Answer:
Explanation:
The centripetal acceleration is given by
[tex]undefined[/tex]A wire passes 45 C of charge every 0.58 s. How many amperes of current is passing through the wire during this time?
We will have the following:
[tex]A=\frac{45C}{0.58s}\Rightarrow Amp=\frac{2250}{29}Amp\Rightarrow A\approx77.59Amp[/tex]So there will be approximately 77.29 amperes passing through.
A freight car with a mass of 10 metric tons is rolling at 108 km/h along a level track when it collides with another freight car, which is initially at rest. If the speed of the cars after they couple together is 36 km/h, what is the mass of the second car?1) 40 metric tons2) 20 metric tons3) 10 metric tons4) 5 metric tons
Given data:
* The mass of the first car is 10 metric tons.
* The initial speed of the first car is 108 km/h
* The speed of couple of both the cars after the collision is 36 km/h.
* The initial speed of the second car is zero. ial
Solution:
By the law of conservation of momentum,
[tex]mu_1+Mu_2=mv_1_{}_{}+Mv_2[/tex]where m is the mass of first car, M is the mass of second car, u_1 is the initial velocity of first car, u_2 is the initial velocity of second car, v_1 is the final velocity of first car, and v_2 is the final velocity of second car,
Substituting the known values,
[tex]\begin{gathered} 10\times108+0=(m+M)\times36 \\ 1080=(10+M)\times36 \\ 10+M=30 \\ M=20\text{ metric tons} \end{gathered}[/tex]Thus, the mass of the second car is 20 metric tons.
Hence, option 2 is the correct answer.
what's force and laws of motion
A force is a push or pull upon an object resulting from the object's interaction with another object.
We have 3 laws of motion or Newton's laws
1. The law of inertia or first law postulates that a body will remain at rest or in straight motion with a constant speed unless an external force is applied.
2. Newton's second law or fundamental law, postulates that the net force that is applied on a body is proportional to the acceleration it acquires on its path.
The formula is F=ma
where F is the force, m is the mass and a is the acceleration
3. Newton's third law postulate says that every action generates an equal reaction but in the opposite direction.
What is the tension in the swing’s chain at this time?
Answer:
A. 325.4 N
Explanation:
At the bottom of the swing, the net force is equal to:
[tex]F_{net}=T-mg=ma[/tex]Where T is the tension, m is the mass, g is the gravity, and a is the centripetal acceleration, so a = v²/r. With v the speed and r is the length of the swing. So, solving for T, we get:
[tex]\begin{gathered} T-mg=m\frac{v^2}{r} \\ \\ T=m\frac{v^2}{r}+mg \end{gathered}[/tex]Now, we can replace m = 27 kg, v = 3 m/s, r = 4 m, and g = 9.8 m/s² to get
[tex]\begin{gathered} T=(27\text{ kg\rparen}\frac{(3\text{ m/s\rparen}^2}{4\text{ m}}+(27\text{ kg\rparen\lparen9.8 m/s}^2) \\ \\ T=60.75\text{ N + 264.6 N} \\ T=325.4\text{ N} \end{gathered}[/tex]Therefore, the answer is
A. 325.4 N
which of the following X-Y tables with the information in this problem?
The initial speed is in Y, is to the north and is positiv
The acceleration is negative in y and positive in x thanks to the angle, so the answer is B.
Remember the 0.35 acceleration is decompossed in 2 accelerations one in x one in y. Thats why C is incorrect
certainSaturated4.What would you call a solution that is full of solute forthat temperature?
Solution that is full of solute at given temperature is known as saturated solution.
Nadia threw a 4.1 kg ball straight upward with a speed of 15m/s. How high is the ball at the very top just before it starts falling back down
ANSWER:
11.48 meters
STEP-BY-STEP EXPLANATION:
Let's establish the equation of the situation:
[tex]\frac{1}{2}\cdot m\cdot(v_i)^2+m\cdot g\cdot h_i=\frac{1}{2}\cdot m\cdot(v_f)^2+m\cdot g\cdot h_f[/tex]We can determine that according to the situation the initial speed is 15 m/s, the final height is 0, the final speed is 0, therefore, we must calculate the final height, like this:
Replacing tue values:
[tex]\begin{gathered} \frac{1}{2}\cdot1\cdot15^2+1\cdot9.8\cdot\cdot0=\frac{1}{2}\cdot1\cdot0+1\cdot9.8\cdot\: h_f \\ 112.5+0=0+9.8\cdot h_f \\ h_f=\frac{112.5}{9.8} \\ h_f=11.48\text{ m} \end{gathered}[/tex]Therefore, the height is 11.48 meters
If i lift the 30kg object to a height of 2 meters in 0.5 seconds, how much power did I lift with?
