Given:
height = 65 m
Given that the object is in free fall, let's solve for the following:
• (a). determine the final speed in m/s.
To find the final velocity, apply the kinematics equation:
[tex]v^2=u^2-2ax[/tex]Where:
v is the final velocity
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/s²
x is the displacement = 65 m
Thus, we have:
[tex]\begin{gathered} v^2=0^2-2(-9.8)(65) \\ \\ v^2=-(-1274) \\ \\ v^2=1274 \\ \\ \text{ Take the square root of both sides:} \\ \sqrt{v^2}=\sqrt{1274} \\ \\ v=35.69\text{ m/s} \end{gathered}[/tex]Therefore the final speed will be -35.69 m/s.
• (c). The distance traveled during the last second of motion before hitting the ground.
To find the distance, apply the formula:
[tex]H=ut+\frac{1}{2}at^2[/tex]Where:
H is the height.
u is the initial velocity = 0 m/s
t is the time
a is acceleration due to gravity.
Let's rewrite the formula to find the time traveled.
[tex]\begin{gathered} H=0t+\frac{1}{2}at^2 \\ \\ H=\frac{1}{2}at^2 \\ \\ t=\sqrt{\frac{2H}{a}} \end{gathered}[/tex]Thus, we have:
[tex]\begin{gathered} t=\sqrt{\frac{2*65}{9.8}} \\ \\ t=\sqrt{\frac{130}{9.8}} \\ \\ t=\sqrt{13.26} \\ \\ t=3.64\text{ s} \end{gathered}[/tex]Therefore, the time is 3.64 seconds.
Now, to find the distance traveled during the last second of motion, apply the formula:
[tex]s=\frac{1}{2}a(t_2^2-t_1^2)[/tex]Where:
t2 = 3.64 seconds
t1 = 3.64 seconds - 1 second = 2.64 seconds
Thus, we have:
[tex]\begin{gathered} s=\frac{1}{2}(9.8)((3.64)^2-(2.64)^2) \\ \\ s=4.9(13.2496-6.9696) \\ \\ s=4.9(6.28) \\ \\ s=30.77 \end{gathered}[/tex]Therefore, the distance in meters, traveled during the last second of motion before hitting the ground is 30.77 meters.
ANSWER:
(A). -35.69 m/s
(C). 30.77 m
If there is a source voltage of 12 volts with a 2.5 volt drop over the LED, and an intended current of 100 mA, what size resistor should be used?
Given:
A voltage source has a voltage of 12 V.
A voltage drop over the LED is 2.5 V.
The current in the circuit is 100 mA.
To find:
The resistance of the resistor should be used.
Explanation:
LED is a forward-biased PN junction that emits light. The voltage drop over the LED is 2.5 V.
The voltage source has a voltage of 12 V.
Hence, the total voltage in the circuit = 12 V - 2.5 V = 9.5 V
The resistance of the circuit in the forward biased is almost equal to zero. Let the resistance in the circuit be R.
Thus, the total resistance in the circuit = 0 + R = R
The current in the circuit is = 100 mA = 0.1 A
The resistance R of the circuit can be calculated as:
[tex]R=\frac{V}{I}[/tex]Here, V is the voltage in the circuit and I is the current in the circuit.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} R=\frac{9.5\text{ V}}{0.1\text{ A}} \\ \\ R=95\text{ }\Omega \end{gathered}[/tex]Final answer:
Hence, a resistor of 95 Ω should be used in the circuit.
What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored in the 2.50 µF capacitor?
The total energy stored in the capacitors is determined as 2.41 x 10⁻⁴ J.
What is the potential difference of the circuit?The potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
where;
C is capacitance of the capacitorV is the potential differenceFor a parallel circuit the voltage in the circuit is always the same.
The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
2U = CV²
V = √2U/C
V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)
V = 12 V
The equivalent capacitance of C1 and C2 is calculated as follows;
1/C = 1/C₁ + 1/C₂
1/C = (1)/(0.9 x 10⁻⁶) + (1)/(16 x 10⁻⁶)
1/C = 1,173,611.11
C = 1/1,173,611.11
C = 8.52 x 10⁻⁷ C
The total capacitance of the circuit is calculated as follows;
Ct = 8.52 x 10⁻⁷ C + 2.5 x 10⁻⁶ C
Ct = 3.35 x 10⁻⁶ C
The total energy of the circuit is calculated as follows;
U = ¹/₂CtV²
U = ¹/₂(3.35 x 10⁻⁶ )(12)²
U = 2.41 x 10⁻⁴ J
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Describe
protons.
