Given data:
* The mass of the box is 2 kg.
* The time taken by the box to travel the given distance is 5 seconds.
* The distance traveled by the box is 10 meters.
* The initial velocity of the box is 0 m/s.
Solution:
By the kinematics equation, the distance traveled by the box in terms of its acceleration is,
[tex]S=ut+\frac{1}{2}at^2[/tex]where u is the initial velocity, t is the time taken, a is the acceleration, and S is the distance traveled,
Substituting the known values,
[tex]\begin{gathered} 10=0+\frac{1}{2}\times a\times(5)^2 \\ 10=\frac{25}{2}\times a \\ a=10\times\frac{2}{25} \\ a=0.8ms^{-2} \end{gathered}[/tex]By the Newton's second law, the force exerted on the box in terms of the acceleration is,
[tex]F=ma[/tex]where m is the mass of the box, a is the acceleration and F is the force,
Substituting the known values,
[tex]\begin{gathered} F=2\times0.8 \\ F=1.6\text{ N} \end{gathered}[/tex]Thus, the force of the push is 1.6 N.
how far will you go (km) in 3 min traveling 60 km/hr?
We will have the following:
First, we transform minutes to hours, that is:
[tex]3\min \cdot\frac{1h}{60\min }=0.05h[/tex]Now, we determine the distance traveled:
[tex]d=(60km/h)(0.05h)\Rightarrow d=3km[/tex]So, you will go 3km.
Two remote control cars with masses of 1.13 kilograms and 1.94 kilograms travel toward each other at speeds of 8.50 meters per second and 3.79 meters per second, respectively. The cars collide head-on, and the less massive car recoils with a speed of 2.02 meters per second. What is the total kinetic energy of the two cars before the collision? Include units in your answer. Answer must be in 3 significant digits.
What is the voltage drop across point A and B?
We are asked to find the voltage drop at point A and B
Notice that point A and B have 3 resistors connected in parallel so the voltage across these 3 resistors will be the same.
First, we have to find the equivalent resistance of these 3 parallel resistors.
[tex]\begin{gathered} R_{AB}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}} \\ R_{AB}=\frac{1}{\frac{1}{120}+\frac{1}{60}+\frac{1}{30}} \\ R_{AB}=17.14\; \Omega \end{gathered}[/tex]So, the resistance of the parallel resistors is 17.14
Now, we can simply use the voltage drop formula to find the voltage drop at point A and B
[tex]\begin{gathered} V_{AB}=\frac{R_{AB}}{R_{total}}\times V_{\text{in}} \\ V_{AB}=\frac{R_{AB}}{R_{AB}+R_{CD}}\times V_{\text{in}} \end{gathered}[/tex]Where Vin is the input voltage that is 100 V
[tex]\begin{gathered} V_{AB}=\frac{17.14}{17.14+100}\times100 \\ V_{AB}=14.63\; V \end{gathered}[/tex]Therefore, there is a 14.63 V drop at point A and B
Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Planck’s constant (6.63E-34) to calculate the momentum of this photon:
The wavelength is divided by Plank's constant to get the momentum equation for photons: p = h /λ.
What is photon?The electromagnetic force is carried by a photon, a basic particle that is a quantum of the electromagnetic field and includes electromagnetic radiation like light and radio waves. Since photons have no mass, they constantly move at the 299792458 m/s speed of light in a vacuum.
Assuming the wavelength determined in a prior issue, = 656 nm = 656 * 10 - 9 m, you get:
p = (6.63 * 10 ^-34) / (656 * 10 ^ -9) kg * m/s
P, which must be rounded to three significant numbers, is equal to 1.01067 * 10 - 27 kg*m/s.
Consequently, p = 1.01 * 10 -27 kg*m/s
Our number, rounded to two significant figures, is 1.0 * 10 - 27 kg*m/s because the answers are only rounded to two significant values.
Therefore, given that the wavelength is 656 nm, the first option—1.0*10-27 kg*m/s—is the correct response.
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Sound is an example of a longitudinal wave because the wave travels by compressions and rarefactions of air particles?True or false
To find
The given statement is true or false.
Explanation
The sound waves propag5ate th6roug5h6 a medium by the compression and rarefraction of the m
Three resistors with resistances of 9 Ω, 18 Ω, and 30 Ω are in a series circuit with a 12 volt battery. What is the total resistance of the resistors in the circuit?