Power = Work / time
Work = Force x distance
Force = mass x gravity
g = gravity = 9.8 m/s^2
m= mass = 30 kg
F = force = 30 kg x 9.8 m/S^2 = 294 N
Work = 294 N x 2 m = 588 J
Power = 588 J / 0.5 s = 11,600 W
A circuit with a battery, a 6 Ω resistor, a 21 Ω resistor, and a 27 Ω resistor in parallel. The total voltage is the system is 7.0 V. What is the total resistance of the circuit?
Given that the resistances are connected in parallel.
[tex]\begin{gathered} R_1=6\text{ ohm} \\ R_2=21\text{ ohms} \\ R_3\text{ = 27 ohms} \end{gathered}[/tex]Also, the total voltage is 7 V.
The formula to find the equivalent resistance connected in parallel is
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]Substituting the values, the resistance will be
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{6}+\frac{1}{21}+\frac{1}{27} \\ =\frac{63+18+14}{378} \\ =\frac{95}{378} \\ R=3.97\text{ ohm} \end{gathered}[/tex]Thus, the total resistance of the circuit is 3.97 ohms.
A tortoise can crawl at 0.10 m/s and crawls for 50 seconds. How far did the tortoise crawl?
ANSWER:
5 meters
STEP-BY-STEP EXPLANATION:
Given:
Speed (v) = 0.10 m/s
Time (t) = 50 s
We can determine the distance traveled as follows:
[tex]\begin{gathered} v=\frac{d}{t} \\ \\ \text{ We replacing:} \\ \\ 0.10=\frac{d}{50} \\ \\ d=0.10\cdot50 \\ \\ d=5\text{ m} \end{gathered}[/tex]The tortoise crawls for about 5 meters.
Which state has particles that are the closest together?O A. LiquidO B. Solid&O c. PlasmaO D. GasSUBM
B. Solid
In solid state, particles have little movement allowing them to be compact.
A block attached to a horizontal spring with a spring constant k = 200 N/m is displaced to x = +6 cm and released from rest. The time needed for the block to reach x = +6 cm for the first time is 0.2 sec. Determine the maximum restoring force applied on the block and its maximum acceleration.
Given:
The spring constant of the spring is k = 200 N/m
The maximum displacement of the spring is x = 6 cm = 0.06 m
The time period is T = 0.2 s
To find the maximum restoring force and the maximum acceleration.
Explanation:
The maximum restoring force applied on the block can be calculated as
[tex]\begin{gathered} F=kx \\ =200\times0.06\text{ } \\ =\text{ 12 N} \end{gathered}[/tex]First, we need to calculate angular frequency.
The angular frequency can be calculated as
[tex]\begin{gathered} \omega=\frac{2\pi}{T} \\ =\frac{2\times3.14}{0.2} \\ =31.4\text{ rad/s} \end{gathered}[/tex]The maximum acceleration can be calculated as
[tex]\begin{gathered} a=x\omega^2 \\ =0.06\times(31.4)^2 \\ =59.2m/s^2 \end{gathered}[/tex]Final Answer: The maximum restoring force is 12 N and the maximum acceleration is 59.2 m/s^2.
A car accelerates from 4 m/s to 16 m/s in 4 seconds. The car's acceleration is:____ meters/second².
In order to calculate the acceleration, we can use the formula below:
[tex]a=\frac{v_f-v_i}{t}[/tex]If the final velocity is 16 m/s, the initial velocity is 4 m/s and the time is 4 seconds, we have:
[tex]\begin{gathered} a=\frac{16-4}{4}\\ \\ a=\frac{12}{4}\\ \\ a=3\text{ m/s^^b2} \end{gathered}[/tex]Therefore the acceleration is 3 m/s².
The Jamaican bobsled team hit the breaks on their sled so that it decelerates at a uniform rate of 0.43m/s^2. How long does it take to stop if it travels 85m before coming to rest?
Given:
The sled decelerates with a uniform rate of: a = -0.43 m/s^2 (The negative sign indicates that the velocity of the sled is decreasing)
The sled covers a distance of 85 m before coming to rest.
To find:
The time sled takes to come to rest
Explanation:
We make use of the following kinematic equation,
[tex]v^2=u^2+2as[/tex]Here, v is the final velocity, u is the initial velocity, a is the deceleration and s is the displacement.