Location:
Charge:
Mass:
The protons in an atom are classified according to their mass, charge, and location as follows:
Particle: Protons
Mass: 1.67262 × 10⁻²⁷ kg
Charge: Positive charge (+e or +1)
Location: Found in the nucleus of every atom
The proton is a stable subatomic particle with a rest mass of 1.67262 x 10⁻²⁷ kg, or 1,836 times the mass of an electron, with a positive charge that is equal to one electron's charge in magnitude.
All atomic nuclei, with the exception of the hydrogen nucleus, are composed of protons and neutrons, which are electrically neutral particles (that consist of a single proton). A given chemical element's nuclei all contain the same number of protons. This number establishes an element's atomic number and establishes the element's position in the periodic table. An atom is electrically neutral when the number of protons in its nucleus equals the number of electrons in its orbit.
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It takes 5 seconds for a 2 kg box to be pushed 10 meters from rest. What was the forceof the push?
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.
A motorcycle has a constant acceleration of 3.74 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a)21.5 to 31.5 m/s, and (b)51.5 to 61.5 m/s?(a)Number ______ Units_________(b)Number ______ Units_________
Since the acceleration is constant we know that it is given by:
[tex]a=\frac{v_f-v_0}{t}[/tex]from where we have that:
[tex]t=\frac{v_f-v_0}{a}[/tex]once we have this equation we can determine the time by plugging the values of the acceleration and velocities.
a)
In this case we have that:
[tex]t=\frac{31.5-21.5}{3.74}=2.67[/tex]Therefore it takes 2.67 s.
b)
In this case we have that:
[tex]t=\frac{61.5-51.5}{3.74}=2.67[/tex]Therefore it takes 2.67 s.
A blink of an eye is a time interval of about 150ms for an average adult. The closure portion of the blink takes only about 55ms. Let us model the closure of the upper eyelid as uniform angular acceleration through an angular displacement of 16.6 degree. What is the value of the angular acceleration the eyelid undergoes while closing 2. What is the tangential acceleration of the edge of the eyelid while closing if the radius of the eyeball is 1.25 cm?
ANSWER:
STEP-BY-STEP EXPLANATION:
The first thing is to convert the time into a second, just like this:
[tex]t=55\text{ ms}\cdot\frac{1\text{ s}}{1000\text{ ms}}=0.055\text{ s}[/tex]Now, convert the angular displacement of the eyelid from degrees to rad:
[tex]\partial\theta=16.6\text{\degree}\cdot\frac{2\pi\text{ rad}}{360\text{\degree}}=0.29\text{ rad}[/tex]We can calculate the angular velocity, dividing the angular momentum by the time, like this:
[tex]w=\frac{0.29}{0.055}=5.27\text{ rad/s}[/tex]The angular acceleration is calculated by means of the quotient of the difference in angular velocity and time, like this:
[tex]a_w=\frac{\delta w}{\delta t}=\frac{5.27-0}{0.15-0.055}=55.47\text{ rad/s}^2[/tex]the tangential acceleration would be:
An object of mass m moves a circular path with a constant speed v. The centripetal force of the object is F. If the objects speed were halved in the mass was tripled, what would happen to the centripetal force?
An object of mass m moves a circular path with a constant speed v. The centripetal force of the object is F. If the object's speed were halved in the mass was tripled, then the centripetal force would be 0.75 times the original centripetal force.
What is a uniform circular motion?It is defined as motion when the object is moving in a circle with a constant speed and its velocity is changing with every moment because of the change of direction but the speed of the object is constant in a uniform circular motion.
A mass m object travels in a circle at a constant speed v. The object's centripetal force is F. The centripetal force would be 0.75 times greater if the object's mass were tripled and its speed was cut in half.
Centripetal force = m × v²/r
=3m × (0.5v)² / r
= 0.75 mv² / r
Thus, the centripetal force would become 0.75 times the original centripetal force.
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What does this image reveal about gravityon the moon compared to Earth?
ANSWER:
The Moon's gravity is less than Earth's.
