ANSWER
• I(toaster) = 15.08 A
,• I(pan) = 11.17 A
,• I(lamp) = 0.71 A
EXPLANATION
Given:
• The toaster's power, P(toaster) = 1810 W
,• The frying pan's power, P(pan) = 1340 W
,• The lamp's power, P(lamp) = 85 W
,• The total current supported by the circuit, I = 15 A
,• The voltage of the circuit, V = 120 V
Find:
• The current across each device
Power is the product of current and voltage,
[tex]P=I\cdot V[/tex]We know that all three devices are connected in parallel, which means that they all have the same voltage, 120 V.
Solving the equation above for I,
[tex]I=\frac{P}{V}[/tex]So, for each device,
[tex]I_{toaster}=\frac{P_{toaster}}{V}=\frac{1810W}{120V}\approx15.08A[/tex][tex]I_{pan}=\frac{P_{pan}}{V}=\frac{1340W}{120V}\approx11.17A[/tex][tex]I_{lamp}=\frac{P_{lamp}}{V}=\frac{85W}{120V}\approx0.71A[/tex]Hence, the current drawn by each device is:
• I(toaster) = 15.08 A
,• I(pan) = 11.17 A
,• I(lamp) = 0.71 A
A train car with a mass of 10kg and speed of 10 m/s is traveling to the right. Another train car with a mass of 20kg is moving to the left at -40 m/s. After the collision, the 10 kg train car is now moving at -20 m/s and we need to find the Velocity of the 20 kg train car.
When two particles collide and the masses of the particles are given, as well as the initial and final velocity of one particle and one of the velocities of the second particle, then the remaining velocity of the second particle is given by the expression:
[tex]v_2^{\prime}=\frac{m_1v_1+m_2v_2-m_1v_1}{m_2}[/tex]Which can be deduced from the Law of Conservation of Linear Momentum.
In the given problem, the initial and final velocities of the train car with mass 10kg are given, as well as the initial velocity of the 20kg car:
[tex]\begin{gathered} m_1=10kg \\ v_1=10\frac{m}{s} \\ v_1^{\prime}=-20\frac{m}{s} \\ \\ m_2=20kg \\ v_2=-40\frac{m}{s} \\ v_2^{\prime}=\text{ unknown} \end{gathered}[/tex]Replace those values into the given equation to find v₂':
[tex]\begin{gathered} v_2^{\prime}=\frac{(10kg)(10\frac{m}{s})+(20kg)(-40\frac{m}{s})-(10kg)(-20\frac{m}{s})}{20kg} \\ \\ \Rightarrow v_2^{\prime}=-25\frac{m}{s} \end{gathered}[/tex]Therefore, the velocity of the 20kg train car after the collision, is: -25 m/s.
IThe alignment of electrons in a magnetic field does not contribute to the properties of magnets true or false?
To determine whether the given statement is true or false.
Explanation:
The alignment of electrons in a particular direction in a magnetic field form magnetic domains.
These magnetic domains are responsible for magnetic properties.
Thus the given statement is true.
A rock is thrown off of a 120 foot cliff with an upward velocity of 20 ft/s. As a result its height after t seconds is given by the formula:h(t) = 120 + 20t - 5t^2What is its height after 2 seconds?___What is its velocity after 2 seconds?____(Positive velocity means it is on the way up, negative velocity means it is on the way down.)
We are given that the height of a rock in terms of the time is given by the following equation:
[tex]h\mleft(t\mright)=120+20t-5t^2[/tex]We are asked to determine the height after two seconds. To do that we will substitute in the equation the value of "t = 2s", like this:
[tex]h(2)=120+20(2)-5(2)^2[/tex]Solving the operations we get:
[tex]h(2)=140[/tex]Therefore, the height after 2 seconds is 140 ft.
Now, to determine an equation for the velocity we will determine the derivative with respect to the time of the equation for the height.