When resistors are in series, the total resistance is equal to the sum of the individual resistances of each resistor. Thus, the total resistance in this circuit is 9+18+30 = 57Ω
For each, find the radius, diameter, and circumference of the circular object (one of these measurements is given in the problem). When working with the circumference, use π and round to the nearest whole number. a. Breaking a cookie in half creates a straight side 10 cm long.radius: cmdiameter: cmcircumference: cm
ANSWER
• radius: ,5 cm
,• diameter: ,10 cm
,• circumference: ,31 cm
EXPLANATION
If we assume that the cookie is round, when we cut it in half we'll have the following shape:
The straight side is the diameter of the cookie, which is 10 cm long.
The radius of a circle is half the diameter, hence the radius of the cookie is 5 cm.
The circumference is,
[tex]C=\pi\cdot d[/tex]Where d is the diameter of the circle. In this case, this is 10cm,
[tex]C=\pi\cdot10\operatorname{cm}\approx31\operatorname{cm}[/tex]The circumference of the cookie is 31 cm, rounded to the nearest whole number.
When considering the Law of Universal Gravitation, the graph of force v. distance is _____.1)linear2)parabolic3)circular4)none of the above
According to Law of Universal Gravitation, the graph between Force and distance is hyperbolic as the force depends sqaure of the distance. Therefore, the correct option is (4).
a) your one friend and their bumper boat (m = 210 kg) are traveling to the left at 3.5 m/s and your other friend and their bumper boat (m = 221 kg) are traveling in the opposite direction at 1.8 m/s. The two boats then collide in a perfectly elastic one-dimensional collision. How fast is your first friend and their boat moving after the collision?b) after the collision, a constant drag force between the water and the boat causes your first friends boat to come to a stop. If the boat travels 2.7 m before stopping, what is the magnitude of the constant drag force?
ANSWER:
a) 1.757 m/s
b) 119.91 N
STEP-BY-STEP EXPLANATION:
Given:
Mass 1 (m1) = 210 kg
Initial speed 1 (u1) = 3.5 m/s
Mass 2 (m2) = 221 kg
Initial speed (u2) = 1.8 m/s
We make a sketch of the situation:
a)
We make a momentum balance by taking into account the conservation of momentum:
[tex]\begin{gathered} m_1u_1-m_2u_2=m_1v_1+m_2v_2 \\ \\ v_2=\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\rightarrow(1) \end{gathered}[/tex]Now an energy balance taking into account the conservation of energy, as follows:
[tex]\begin{gathered} \frac{1}{2}m_1(u_1)^2+\frac{1}{2}m_2(u_2)^2=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\lparen v_2)^2\rightarrow(2) \end{gathered}[/tex]Now, we substitute equation (1) in (2) and we get the following:
[tex]\begin{gathered} m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\right)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{(m_2)^2}\right) \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{m_2}\right) \\ \\ v_1=u_1\frac{m_1-m_2}{m_1+m_2}+u_2\frac{2m_2}{m_1+m_2} \\ \\ \text{ Now, we substitute each value, like so:} \\ \\ v_1=3.5\cdot\frac{210-221}{210+221}+1.8\cdot\frac{2\cdot221}{210+221} \\ \\ v_1=1.757\text{ m/s} \end{gathered}[/tex]b)
We use the following formula to determine the force:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ Wee replacing:} \\ \\ (1.756)^2=0^2+2\cdot a\cdot2.7 \\ \\ a=0.003375\text{ m/s}^2 \\ \\ \text{ Therfore:} \\ \\ F=m\cdot a=210\cdot0.571=119.91\text{ N} \end{gathered}[/tex]Problem 42 (p.228) Different radioisotopes have different half-lives. For example, the half-life of carbon-14 is 5700 years, the half-life of uranium-235 is 704 million years , the half-life of potassium-40 is 1.3 billion years, and the half-life of rubidium-87 is 49 billion years.
Why would an isotope with a half life like that of carbon-14 be a poor choice to get the age of the Solar System?
b) The age of the universe is approximately 14 billion years. Does that mean that no
rubidium-87 has decayed yet?
a) The choice of carbon - 14 is not pat because it has a very small half life and would vanish quickly
b) The rubidium-87 is yet to decay since it is older than the universe.
What is the solar system?The solar system comprises of the sun and all the planets that move round the sun. Recall that the sun lies at the center of the solar system. This implies that we have the planets as they move round the sun in concentric circles.