STEP-BY-STEP EXPLANATION:
When you jump, you fall back to the ground. Apples or leaves also fall: we are all attracted to the Earth. It is the terrestrial attraction due to the force of gravity.
The force of gravity also exists on the Moon. But since the Moon is smaller than the Earth, the attraction felt on the Moon is smaller than the Earth's attraction.
As the gravity is less, you can do things such as the one shown in the image.
As the force of attraction is less, the weight is less on the Moon, which can cause things that would be impossible on Earth.
What momentum does a car of mass 1,116 kg have if it’s traveling at 18m/s? Submit anwser in exponential form.
ANSWER:
20088 kg*m/s
STEP-BY-STEP EXPLANATION:
The momentum is given by the following formula:
[tex]p=m\cdot\Delta v[/tex]We know the mass and also the speed, therefore:
[tex]\begin{gathered} p=1116\cdot18 \\ p=20088\text{ kg}\cdot\frac{m}{s} \end{gathered}[/tex]Por lo tanto, el momento es igual a 20088 kg*m/s
25. A student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force?
Work = change in energy
Fd = 1/2 mv^2
F = 1/2 x 120 x 5^2 / 10
F = 150 N
If a student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m, then the average braking force would be 150 Newtons.
What is power?The rate of doing work is known as power. The Si unit of power is the watt.
Power =work / time
Work done by the braking force = change in kinetic energy
F × s = 1/2 × m × v²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force would be 150 Newtons.
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A golf ball is initially on a tee when it is
struck by a golfer. The ball is given an
initial velocity of 50 m/s at a 37° angle. The
ball hits the side of a building that is 200
meters horizontally away from the golfer.
(a) What are the horizontal and vertical
components of the ball's initial
velocity?
(b) How much time elapses before the
ball strikes the side of the building?
(c) How far from the ground does the ball
strike the building?
Answer:
a.
[tex]horizontal=39.9[/tex] m/s
[tex]vertical=30.1[/tex] m/s
b.
[tex]t=5.009[/tex]
c.
[tex]y=27.7[/tex]
Explanation:
Lets write down what we were given.
Angle = 37°
Initial Velocity = 50 m/s
Displacement in x direction = 200 m
Take note:
I am having some trouble with the theta symbol so let theta = [tex]N[/tex]
Lets do question C first.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
Rearranging terms, we have
[tex]y=(tan(N)*x)-[\frac{g}{2(v_{0}cos(N))^{2} } ]x^{2}[/tex]
Now lets substitute our numbers in for the variables. Then simplify.
[tex]y=(tan37*200)-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[0.0030761]200^{2}[/tex]
[tex]y=150.7108-(0.0030761*40000)[/tex]
[tex]y=150.7108-123.0444[/tex]
[tex]y=27.7[/tex]
Now lets do question B.
Lets steal this from the last question.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
We can substitute [tex]t[/tex] for [tex]\frac{x}{v_{0}cos(N) }[/tex]
[tex]y=(v_{0}sin(N))*(t)-\frac{1}{2}g(t) ^{2}[/tex]
We can rewrite the equation as
[tex](v_{0}sin(N)(t)-\frac{1}{2}*(g(t)^{2})=y[/tex]
Now lets substitute our numbers in for the variables.
[tex](50sin(37)(t)-\frac{1}{2}*(9.81(t)^{2})=27.7[/tex]
After some painful algebra and factoring we get
[tex]30.09075115t-4.905t^{2}=27.6664[/tex]
Subtract [tex]27.6664[/tex] from both sides.
[tex]30.09075115t-4.905t^{2}-27.6664=0[/tex]
Use the quadratic formula to find the solutions.
[tex]\frac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]
After some more painful algebra we get
[tex]t=5.00854263, 1.12616708[/tex]
1.126 does not make any sense so.
[tex]t=5.009[/tex]
Finally lets do question A.
Lets draw a triangle. We have the velocity which is the hypotenuse and we have the angle. From there we can solve for the opposite and adjacent sides.
Let [tex]A=horizontal[/tex] and [tex]O=vertical[/tex]
[tex]cos(37)=\frac{A}{50}[/tex]
[tex]A=39.9[/tex]
[tex]sin37=\frac{O}{50}[/tex]
[tex]O=30.1[/tex]
Two 4.587 cm by 4.587 cm plates that form a parallel-plate capacitor are charged to +/- 0.671 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.257 mm?