[tex]\frac{dh}{dt}=\frac{d}{dt}(120+20t-5t^2)[/tex]Now, we distribute the derivative:
[tex]\frac{dh}{dt}=\frac{d}{dt}(120)+\frac{d}{dt}(20t)-\frac{d}{dt}(5t^2)[/tex]For the first derivative we will use the following rule:
[tex]\frac{d}{dt}(a)=0[/tex]Where "a" is a constant. Applying the rule we get:
[tex]\frac{dh}{dt}=\frac{d}{dt}(20t)-\frac{d}{dt}(5t^2)[/tex]For the second derivative we will use the following rule:
[tex]\frac{d}{dt}(at)=a[/tex]Where "a" is a constant. Applying the rule we get:
[tex]\frac{dh}{dt}=20-\frac{d}{dt}(5t^2)[/tex]For the last derivative we will use the following rule:
[tex]\frac{d}{dt}(at^n)=\text{nat}^{n-1}[/tex]Applying the rule we get:
[tex]\frac{dh}{dt}=20-10t[/tex]Since the derivative of the position with respect to time is the velocity we have:
[tex]\frac{dh}{dt}=v=20-10t[/tex]Now, we substitute the value of "t = 2s":
[tex]v=20-10(2)[/tex]Now, we solve the operations:
[tex]\begin{gathered} v=20-20 \\ v=0 \end{gathered}[/tex]Therefore, the velocity after 2 seconds is 0.
A wave traveling on a Slinky® that is stretched to 4 m takes 4.97 s to travel the length of the Slinky and back again.(a) What is the speed (in m/s) of the wave? m/s
ANSWER
1.61 m/s
EXPLANATION
Given:
• The distance the Slinky is stretched, d = 4 m
,• The time it takes for the wave to travel the length mentioned before and back, t = 4.97 s
Find:
• The speed of the wave, v
We know that the wave takes 4.97 seconds to travel the 4 meters the Slinky was stretched and back, so it travels a total of 8 meters in that time. This means that the speed of the wave is,
[tex]v=\frac{2d}{t}=\frac{8m}{4.97s}\approx1.61m/s[/tex]Hence, the speed of the wave is 1.61 m/s.
Object a attracts objects be the gravitational force of 10 N from a given distance the distance between the two objects is doubled what is the new force of attraction between
We are given that two objects are being attracted by a gravitational force between each other of 10N. The gravitational force between two masses is given by the following equation:
[tex]F_g=G\frac{m_Am_B}{r^2}[/tex]Where:
[tex]\begin{gathered} F_g=\text{ Gravitational force} \\ m=\text{mass} \\ G=\text{ Gravitational constant} \\ r=\text{ distance between the masses} \end{gathered}[/tex]Replacing the given values for the 10N force:
[tex]10=G\frac{m_Am_B}{r^2_1}[/tex]Where:
[tex]r_1=\text{ initial distance}[/tex]Now we will solve for the product of the masses and the gravitational constant by multiplying both sides by the distance squared:
[tex]10r^2_1=Gm_Am_B[/tex]Now, the product of the masses and the gravitational constant won't change if we double the distance, therefore, if we apply the equation for the gravitational force for the new distance we get:
[tex]F_{g2}=G\frac{m_Am_B}{r^2_2}[/tex]We can replace the value we determined earlier:
[tex]F_{g2}=\frac{10r^2_1}{r^2_2}[/tex]Since the distance is double, we have:
[tex]r_2=2r_1[/tex]Replacing in the previous equation:
[tex]F_{g2}=\frac{10r^2_1}{(2r_1)^2}[/tex]Solving the square:
[tex]F_{g2}=\frac{10r^2_1}{4r^2_1}[/tex]Now we can cancel out the distances squared:
[tex]F_{g2}=\frac{10}{4}[/tex]Solving the operation:
[tex]F_{g2}=2.5[/tex]Therefore, doubling the distance the new gravitational force is 2.5N.
The current in the circuit shown is 5.0 A, and the resistor is 2.0 Ω. What is the potential across the battery? A.8.0 VB.2.5 VC.10 VD.3.0 V
Given:
The current is,
[tex]i=5.0\text{ A}[/tex]The resistance is,
[tex]R=2.0\text{ ohm}[/tex]To find:
The potential across the battery
Explanation:
Using Ohm's law, we can write the potential across the battery is,
[tex]\begin{gathered} V=iR \\ =5.0\times2.0 \\ =10\text{ V} \end{gathered}[/tex]Hence, the potential across the battery is 10 V.
Identify the following examples of forces as either a contact force or a non-contact force:
Given:
Different kinds of forces
To find:
Identify the contact forces and non-contact forces
Explanation:
A force that acts between two surfaces in contact is a contact force, while the force that acts between the two surfaces, not in contact, is a non-contact force.
The contact forces here are:
Sitting on a chair (friction)
Skateboard slows down on a rough surface (friction)
Slamming the door (friction)
Non-contact forces:
Magnet attracting paper clips (magnetic force)
Two like charges repel each other (electrostatic force)
When A peach falls from a tree (gravitational force)
Question 17 of 25A conductor is a material that:O A. allows easy movement of charge.B. is never made of metal,C. hinders the passage of electricity.O D. is made of glass.