Now we can see that the half lives of all the isotopes that were listed in the question are;
carbon-14 - 5700 yearsuranium-235 - 704 million yearspotassium-40 - 1.3 billion yearsrubidium-87 - 49 billion yearsGiven that the estimated age of the sun is 4.603 billion years, it is clear that the carbon-14 would be a poor choice for the dating of the sun since it has a half life of only a few thousand years.
b) Given the fact that the half life of the rubidium-87 isotope is 49 billion years, it follows that none of it has decayed as yet.
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The distance and direction in which an object travels per unit of time. A. velocity B. magnitude C. speed
Answer:
Answer is C. Speed
Yea thanks thank you for the info thanks
To help us solve this problem let's plot the points given in the table:
From the graph we notice that this the position can be modeled by a sine function, we also notice that the period of this function is 8. We know that a sine function can be modeled by:
[tex]A\sin(B(x+C))+D[/tex]where A is the amplitude, C is the horizontal shift, D is the vertical shift and
[tex]\frac{2\pi}{B}[/tex]is the period.
From the graph we have we notice that we don't have any horizontal or vertical shift, then C=0 and D=0. We also notice that the amplitude is 15, then A=15. Finally, as we said, the period is 8, then:
[tex]\begin{gathered} 8=\frac{2\pi}{B} \\ B=\frac{2\pi}{8} \\ B=\frac{\pi}{4} \end{gathered}[/tex]Plugging these values in the sine function we have:
[tex]x(t)=15\sin(\frac{\pi}{4}t)[/tex]If we graph this function along the points on the table we get the following graph:
We notice that we don't get an exact fit but we get a close one.
Now, that we have a function that describes the position we can find the velocity by taking the derivative:
[tex]\begin{gathered} x^{\prime}(t)=\frac{d}{dt}\lbrack15\sin(\frac{\pi}{4}t)\rbrack \\ =\frac{15\pi}{4}\cos(\frac{\pi}{4}t) \end{gathered}[/tex]Therefore, the velocity is:
[tex]x^{\prime}(t)=\frac{15\pi}{4}\cos(\frac{\pi}{4}t)[/tex]Once we have the expression for the velocity we can find values for the times we need, they are shown in the table below:
From the table we have that:
[tex]x^{\prime}(0.5)=10.884199\text{ cm/s}[/tex]And that:
• The earliest time when the velocity is zero is 2 s.
,• The second time when the velocity is zero is 6 s.
,• The minimum velocity happens at 4 s.
,• The minimum velocity is -11.780972 cm/s
How much force must be applied to push a 253.2 kg crate across the floor at a constant velocity if the coefficient of kinetic friction is 0.55?
Given,
The mass of the crate, m= 253.2 kg
The coefficient of the kinetic friction between the floor and the crate, μ=0.55
Given that the crate is pushed with a constant velocity. That is the net force on the crate is zero.
The only two forces acting on the crate are the force with which it is being pushed and the friction that is opposing the applied force.
The net force on the crate is given by,
[tex]\begin{gathered} F_n=0=F-f \\ =F-mg\mu \end{gathered}[/tex]Where f is the frictional force between the floor and the crate, F is the applied force, and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} F-253.2\times9.8\times0.55=0 \\ \Rightarrow F=253.2\times9.8\times0.55 \\ =1364.75\text{ N} \end{gathered}[/tex]Thus the force with which the crate must be pushed is 1364.75 N
A 12 ohm hair dryer is plugged into a 240 V power supply. What is the current?
Given,
The resistance of the hairdryer, R=12 Ω
The voltage of the power supply, V=240 V
From Ohm's law,
[tex]V=IR[/tex]Where I is the current.
On substituting the known values,
[tex]\begin{gathered} 240=I\times12 \\ \Rightarrow I=\frac{240}{12} \\ =20\text{ A} \end{gathered}[/tex]Thus the current is 20 A.
A person takes a trip, driving with a constant speed of 93.5 km/h, except for a 22.0-min rest stop. If the person's average speed is 71.6 km/h, find the following.(a) How much time is spent on the trip? h(b) How far does the person travel? km
We will have the following:
First, we determine the time:
[tex]\begin{gathered} \frac{93.5km/h+0km/h}{H}=71.6km/h\Rightarrow H=\frac{935}{716}h \\ \\ \Rightarrow H\approx1.3h \end{gathered}[/tex]So, the time spent on the trip was 935/716 h; that is approximately 1.3 hours.
Now, we deteremine the distance traveled:
[tex]\begin{gathered} d=(93.5km/h)(\frac{935}{716}-\frac{11}{30}h)\Rightarrow d=87.81513035...km \\ \\ \Rightarrow d\approx87.82km \end{gathered}[/tex]So, the distane traveled was approximately 87.82 km.