ANSWER:
3.6 x 10^6 N/C
STEP-BY-STEP EXPLANATION:
Given:
Charge (q) = 0.671 nC = 0.671 x 10^-9 C
Side (s) = 4.587 cm = 4.587 x 10^-3 m
Vacuum permittivity (ε0) = 8.85 x 10^-12 F/m
We can calculate the electric field using the following formula:
[tex]\begin{gathered} E=\frac{q}{ε_0\cdot A} \\ \\ \text{ We replacing:} \\ \\ E=\frac{0.671\cdot10^{-9}}{(8.85\cdot10^{-12})(4.587\cdot10^{-3})(4.587\cdot10^{-3})} \\ \\ E=\:3603477.12=3.6\cdot10^6\text{ N/C} \end{gathered}[/tex]The electric field is equal to 3.6 x 10^6 N/C
Determine the resistance, in milliOhms, of a metal rod 2.96 m long, 0.89cm diameter and composed of aluminum of resistivity 2.8 x 10-8 Ωm .
The resistance R of a rod with length L, cross-sectional area A and resistivity ρ is given by:
[tex]R=\frac{\rho L}{A}[/tex]On the other hand, the area of a circle with diameter D is given by:
[tex]A=\frac{\pi}{4}D^2[/tex]Then, the resistivity of the rod in terms of its diameter is:
[tex]R=\frac{4\rho L}{\pi D^2}[/tex]Replace L=2.96m, D=0.89cm and ρ=2.8×10^(-8)Ωm to find the resistance of the metal rod:
[tex]\begin{gathered} R=\frac{4\rho L}{\pi D^2} \\ \\ =\frac{4(2.8\times10^{-8}\Omega m)(2.96m)}{\pi(0.89cm)^2} \\ \\ =\frac{4(2.8\times10^{-8}\Omega m)(2.96m)}{\pi(0.89\times10^{-2}m)^2} \\ \\ =1.332232...\times10^{-3}\Omega \\ \\ \approx1.33m\Omega \end{gathered}[/tex]Therefore, the resistance of the metal rod is approximately 1.33 miliOhms.
Harry Hand can run 100m in 20s. His daughter, Linda Hand, canrun 50m in 8.5s. Who was faster
The speed is given as;
[tex]v=\frac{d}{t}[/tex]Here, d is the distance covered and t is the time.
Harry Hand covers a distance of 100 m (d_H) in 20 s (t_H). Therefore, the speed of the Harry Hand is,
[tex]v_H=\frac{d_H}{t_H}[/tex]Substituting all known values,
[tex]\begin{gathered} v_H=\frac{100\text{ m}}{20\text{ s}} \\ =5.0\text{ m/s} \end{gathered}[/tex]Now, Linda Hand covers a distance of 50 m (d_L) in 8.5 s (t_L). Therefore, the speed of Linda Hand is given as,
[tex]v_L=\frac{d_L}{t_L}[/tex]Substituting all known values,
[tex]\begin{gathered} v_L=\frac{50\text{ m}}{8.5\text{ s}} \\ \approx5.88\text{ m/s} \end{gathered}[/tex]Since the speed of Linda Hand is greater than Harry Hand (v_L>v_H). Therefore, Linda Hand is faster.
A light, inextensible cord passes over alight, frictionless pulley with a radius of15 cm. It has a(n) 18 kg mass on the left and a(n) 2.6 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1.5 m apart.The acceleration of gravity is 9.8 m/s².
At what rate are the two masses accelerating when they pass each other answer in units of m/s^2
Answer:
quizlet
Explanation:
they help
Two railcars have a head-on collision, couple together, and stop dead. If Car A was moving four times as quickly as Car B was, and the total mass of both cars together is 90,000 kg, what are the masses of each car individually?
If Car A was moving four times as quickly as Car B was, and the total mass of both cars together is 90,000 kg, then the mass of each car would have been 18000 kilograms and 72000 kilograms respectively.
What is momentum?It can be defined as the product of the mass and the speed of the particle.