From the given list, let's select the statement that best defines a conductor.
A conductor can be defined as any material that allows the flow of electric charge.
Examples of conductors are:
• Copper
,• Aluminium
,• Silver...
Therefore, the best statement that defines a conductor is that a conductor is a material that allows easy movement of charge.
• Option B is wrong because most conductors are made of metal.
,• Option C is wrong because it is an ,insulator ,hinders the passage of electricity.
,• Option D is wrong because a glass is an insulator not a conductor.
ANSWER:
A. allows easy movement of charge
Apassenger in a drops a and it freely from rest to the groundWhat is the speed of the ball a distance of 20 massuming the acceleration due to gravity equal to 9.01 m/s ^ 2
Given data
*The given initial speed is u = 0 m/s
*The given distance is s = 20 m
*The value of the acceleration due to gravity is g = 9.01 m/s^2
The formula for the final speed of the ball at a distance of 20 m is given by the kinematic equation of motion as
[tex]v^2=u^2+2gs[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v^2=(0)^2+2(9.01)(20) \\ v^2=\text{36}0.04 \\ v=18.98\text{ m/s} \end{gathered}[/tex]Hence, the speed of the ball at a distance of 20 m is v = 18.98 m/s
__________ energy is the energy a body has as a function of its temperature.A: potentialB:thermalC:internaD:electrical
Internal energy is a function of temperature.
When the temperature is provided to a body, it results in an increase in internal energy.
Thus, internal energy is a function of temperature.
If you have 100.g of a radioactive isotope with a half-life of 10. years, how much of the isotopewill you have left after 20. years?
Use the radioactive decay formula using t_0 as the half life:
[tex]A=A_0\cdot2^{-t/t_0}[/tex]Substitute the initial amount of radioactive element A_0=100g, the half life t_0=10y and the time period t=20y to find the remaining amount after 20 years:
[tex]\begin{gathered} A=100g\times2^{-20y/10y} \\ =100g\times2^{-2} \\ =100g\times\frac{1}{4} \\ =25g \end{gathered}[/tex]Therefore, after 20 years, there are 25 grams left.
I need help on this science homework I forgot on what to do through number 4-7 including a,b,c,d, and e.
Answer:
[tex]\begin{gathered} (a)\Rightarrow v=2ms^{-1} \\ (b)\Rightarrow v=1ms^{-1} \\ (c)\Rightarrow v=6.67ms^{-1}_{} \\ (d)\Rightarrow v=1.2ms^{-1} \\ (e)\Rightarrow v=0ms^{-1} \end{gathered}[/tex]Explanation: We need to calculate the speed on intervals a b c d and e, the speed can be calculated with the following formula:
[tex]v=\frac{\Delta S}{\Delta t}\Rightarrow(1)[/tex](a) 0-5 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(10m-0m)}{(5s-0s)}=\frac{10m}{5s}=2ms^{-1} \\ v=2ms^{-1} \end{gathered}[/tex](b) 5-15 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(20m-10m)}{(15s-5s)}=\frac{10m}{10s}=1ms^{-1} \\ v=1ms^{-1} \end{gathered}[/tex](c) 15-18 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(40m-20m)}{(18s-15s)}=\frac{20m}{3s}=6.67ms^{-1} \\ v=6.67ms^{-1} \end{gathered}[/tex](d) 18-23 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(46m-40m)}{(23s-18s)}=\frac{6m}{5s}=1.2ms^{-1} \\ v=1.2ms^{-1} \end{gathered}[/tex](e) 23-25 seconds Intervals:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(40m-40m)}{(25s-23s)}=\frac{0m}{2s}=0ms^{-1} \\ v=0ms^{-1} \end{gathered}[/tex]Every 5 seconds, the crest of a wave in the ocean travels 25 meters. What is the speed of the wave?
Given
Every 5 seconds
One wave travel 25m
Speed of waves
Explanation
[tex]\begin{gathered} v=\frac{25m}{5s} \\ v=5\text{ m/s} \end{gathered}[/tex]
The answer would be 5 m/s
The potential energy of a system can be changed by varying the position of objects in the system. At which point do the coaster cars have the most potential energy?
To find:
At which point the potential energy of the coaster car is the highest?