Directions: Solve the following problems. Show your solutions.2. A circuit has three resistors connected in parallel. Their resistances are 11 Ω, 17 Ω, and 12 Ω as shown on the figure below. Find for: a. Voltage in R1 (V1)b. Voltage in R2 (V2)c. Voltage in R3 (V3)d. Total Resistance (RT)e. Total Current (IT)f. Current in R1 (I1)g. Current in R2 (I2)h. Current in R3 (I3)
Since the resistances are in parallel, the voltage in each one is the same, so:
a. V1 = 60 V
b. V2 = 60 V
c. V3 = 60 V
d.
The total resistance of parallel resistances can be calculated with the formula below:
[tex]\begin{gathered} \frac{1}{RT}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}\\ \\ RT=\frac{R1\cdot R2\cdot R3}{R1R2+R2R3+R1R3}\\ \\ RT=\frac{11\cdot17\cdot12}{11\cdot17+17\operatorname{\cdot}12+11\operatorname{\cdot}12}\\ \\ RT=\frac{2244}{523}\\ \\ RT=4.29\text{ ohms} \end{gathered}[/tex]e.
The total current is given by the voltage divided by the total resistance:
[tex]IT=\frac{V}{RT}=\frac{60}{4.29}=13.99\text{ A}[/tex]The current in each resistor is given by the voltage divided by the resistance:
f.
[tex]I1=\frac{V1}{R1}=\frac{60}{11}=5.45\text{ A}[/tex]g.
[tex]I2=\frac{V2}{R2}=\frac{60}{17}=3.53\text{ A}[/tex]h.
[tex]I3=\frac{V3}{R3}=\frac{60}{12}=5\text{ A}[/tex]A ball is equipped with a speedometer and launched straight upward. The speedometer reading two seconds after launch is shown at the right; the ball is moving upward. At what approximate times would the ball display the following speedometer readings?
The time read by the speedometer is, t = 4 s.
The time displayed by the speedometer for a speed of 10 m/s is one second.The time displayed by the speedometer for a speed of 20 m/s is one second.The time displayed by the speedometer for a speed of 30 m/s is one second.What is Gravitational acceleration?The strength of a gravitational field is denoted by gravitational acceleration (symbolized g). It is measured in meters per second (meters per second squared). At the earth's surface, 1 g equals 9.8 m/s2.
Therefore,
The time read by the speedometer is, t = 4 s.
Because the speedometer has a low precision, value approximation is possible.
For the speedometer showing the speed 20 m/s. The time is calculated as,
v = u + gt1
Here, u is the initial speed and g is the gravitational acceleration and its approximate value is, g ≈ 10 m/s²
Solving as,
10 = 0 + 10 t1
t1 = 1s
Thus, the time shown by the speedometer corresponding to speed of 10 m/s is 1 s.
For the speedometer showing the speed 20 m/s. The time is calculated as,
v = u + gt2
Solving as,
20 = 10 + 10 t2
t2 = 1s
Thus, the time shown by the speedometer corresponding to speed of 20 m/s is 1 s.
For the speedometer showing the speed 30 m/s. The time is calculated as,
v = u + gt3
Solving as,
30 = 20 + 10 t3
t3 = 1s
Thus, the time shown by the speedometer corresponding to speed of 30 m/s is 1 s.
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Power equals work multiplied by time.Question 9 options:TrueFalse
The power in terms of work done and time is defined as,
[tex]P=\frac{W}{t}[/tex]Thus, power is is workdone per unit time.
Hence, given statement is false.
Suppose that the Towngas supply pressure is 8.5 kPa (gauge pressure) andthe total volume is 2.4 m? enters a building from outside where thetemperature is 10 °C and passes into a building where the temperature is 35°C, if the pressure was reduced to 2 kPa. What would be the new volume ofthe gas?
Given,
The initial pressure of the gas, P₁=8.5 kPa
The initial volume of the gas, V₁=2.4 m³
The initial temperature of the gas, T₁=10 °C=282.15 K
The temperature of the building, i.e., the temperature of the gas after entering the building, T₂=35 °C=308.15 K
The pressure of the gas after entering the building, P₂=2 kPa
Let us assume the new volume of the gas is V₂
From the combined gas law,
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2} \end{gathered}[/tex]On substituting the known values,
[tex]undefined[/tex]A 590 kg elevator accelerates upward at 1.1 m/s2 for the first 15 m of its motion. How much work is done during this part of its motion by the cable that lifts the elevator? Neglect any friction.