The final velocity v is zero as the sled is at rest. Thus, v = 0 m/s
The displacement of the sled after hitting the breaks is: s = 85 m
Substituting the values in the above equation, we get:
[tex]\begin{gathered} 0=u^2-2\times0.43\text{ m/s}^2\times85\text{ m} \\ \\ u^2=73.1\text{ m}^2\text{/s}^2 \\ \\ u=\sqrt{73.1\text{ m}^2\text{/s}^2} \\ \\ u=8.55\text{ m/s} \end{gathered}[/tex]The initial velocity of the sled when breaks are applied is 8.55 m/s.
Now, consider another kinematical equation,
[tex]v=u+at[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} 0=8.55\text{ m/s}-0.43\text{ m/s}^2\times t \\ \\ t=\frac{8.55\text{ m/s}}{0.43\text{ m/s}^2} \\ \\ t=19.88\text{ s} \end{gathered}[/tex]Final answer:
The sled takes 19.88 seconds before it comes to rest.
A brick is thrown upward from the top of a building at an angle of 10° to the horizontal and with an initial speed of 16 m/s. If the brick is in flight for 3.1 s, how tall is the building?_____ m
Answer:
38.47 m
Explanation:
To find the height of the building, we will use the following equation
[tex]y_f=y_i+v_{iy}t+\frac{1}{2}at^2[/tex]Where yf is the final height, yi is the initial height, viy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.
If the brick is in flight for 3.1 s, we can say that when t = 3.1s, yf = 0 m. So, replacing
viy = (16 m/s)sin(10) = 2.78 m/s
a = -9.8 m/s²
we get
[tex]0=y_i+2.78(3.1)+\frac{1}{2}(-9.8)(3.1)^2[/tex]Solving for yi
[tex]\begin{gathered} 0=y_i-38.48 \\ y_i=38.48\text{ m} \end{gathered}[/tex]Therefore, the height of the building is 38.48 m
What current (in A) flows through the bulb of a 3.00 V flashlight when its hot resistance is 3.90 Ω?
Given:
The voltage across the bulb of the flashlight is: V = 3.00 v
The resistance offered by the bulb of the flashlight is: R = 3.90 Ω
To find:
The amount of current flowing through the bulb.
Explanation:
The relation between voltage V, current I, and the resistance R is given by Ohm's law. The expression for Ohm's law is:
[tex]V=IR[/tex]Rearranging the above equation, we get:
[tex]I=\frac{V}{R}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} I=\frac{3.00\text{ v}}{3.90\text{ ^^^^2126}} \\ \\ I=0.77\text{ A} \end{gathered}[/tex]Final answer:
The current flowing through the bulb of the flashlight is 0.77 Ampere.
Artificially produced radioactive isotopes are used in medicine and medical research. Is this true or false?
The radioactive isotopes are used in the medicines like Tc-99.
These isotopes are used in the nuclear medicines like Iodine-131 and phosphorus-32 are used for the thyroid or cancer treatments.
Thus, the given statement is true.
A sea otter swims toward an inlet with a speed of 5.0 m/s as a current of1.0 m/s flows in the opposite direction. How long will it take the sea otterto swim 100. m?*
Given:
The speed of sea otter is
[tex]v_s=5\text{ m/s}[/tex]The speed of the current is
[tex]v_c=\text{ 1 m/s}[/tex]in the opposite direction.
The distance is d = 100 m
To find the time taken by sea otter to cover a distance d = 100 m.
Explanation:
The total speed of sea otter is
[tex]\begin{gathered} v_=v_s-v_c \\ =5-1 \\ =4\text{ m/s} \end{gathered}[/tex]The time taken will be
[tex]\begin{gathered} t=\frac{d}{v} \\ =\frac{100\text{ m}}{4\text{ m/s}} \\ =25\text{ s} \end{gathered}[/tex]Thus, the time taken is 25 s to swim 100 m.
During times of dire emergency, people have been known to lift tremendous weights, such as the rear of a car to free a mechanic on whom the car has fallen. Is greater power necessary to perform such feats than when lifting the same car using a jack? Explain.
If any object is manually lifted, then greater power is required to lift the object.
But by using a jack, lesser power is required.
Work done = Force X displacement
Greater force is required to displace any object.
For a screw jack, rotational force (torque) is used.
The formula for torque is
[tex]\tau=Force\times\text{ distance from the axi}s[/tex]Here, when we apply force by rotating such that its distance from the axis of rotation is maximum.
The concept here is torque is directly proportional to force applied and distance from the axis.
So, in order to minimize force, we can increase the distance from the axis.