Explanation:
The potential energy of an object is the energy possessed by the object due to the position of the object. The gravitational potential energy of the object is directly proportional to the height of the object.
Thus an object will have the highest potential energy when it is at the highest point.
Final answer:
Thus the coaster cars will have the highest potential energy when it is at the highest point on the roller coaster.
Therefore, the coaster cars will have the most potential energy at point A.
Two drops of mercury each has a charge on 2.42 nC and a voltage of 293.97 V. If the two drops are merged into one drop, what is the voltage on this drop?
The electric potential is given by:
[tex]\begin{gathered} V=\frac{Kq}{r} \\ \end{gathered}[/tex]Let's find r first:
[tex]\begin{gathered} r=\frac{Kq}{V}=\frac{8.988\times10^9\cdot2.42\times10^{-9}}{293.97} \\ r\approx0.074m \end{gathered}[/tex]Now we can find the radius of the new drop:
[tex]r_t=2(r)=2(0.074)=0.148[/tex]So:
[tex]\begin{gathered} V=\frac{K(2q)}{r_t}=\frac{8.988\times10^9\cdot2(2.42\times10^{-9})}{(0.148)} \\ V=293.93V \end{gathered}[/tex]You see a blue star directly over Avalon's equator. It appears to be moving north at 5.0 arcseconds per year. How far away is this star from Avalon (in parsecs - a parsec is ).
The distance of the star from the sun is 1.25 parsecs.
What quantity is measured in arcseconds?
The distance of the sun to other celestial bodies is measured in arcseconds.
The distance from the sun to a celestial object is the reciprocal of the angle, measured in arcseconds, of the object's apparent movement caused by parallax.
2 arcseconds = 0.5 parsecs
5 arcseconds = ?
= (5 arcseconds x 0.5 parsecs) / (2 arcseconds)
= 1.25 parsecs
Thus, the star appears to be moving north at 1.25 parsecs per year.
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Using the work energy theorem, what is the final velocity of a roller coaster at the bottom of the hill. The coaster has a mass of 839 kg and starts at rest from the top of a hill that is 75 meters tall.
ANSWER
38.34 m/s
EXPLANATION
Given:
• The mass of the coaster, m = 839 kg
,• The initial height of the coaster, h = 75 m
,• The acceleration due to gravity, g = 9.8 m/s²
Find:
• The final velocity of the roller coaster at the bottom of the hill, v.
The roller coaster starts from rest, so at the top of the hill, it only has gravitational potential energy and no kinetic energy. Then, at the bottom of the hill, the roller coaster is in motion, to it has kinetic energy, and, because the difference of height with the reference - which is the bottom of the hill, is zero, it has no potential energy,
By the work-energy theorem, we have the equation,
[tex]KE_i+PE_i+W_{nc}=KE_f+PE_f_{}[/tex]As explained above, the initial kinetic energy is 0 and the final potential energy is also 0. If we assume that there is no friction, air resistance, or other external forces, then the work done by non-conservative forces is also 0,
[tex]PE_i=KE_f[/tex]Replace each kind of energy with the expression to obtain them,
[tex]m\cdot g\cdot h=\frac{1}{2}\cdot m\cdot v^2[/tex]The mass cancels out,
[tex]g\cdot h=\frac{1}{2}\cdot v^2[/tex]Solving for v,
[tex]v=\sqrt[]{2\cdot g\cdot h}[/tex]Replace with the known values and solve,
[tex]v=\sqrt[]{2\cdot9.8m/s^2\cdot75m}=\sqrt[]{1470m^2/s^2}\approx38.34m/s[/tex]Hence, the velocity of the roller coaster at the bottom of the hill is 38.34 m/s, rounded to the nearest hundredth.
this is a 3 part question20) A 9.50-g bullet has a speed of 1.30 km>s. (a) What is its kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled?
Given that mass of bullet, m = 9.50 g = 0.0095 kg
speed of bullet, v is 1.30 km/s
(a) Kinetic energy is given by the formula
[tex]K\mathrm{}E\text{. = }\frac{1}{2}mv^2[/tex]Substituting the values in the above formula, we get
[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(0.0095)(1.30)^2=8.0275\times10^{-3}\text{ J}[/tex](b) Speed of bullet is v/2
Sustituting this value in the formula of kinetic energy, we get
[tex]\begin{gathered} K.E._1=\text{ }\frac{1}{2}m(\frac{v}{2})^2 \\ =\frac{1}{2}\times(0.0095)\times(\frac{1.30}{2})^2 \\ =2.0068\times10^{-3\text{ }}J \end{gathered}[/tex](c) Speed of bulllet becomes 2v
Sustituting this value in the formula of kinetic energy, we get
[tex]\begin{gathered} K.E._2=\text{ }\frac{1}{2}m(2v)^2 \\ =\frac{1}{2}\times(0.0095)\times(2\times1.30)^2 \\ =0.03211J \end{gathered}[/tex]A 1 kg apple and a 2 kg grapefruit are 2 m apart. What is the force of gravity between each fruit?
