\right]\right)L_0 \\ \\ =\left(\frac{v}{c}\right)^2\frac{L_0}{2} \end{gathered}[/tex]Replace v=51.35km/h, c=300,000km/s and L_0=3.133m:
[tex]\begin{gathered} \Delta L=\left(\frac{v}{c}\right)^2\frac{L_0}{2} \\ \\ =\left(\frac{51.35\frac{km}{h}\times\frac{1h}{3600s}}{300,000\frac{km}{s}}\right)^2\frac{3.133m}{2} \\ \\ =3.54...\times10^{-15}m \\ \\ =3.54...fm \end{gathered}[/tex]The diameter of a hydrogen atom is approximately 10.6*10^-11m. Then, the length contraction of th ecar is much less than the diameter of a hydrogen atom.
Therefore, the length contraction of the automobile is approximately 3.54 femtometers.