Hence, we get more work done with lesser force.
Uranium 238 has 92 protons. It undergoes an alpha decay, then a beta decay, and then another beta decay. How many protons remain?90928991
Alpha decay is when nuclei emits an alpha particle, for this case
[tex]undefined[/tex]Using the momentum formula, calculate the momentum of a 325 kg rock, rolling down a hill at 1.75 m/s.
We will have the following:
First, we recall that momentum is given by:
[tex]\tau=m\ast v[/tex]So:
[tex]\tau=(325kg)(1.75m/s)\Rightarrow\tau=568.75kg\ast m/s[/tex]So, the momentum of the rock is 568.75 kg*m/s.
A force of 60. newtons is applied to a rope to pull a sled across a horizontal surface at a constant velocity. The rope is at an angle of 30. Degrees above the horizontal.8-9. Calculate the magnitude of the component of the 60.-newton force that is parallel to the horizontal surface. [Show all work, including the equation and substitution with units.] [2]
ANSWER:
51.96 N
STEP-BY-STEP EXPLANATION:
The horizontal component of the pulling force of Force is FAx since the axis is horizontal, therefore:
[tex]\begin{gathered} F_x=F\cdot\cos _{}\theta \\ \text{ replacing} \\ F_x=60\cdot\cos 30 \\ F_x=51.96\text{ N} \end{gathered}[/tex]The magnitude of the component of the 60 newton is 51.96 N
Determine from the following diagram a. The spring constant of the bowb. The Elastic Potential Energy storedc. What will be the initial velocity of the arrow once the bow is released?
F= 300 N
d = 0.6m
m = 30g = 0.03 kg
• a.
Apply Hooke's law formula:
F = k x
k= F/ x
Where:
k= spring constant
F= force = 300N
x = distance = 0.6 m
k = 300N/0.6m = 500N/m
• b.
PE = 1/2 k (x)^2
PE = 1/2 (500 N/m) (0.6)^2 = 90 J
• c.
KE = 1/2 m v^2
KE = kinetic energy = PE = potential energy
v = √KE/ (1/2*m) = √2KE/m = √ (2* 90 / 0.03 ) = 77.46 m/s
Answers:
a = 500 N/m
b= 90 J
C = 77.46 m/s
2a. Based on the information presented in Diagram 2A/2B and Table 2 above, does brass orsteel expand/contract more at a given change of temperature? Explain why.
Change in the length of a material due to a change in temperature, ΔT is given as,
[tex]\Delta L=\alpha L\Delta T[/tex]Where L is the initial length of the material and α is the coefficient of linear expansion.
It is clear from the above equation that the expansion of a material is proportional to its coefficient of linear expansion.
From the table, Coefficient of expansion of steel,α₁=12.00
and, Coefficient of expansion of brass is, α₂=19.00
As the coefficient of expansion of brass is more than that of steel, brass will expand more than steel.
How much kinetic energy does the 75 kg skater have at 0 m/s
Given:
Mass, m = 75 kg
Velocity, v= 0 m/s
Let's find the kinetic energy.
To find the kinetic enefgy, apply the formula:
[tex]KE=\frac{1}{2}mv^2[/tex]Where:
m is the mass = 75 kg
v is the velocity = 0 m/s
Hence, we have:
[tex]\begin{gathered} KE=\frac{1}{2}\ast75\ast0 \\ \\ KE=0\text{ Joules} \end{gathered}[/tex]The kinetic energy is 0 Joules. This means there is no kinetic energy.
ANSWER:
You and your buddy are scuba diving and notice that air bubbles triple in volume as they rise to the surface from where you are in the ocean. Ignoring any temperature changes, how far below the surface of the water (in m) are you when these bubbles are released? The ocean has a density of 1,031 kg/m3.
ANSWER
[tex]h=20.06\text{ m}[/tex]EXPLANATION
Since the temperature is constant, applying the ideal gas equation, we have that the product of volume and pressure remains constant:
[tex]P_1V_1=P_2V_2[/tex]When they are submerged in water, their pressure is:
[tex]P_1=P_{atm}+h\rho g[/tex]where Patm = atmospheric pressure
h = depth below the surface
ρ = density
g = acceleration due to gravity
At the surface of the water:
[tex]P_2=P_{atm}[/tex]Applying the ideal gas equation, we have that:
[tex]P_1V_1=P_2V_2[/tex]But we have that V2 = 3V1. Substituting that into the equation:
[tex]\begin{gathered} P_1V_1=P_2(3V_1) \\ \\ \Rightarrow P_1=3P_2 \\ \\ P_{atm}+h\rho g=3P_{atm} \\ \\ h\rho g=3P_{atm}-P_{atm}=2P_{atm} \\ \\ h=\frac{2P_{atm}}{\rho g} \end{gathered}[/tex]Substitute the given values into the equation and solve for h:
[tex]\begin{gathered} h=\frac{2*101325}{1031*9.8} \\ \\ h=20.06\text{ m} \end{gathered}[/tex]That is the height when the bubbles are released.