According to Newton's law of gravitation, any particle of matter in the universe attracts any other with a force varying directly as the product of their masses and inversely as the square of the distance between them. It is expressed as
F = Gm1m2/r^2
where
G = gravitational constant and its value is G = 6.673 x 10^-11 Nm^2/kg^2
m1 and m2 are the masses of the bodies
r is the distance between the bodies
From the information given,
m1 = 1
m2 = 2
r = 2
By substituting these values into the formula,
F = (6.7 x 10^-11 x 1 x 2)/2^2
F = 33.5 x 10^- 12
F = 3.35 x 10^-11
The second option is correct
As the speed of an object falling toward Earth increases, the gravitational potential energy of the object with respect to EarthA. IncreasesB. DecreasesC. Remains the same
Answer:
B. Decreases
Explanation:
When an object is falling toward Earth, the height of the object decreases, and the speed increases. Then, the gravitational potential energy decreases, and the kinetic energy increase because the first one depends on the height and the second one depends on the speed. Therefore, the answer is:
B. Decreases
Mass/energy equivalence is expressed mathematically in which of the following expressions?
Check each option to see how it relates to different concepts.
Option 1: E=hf
This equation tells the energy carried by an electromagnetic wave with frequency f.
Option 2: E=mc
This equation is not correct, since the left member is measured in units of energy and the right member does not.
Option 3: E=(1/2)mv^2
This equation relates the energy of a moving object with its mass and its velocity. It is known as kinetic energy.
Option 4: E=mc^2
Since c is a constant (the speed of light), this equation relates the energy of an object with its mass.
Therefore, the mass/energy equivalence is expressed mathematically in the equation:
[tex]E=mc^2[/tex]A 4000-kg truck traveling with a velocity of 20 m/s due south collides head-on with a 1320- kg car traveling with a velocity of 10 m/s due north. The two vehicles stick together after the collision.A. What are the magnitude and the direction of the momentum of each vehicles Prior to the collision?B. What are the magnitude and the direction of the velocity of both vehicles after their collide?
Before we begin, we will establish that the north direction is the positive direction which means that the south direction is negative.
A.
The momentum of an object is given by:
[tex]p=mv[/tex]For the truck we know that the velocity is -20 m/s (since it is traveling south) and its mass is 4000 kg, then its momentum is:
[tex]p=(4000)(-20)=-80000[/tex]Therefore, the momentum of the truck is -80000 kg m/s; this means that its magnitude is 80000 kg m/s and its direction is south.
For the car we know that the velocity is 10 m/s and its mass is 1320 kg, then its momentum is:
[tex]p=(1320)(10)=13200[/tex]Therefore, the momentum of the car is 13200 kg m/s which means that its magnitude is 13200 kg m/s and its direction is north.
B.
In a collision the momentum is conserved, that is, the total initial and final momentum is equal, that is:
[tex]p_i=p_f[/tex]In this case, we know that the vehicles stick together after they collide, then we have:
[tex]m_tv_t+m_cv_c=(m_t+m_c)u[/tex]where u is the velocity of the vehicles after they collide. Plugging the values we know, we have that:
[tex]\begin{gathered} -80000+13200=(4000+1320)u \\ u=\frac{-80000+13200}{4000+1320} \\ u=-12.56 \end{gathered}[/tex]Therefore, the final velocity of the system is -12.56 m/s which means that the magnitude of the velocity is 12.56 m/s and its direction is south.
What is troubling about the circumstance when a coil is stationary and a magnet moves as compres to when a magnet is stationary and the coil moves?
When the magnet is moved, the galvanometer needle will deflect. It shows that the current is flowing in the coil. When the magnet moves into the coil, the needle deflects into one way, and if the magnet moves out of the coil, the needle deflects into the other way.