As given in the problem Two railcars have a head-on collision, couple together, and stop dead. If Car A was moving four times as quickly as Car B was, and the total mass of both cars together is 90,000 kg,
Let us suppose the mass of Car A would have been X kilograms
Mass of car B = ( 90000 - X ) kilograms
Given the final momentum of the system is zero, therefore by using the conservation of the momentum
4 × X + -1 ( 90000 - X ) = 0
5X = 90000 kilograms
X = 18000 Kilograms
Thus, the mass of each car would be 18000 kilograms and 72000 kilograms respectively.
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part B:
Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
The magnitude of the acceleration of the box is 9.65 m/s².
What is the net force of the box?
The net force on the box is calculated as follows;
F(net) = F - Ff
where;
F is the applied forceFf is the force of frictionF(net) = F - μmgcosθ
where;
μ is the coefficient of friction given as 0.3θ is the angle of inclination of the plane = 55⁰m is the mass of the box = 15 kgF(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.71 N
The magnitude of the acceleration of the box is calculated as;
a = F(net) / m
a = (144.71) / (15)
a = 9.65 m/s²
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A negative charge of .30 c and a positive charge of .50 are separated by .40What is the force between the charges?
Given:
The negative charge q1 = 0.3 C
The positive charge q2 = 0.5 C
The distance between the charges is 0.4 m
To find the magnitude of the force between them.
Explanation:
The formula to calculate the magnitude of the force is
[tex]F=\frac{kq1q2}{r^2}[/tex]Here, k is Coulomb's constant whose value is
[tex]k=\text{ 9}\times10^9Nm^2C^{-2}[/tex]On substituting the values, the magnitude of force will be
[tex]\begin{gathered} F=\frac{9\times10^9\times0.3\times0.5}{(0.4)^2} \\ =8.4375\text{ }\times10^9\text{ N} \end{gathered}[/tex]Final Answer: The magnitude of the force is 8.4375 x 10^(9) N.
How much work is done on a medicine ball with a force of 29 newtons when you lift it 5 meters?
Given data
*The given force is F = 29 N
*The given distance is s = 5 m
The formula for the work is done on a medicine ball is given as
[tex]W=F\mathrm{}s[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} W=(29)(5) \\ =145\text{ J} \end{gathered}[/tex]Hence, the work is done on a medicine ball is W = 145 J
9. A yo-yo is moving in a horizontal circle of radius R. the yo-yo has a mass of 0.250 kg has a speed of 9 m/s and experience this a centripetal force of 26.6 N what is the radius of the circle that the yo-yo is moving in?
ANSWER:
B. 0.761 meters
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 0.250 kg
centripetal force (Fc)= 26.6 N
Speed (v) = 9m/s
We have that the centripetal force can be calculated using the following formula:
[tex]F_c=\frac{m\cdot v^2}{r}[/tex]We substitute each value and solve for the radius, just like this:
[tex]\begin{gathered} r=\frac{m\cdot v^2}{F_c} \\ r=\frac{0.25\cdot9^2}{26.6} \\ r=\frac{0.25\cdot81}{26.6} \\ r=0.761\text{ m} \end{gathered}[/tex]The radius is equal to 0.761 meters
An electron is in an infinite one dimension well that is 8.9 nm wide. What is the ground state energy of the electron?
The ground state energy of the electron is 2.23 x 10⁻¹⁷ J.
What is the ground state energy of the electron?The ground state energy of the electron is calculated by applying the formula for energy of photons.
E = hf
E = hc/λ
where;
h is Planck's constantc is speed of lightλ is the wavelengthE = (6.626 x 10⁻³⁴ x 3 x 10⁸) / (8.9 x 10⁻⁹)
E = 2.23 x 10⁻¹⁷ J
Thus, the ground state energy of the electron is determined by applying the principle or formula for energy of a single photon at the given wavelength of 8.9 nm.
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Kelly uses kinetic energy from her body to lift a heavy box from the floor up onto a table. What type of energy does the box now
have?
A. light energy
B. kinetic energy
C. electrical energy
D. potential energy
Answer: The answer is potential energy!
Explanation:
since kelly is using her own energy it would be potential.
A 0.327-kg model rocket accelerates at 35.7 m/s/s on takeoff. Determine the upward thrust experienced by the rocket
The upward thrust of the rocket is determined as 11.67 N.