The work done by the elevator is - 86.730kj
It should be converted to energy in order to move an object. Energy can be transferred by the use of force. Work done refers to the amount of energy used by a force to move an object.
We are given that ,
The mass of the elevator = m = 590 kg
The acceleration of the elevator upward = 1.1m/s²
The height of the elevator = d = 15m
Therefore, T is the tension of the cable pulling the elevator upwards then the formulation of the work done of the elevator may be given as,
Work done = Force × distance
W = F × d
i.e. F = - mg
Above negative force due to gravity is acting opposite direction to the motion having an upward acceleration.
Thus , from above two equations we can write as,
W = - (mg)(d)
W = -(590kg)(9.8m/s²)(15m)
W = - 86730j
W = - 86.730kj
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I need help with the following question:
USING THE FOLLOWING CONVERSION FACTOR:
Conversions Longitude:
1 degree= 52.505 miles
1 minute= 4620.5 feet
Conversions for Latitude:
1 degree= 69.005 miles
1 minute= 6072.5 feet
How Do I convert these coordinates into feet?
The length of side A (40° 51.485' N, 74° 12.080' W & 40° 51.485' N, 74° 11.883' W) = ------------- ft
The length of side B (40° 51.485' N, 74° 12.080' W & 40° 51.800' N, 74° 11.883' W) = ----------- ft
The length of the side A is 1,13,25,866.44 ft N, 2,70,34,989.4 ft W
The length of the side B is 1,13,25,866.44 ft N, 2,70,34,989.4 ft W
Length is the measurement of anything from end to end.
Conversions Longitude:
Longitude is the angular distance of a location east or west of the meridian in Greenwich, England, or west of the standard meridian of a celestial object.
1 degree= 52.505 miles
1 minute= 4620.5 feet
Conversions for Latitude:
Latitude is the angular distance of a location north or south of the Earth's equator.
1 degree= 69.005 miles
1 minute= 6072.5 feet
Also 1 mile = 5280 feet
The length of side A (40° 51.485' N, 74° 12.080' W)
40° 51.485' N = 40 * 52.505 * 5280 + 51.485 * 4620.5
40° 51.485' N = 1,10,88,000 + 2,37,866.44
40° 51.485' N = 1,13,25,866.44 feet.
74° 12.080' W = 74 * 69.005 * 5280 + 12.080 * 6072.5
74° 12.080' W = 2,69,61,633.6 + 73,355.8
74° 12.080' W = 2,70,34,989.4 feet
The length of side B (40° 51.485' N, 74° 12.080' W)
40° 51.485' N = 1,13,25,866.44 feet.
74° 12.080' W = 2,70,34,989.4 feet.
Side A = Side B = 1,13,25,866.44 ft N, 2,70,34,989.4 ft W.
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A __________ is described as a device with specific resistance and is used to control current.batteryresistorparallel connectionseries connection
ANSWER:
resistor
STEP-BY-STEP EXPLANATION:
A device that has a specific resistance and is also used to control current is the resistor.
Therefore:
A resistor is described as a device with specific resistance and is used to control current.
The ammeter in the figure below reads 1.0A. Calculate the magnitude of the currents I1 and I2.
Given:
• Ammeter reading, I = 1.0 A
,• R1 = 2.0 Ω
,• R2 = 4.0 Ω
,• R3 = 5.0 Ω
Let's find the magnitude of the currents in I1 and I2.
Apply Kirchhoff's Current Law.
We have:
[tex]I_1+I_2+I_3=0[/tex]From Kirchoff's junction rule:
[tex]I_1-I_2-I_3=1A[/tex]Thus, we have:
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5} \\ \\ \frac{1}{R_{eq}}=\frac{10+5+4}{20}=\frac{19}{20} \\ \\ R_{eq}=\frac{20}{19}=1.1\text{ ohms} \end{gathered}[/tex]Apply Ohm's law to find the voltage in the circuit:
[tex]\begin{gathered} V=IR \\ \\ V=1.0*1.1 \\ \\ V=1.1\text{ V} \end{gathered}[/tex]To find I1, apply Kirchoff's loop law:
[tex][/tex]carts, bricks, and bands
2. Which of the following conclusions are specifically supported by the data in Table 1?
a. A constant mass causes the acceleration value to increase.
b. An increase in the number of bricks causes the acceleration to decrease.
c. An increase in the length of the rubber band causes the acceleration to increase.
d. An increase in the number of rubber bands causes an increase in the acceleration.