When the magnet is held stationary near, or even inside, the coil, no current will flow through the coil.
carts, bricks, and bands
6. What acceleration results when four rubber bands stretched to 20 cm is used pull a cart with one brick?
a. About 0.25 m/s2
b. About 0.50 m/s2
c. About 0.75 m/s2
d. About 1.00 m/s2
The acceleration that results when four rubber bands stretched to 20 cm is used pull a cart with one brick is about 1.00 m/s². That is option D
What is acceleration?Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 1 ----> 1 band = 0.24m/s²
Trial 2 ----> 2 bands = 0.51 m/s²
Trial 3 ----> 3 bands = 0.73 m/s²
Trial 4 -----> 4 bands = 1.00 m/s²
This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.
It can clearly be observed that trial 4 that made use of 4 bands resulted in an acceleration of 1.00 m/s² which is the highest observed acceleration.
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7/21/22, 7:37 AMProblem Set ThreeNotes: Use 9.8 m/s 2 for the acceleration due to gravity. Formust be expressed in m/sLaw calculations, mass must be expressed in kg and velocity.A steady 45 N horizontal force is applied to a 15kg object on a table. The object slides against a friction force of 30 N. Calculate the acceleration of the object in m/s.
Given:
The mass of the object is
[tex]m=15\text{ kg}[/tex]The applied force on the object is
[tex]F=45\text{ N}[/tex]The frictional force on the object is
[tex]f=30\text{ N}[/tex]To find:
The acceleration of the object
Explanation:
The net force on the object is
[tex]\begin{gathered} F_{net}=F-f \\ =45-30 \\ =15\text{ N} \end{gathered}[/tex]The acceleration of the object is,
[tex]\begin{gathered} a=\frac{F_{net}}{m} \\ =\frac{15}{15} \\ =1\text{ m/s}^2 \end{gathered}[/tex]Hence, the acceleration is
[tex]1\text{ m/s}^2[/tex]A 22-pound force that makes an angle of 12° with an inclined plane is pulling a box up the plane. The inclined plane makes a 25° angle with the horizontal. What is the magnitude of the effective force pulling the box up the plane?
We will ave that the effective force is:
[tex]\begin{gathered} F=(22Lb)cos(12)\Rightarrow F=21.51924722...Lb \\ \\ \Rightarrow F\approx21.52Lb \end{gathered}[/tex]So, the effective force is 21.52 Pounds.
If a 4 kg ball is dropped from rest and falls without air resistance, what is its speed after 0.5 seconds?
Answer:
4.9 m/s
Explanation:
The velocity of the ball after 5 seconds can be calculated using the following equation
[tex]v_f=v_i+at[/tex]vi = the initial velocity, in this case, it is equal to 0 because the ball is dropped from the rest
a = acceleration, this is the acceleration due to gravity so it is -9.8 m/s²
t = time, it is equal to 0.5 s
So, replacing the values, we get:
[tex]\begin{gathered} v_f=0-9.8(0.5) \\ v_f=-4.9\text{ m/s} \end{gathered}[/tex]Therefore, the speed after 0.5 seconds is 4.9 m/s
A car travels at an average speed of 60 km/h for 15 minutes.How far does the car travel in 15 minutes?D900 kmC 240 kmA4.0 km B 15 km
Notice that the speed is written using units of km/h and the time is written using units of minutes.
Convert the time interval to hours. Remember that 1 hour equals 60 minutes:
[tex]15\min =15\min \times\frac{1h}{60\min }=0.25h[/tex]The distance d that a particle travels during a time t if it moves at an average speed v is given by the formula:
[tex]d=v\cdot t[/tex]Replace v=60km/h and t=0.25h to find the distance traveled by the car:
[tex]d=(60\frac{km}{h})\times(0.25h)=15km[/tex]Therefore, the car travels 15km in 15 minutes.
If you see an object following a curve path, you can safely assume:There is a net force being applied The net force is zero
If an object is following a curve, it means that it is changing direction. This change in direction will cause a change in velocity. Change in velocity causes acceleration. Since the velocity keeps changing, the acceleration is not zero. This also means that the net force is not zero. Therefore, you can safely assume:
There is a net force being applied
Every time a cycle of convection currents occur, the temperature of the water ________.A. IncreasesB. Stays the sameC. DecreasesD. Goes to zero
The answer to this question is B
The reason the answer is B is because of how the cycle works. In the beggining, water is heated up and evaporates into the atmosphere as water vapor. Then the water cools down and condensates, coming back down. The water then ends up back to the same temperature as the earth. So in the end, the entire cycle doesn't add or remove heat from the water.