What is the upward thrust of the rocket?The upward thrust of the rocket is calculated by applying Newton's second law of motion as shown below;
F = ma
where;
m is the mass of the rocketa is the upward acceleration of the rocketSubstitute the given parameters and solve for the upward thrust of the rocket.
F = (0.327 kg) x (35.7 m/s²)
F = 11.67 N
Thus, the upward thrust of the rocket is determined as 11.67 N.
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What is the speed, in m/s, of a wave on a cord if it has a wavelength of 3.5 m and a period of 0.5 s?
The wavelength, period and velocity are related by the equation:
[tex]v=\frac{\lambda}{T}[/tex]where λ is the wavelength and T is the period. In this case the wavelength is 3.5 m and the period is 0.5 s; plugging these values we have:
[tex]\begin{gathered} v=\frac{3.5}{0.5} \\ v=7 \end{gathered}[/tex]Therefore, the speed of the wave is 7 m/s
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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The couple required to hold a triple turn of 1.5cm² area in equilibrium when carrying a current 2A at 70° to a field with 0.15T is?
The couple or torque required to hold the triple turn is 1.27 x 10⁻⁴ Nm.
What is the couple or torque required?
The couple required to hold the triple turn is calculated as follows;
τ = M x Bsinθ
where;
M is the magnetic moment B is the magnetic field strengthThe magnetic moment is calculated as follows;
M = NIA
where;
N is number of turns = 3I is current = 2 AA is the area of the loop = 1.5 cm² = 0.00015 m²M = (3) x (2) x (0.00015)
M = 0.0009 m²A
The torque or couple required is calculated as;
τ = (0.0009) x (0.15 x sin70)
τ = 1.27 x 10⁻⁴ Nm
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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.
Find the magnitude of the sumof these two vectors:B63.5 m101 m57.0°
Vector diagram:
The resultant vector is given as,
[tex]R=\sqrt[]{A^2+B^2+2AB\cos \theta}[/tex]Here, θ is the angle between vector A and B.
Substituting all known values,
[tex]\begin{gathered} R=\sqrt[]{(63.5)^2+(101)^2+2\times101\times63.5\times\cos (33^{\circ})} \\ =158.08\text{ m} \end{gathered}[/tex]Therefore, the resultant magnitue of the sum of these two vectors are 158.08 m.
The x-component of the magnitude is given as,
[tex]\begin{gathered} R_x=101\cos (57^{\circ})+63.5\cos (90^{\circ}) \\ =55.0\text{ m} \end{gathered}[/tex]The y- component of the magnitude is given as,
[tex]\begin{gathered} R_y=63.5\sin (90^{\circ})+101\sin (57^{\circ}) \\ =148.2\text{ m} \end{gathered}[/tex]Therefore, the direction is given as,
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{R_y}{R_x}) \\ =\tan ^{-1}(\frac{148.2\text{ m}}{55.0\text{ m}}) \\ =69.63^{\circ} \end{gathered}[/tex]Therefore, the direction of the resultant vector is 69.63°.
The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.60 m/s
at an angle of 50.0 ∘
above the horizontal, and aims for a beetle on a leaf 2.30 cm above the water's surface.
The maximum height reached by the water is 20.2 cm and it will dislodge the beetle.
What is the maximum height reached by the water?
The maximum height reached by the water squirted by the arch fish is calculated by applying the following kinematic equation.
H = (v² sin²θ) / 2g
where;
v is the speed of the waterθ is the angle of projection of the waterg is acceleration due to gravityH = (2.6² x (sin50)² ) / (2 x 9.8)
H = 0.202 m
H = 20.2 cm
Thus, the water squirted by the arch fish is dislodge the beetle.
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The complete question is below:
The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.60 m/s
at an angle of 50.0 ∘ above the horizontal, and aims for a beetle on a leaf 2.30 cm above the water's surface. Will the water squirted by the arch fish dislodge the beetle?
Three resistors having values of 4 Ω, 6 Ω , and 8Ω are connected in series. Their equivalent resistance is ______.Group of answer choices18 Ω8 Ω6 Ω1.80 Ω
Answer:
18Ω
Explanation:
If the resistors are connected in series, the equivalent resistance is the sum of each resistance, so
Equivalent resistance = 4Ω + 6Ω + 8Ω
Equivalent resisteance = 18Ω
Therefore, the answer is 18Ω