11. An object of mass 3.123 kg is being pulled by ahorizontal force of 210 N on a horizontal force.The coefficient of kinetic friction betwedsthe object and the surface is 0.1. Calculate theacceleration of the object. (1 point)30.373 m/s²A.B. O 66.263 m/s²2C. O 122.405 m/sD. 110.843 m/sE. 088.439 m/s²Submit Query
11.
Apply Newton's law:
∑F = m a
Where:
F = forces
F = 210 N
Fr = friction force = N u
u= coefficient of friction = 0.1
m= mass = 3.123 kg
a = acceleration
N = normal force = mg
g= gravity = 9.8 m/s^2
N = 3.123 kg x 9.8 m/s^2 = 30.6054 N
FR = N u = 30.6054 N x 0.1 = 3.06054 N
∑F = m a
F - Fr = ma
210 N - 3.06054 N = 3.123 kg (a)
206.93946 = 3.123 kg (a)
a = 206.93946 N / 3.123 kg
a = 66.263 m/s^2 (option B)
#93- compare the magnitude of the centripetal acceleration of the car at A to the magnitude of the car’s centripetal acceleration at A if additional passengers were riding in the car.
ANSWER:
The magnitude of the centripetal acceleration in both cases is the same
STEP-BY-STEP EXPLANATION:
We have that the centripetal acceleration is equal to:
[tex]a=speed^{2}/radio[/tex]Therefore, it is not dependent on mass and therefore the centripetal acceleration will be the same in both cases, that is, with additional passengers or without additional passengers.
A student pulls his younger sister in a sled (combined mass 31 kg) with a rope that is angled at = 13° with a force of 290 N. The coefficient of friction between the sled and the ground is 0.28.(a) What is the normal force acting on the sled?(b) What is the acceleration of the sled?
Given,
The combined mass of the sister and the sled, m=31 kg
The angle made by the rope, θ=13°
The force applies by the student, F=290 N
The coefficient of the friction between the sled and the ground, μ=0.28
(a)
The forces that are acting on the sled in the vertical direction are upward normal force, the downward weight of the sled and the sister, and the upward vertical component of the force applied by the student.
The normal force acting on the sled is given by,
[tex]N=mg-F\sin \theta[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} N=31\times9.8-290\sin 13^{\circ} \\ =238.56\text{ N} \end{gathered}[/tex]The normal force acting on the sled is 238.56 N
(b)
The net force acting on the sled in the horizontal direction is given by,
[tex]\begin{gathered} F_N=ma=F\sin \theta-f \\ =F\cos \theta-N\mu \end{gathered}[/tex]Where a is the acceleration of the sled and f is the frictional force between the sled and the ground.
On substituting the known values in the above equation,
[tex]\begin{gathered} 31\times a=290\times\cos 13^{\circ}-238.56\times0.28 \\ \Rightarrow a=\frac{184.35}{31} \\ =5.95m/s^2 \end{gathered}[/tex]Thus the acceleration of the sled is 5.95 m/s².
what is the gravitational potential energy of each of the 2kg books in the image?
Given data:
* The mass of the book is m = 2 kg.
* The height of book A is h_A = 1 m.
* The height of book B is h_B = 1.5 m.
* The height of book C is h_C = 2 m.
Solution:
(a). The gravitational potential energy of book A is,
[tex]U_A=\text{mgh}_A[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} U_A=2\times9.8\times1 \\ U_A=19.6\text{ J} \end{gathered}[/tex]Thus, the gravitational potential energy of book A is 19.6 J.
(b). The gravitational potential energy of book B is,
[tex]\begin{gathered} U_B=\text{mgh}_B \\ U_B=2\times9.8\times1.5 \\ U_B=29.4\text{ J} \end{gathered}[/tex]Thus, the gravitational potential energy of book B is 29.4 J.
(c). The gravitational potential energy of book C is,
[tex]\begin{gathered} U_C=\text{mgh}_C \\ U_C=2\times9.8\times2 \\ U_C=39.2\text{ J} \end{gathered}[/tex]Thus, the gravitational potential energy of book C is 39.2 J.