The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. 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.
This is because increasing the number of rubber bands has the effect of doubling the force leading to an effective increase in velocity of the moving cart.
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The fact that light can be polarized and sound cannot be polarized can be best explained by which?a) Light is a transverse wave and sound is not.b) Light travels much faster than sound.c) Light is a longitudinal wave, and sound is not.d) Light travels much slower than sound.
a) Light is a transverse wave and sound is not.
Explanation:
Light waves are transverse waves, their direction of propagation is perpendicular to the direction of vibration. Transverse waves can be polarized.
In
A boy walks 30m [E25°S] then 60m [E40°N]. Determine his net displacement.
Given,
The distances; a=30 m
b=60 m
Angles; θ=E25°S
α=E40°N
From the diagram, ∠A is given by,
[tex]\angle A=180\degree-\theta-\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \angle A=180\degree-25\degree-40\degree \\ =115\degree^{} \end{gathered}[/tex]From the cos rule,
[tex]d^2=a^2+b^2-2ab\cos A[/tex]On substituting the known values,
[tex]\begin{gathered} d^2=30^2+60^2-2\times30\times60\times\cos 115\degree \\ \Rightarrow d=\sqrt[]{30^2+60^2-2\times30\times60\times\cos 115\degree} \\ =77.6\text{ m} \end{gathered}[/tex]Thus the total displacement of the boy is 77.6 m
if i kicked a empty soda can would it travel further than a filled up soda can, if so why? & what newton law would this be ?
If we kick both sodas with the same force, the soda that has a higher weight will have a lower acceleration. This is explained by Newton's second law of motion
The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
F = ma
where
m = mass
a = accelaration
F = Force
Starting from rest, billy goes down a slide (height of 2.5m above ground) and billy’s mass is 35kg. There is friction acting on billy, when he reaches the ground his speed is 3.5 m/s. How much mechanical energy was lost due to friction (give answer in joules)
In order to determune the amount of mechanicak energy lost, consider that the potential energy at the starting point must be equal to the sum of the energy lost due to the friction, plus the kinetic energy on the ground.
Then, you can write:
[tex]U=E_r+K[/tex]the amount of energy lost is energy associated to the friction.
Solve the previou equation for Er, replace the expressions for U and K, and replace the values of the given parameters, as follow:
[tex]\begin{gathered} E_r=U-K=(35kg)(9.8\frac{m}{s^2})(2.5m)-\frac{1}{2}(35kg)(3.5\frac{m}{s})^2 \\ E_r=643.12J \end{gathered}[/tex]Hence, the amoun of the energy lost was approximately 643.12J
sin ([3pi over 2] + x) + sin ([3pi over 2] + x) = -2must show work also
ANSWER:
[tex]x=0\text{\degree}+360\text{\degree{}n}[/tex]STEP-BY-STEP EXPLANATION:
We have the following equation:
[tex]sin\: \mleft(\mleft[\frac{3\pi}{2}\mright]+x\mright)\: +\: sin\: \mleft(\mleft[\frac{3\pi\:}{2}\mright]+x\mright)\: =\: -2[/tex]Solving for x:
[tex]\begin{gathered} 2\cdot sin\: (\lbrack\frac{3\pi}{2}\rbrack\: +\: x)\: \: =\: -2 \\ sin\: (\lbrack\frac{3\pi}{2}\rbrack\: +\: x)=-\frac{2}{2} \\ sin\: (\lbrack\frac{3\pi}{2}\rbrack\: +\: x)=-1 \\ \frac{3\pi}{2}+\: x=\arcsin (-1) \\ \frac{3\pi}{2}+\: x=\frac{3\pi}{2}+\: 2\pi n \\ x=2\pi n \\ x=0\text{\degree}+360\text{\degree{}n} \\ \text{for n = 0} \\ \end{gathered}[/tex]A +10.31 nC charge is located at (0,8.47) cm and a -2.09 nC charge is located (3.91, 0) cm. Where would a -14.84 nC charge need to be located in order that the electric field at the origin be zero? Express your answer, in cm, as the magnitude of the distance of q3 from the origin.
To make the E-field at the origin become 0, we need to find the E-field at the origin before
E = kq/r^2
E1 = 12934 V/m
E2 = 12303.69 V/m
Etotal = 17851.34 V/m
17851.34 = kq/r^2
r = 8.6526 cm