Answers
Answer:
4 m/s
Explanation:
To find the velocity at t = 8 seconds, we will use the following equation:
[tex]v_f=v_i+at[/tex]Where vf is the final velocity, vi is the initial velocity, a is the acceleration and t is the time.
From t = 0 seconds to t = 3 seconds, we have an acceleration of 6 m/s², so we can calculate the velocity at t = 3 seconds as:
[tex]\begin{gathered} v_f=-4m/s+6m/s^2(3\text{ s)} \\ v_f=-4\text{ m/s + 18 m/s} \\ v_f=14\text{ m/s} \end{gathered}[/tex]Now, from t = 3 seconds to t = 8 seconds, the acceleration is equal to -2 m/s². So we need to use the same equation but this time, the initial velocity will be 14 m/s and the time will be 5 seconds because t = 8 s - 3 s = 5s. Then, we get:
[tex]\begin{gathered} v_f=14m/s-2m/s^2(5s) \\ v_f=14\text{ m/s - 10 m/s} \\ v_f=4\text{ m/s} \end{gathered}[/tex]Therefore, the velocity at t = 8 seconds is 4 m/s
Related Questions
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?
Answers
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 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?
Answers
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
Answers
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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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.
Answers
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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What momentum does a car of mass 1,116 kg have if it’s traveling at 18m/s? Submit anwser in exponential form.
Answers
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
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?
Answers
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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An electron is in an infinite one dimension well that is 8.9 nm wide. What is the ground state energy of the electron?
Answers
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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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_________
Answers
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.
An object on a horizontal, frictionless surface is attached to a spring, displaced, and then released. If it isdisplaced 0.12m from its equilibrium position and released after 0.8s its displacement is found to be 0.12m onthe opposites side, and passed the equilibrium position once during the interval. Find:
Answers
Given:
The maximum displacement from the equilibrium position is A = 0.12 m.
Half of the time period is
[tex]T_{\frac{1}{2}}=\text{ 0.8 s}[/tex]To find the amplitude, time period, and frequency.
Explanation:
Amplitude is the maximum displacement from the equilibrium position.
Thus, the amplitude is A = 0.12 m.
One time period is the time taken from maximum displacement on one side(say A) to maximum displacement on the opposite side and back to the maximum displacement on the same side(A).
Thus, the time period is
[tex]\begin{gathered} T=2T_{\frac{1}{2}} \\ =2\times0.8 \\ =1.6\text{ s} \end{gathered}[/tex]The frequency will be
[tex]\begin{gathered} f=\frac{1}{T} \\ =\frac{1}{1.6} \\ =0.625\text{ Hz} \end{gathered}[/tex]What does this image reveal about gravityon the moon compared to Earth?
Answers
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.
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
Answers
Answer:
quizlet
Explanation:
they help
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?
Answers
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
Find the magnitude of the sumof these two vectors:B63.5 m101 m57.0°
Answers
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°.
A 0.327-kg model rocket accelerates at 35.7 m/s/s on takeoff. Determine the upward thrust experienced by the rocket
Answers
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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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?
Answers
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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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
Answers
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
It takes 5 seconds for a 2 kg box to be pushed 10 meters from rest. What was the forceof the push?
Answers
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.
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 .
Answers
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.
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?
Answers
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]
A negative charge of .30 c and a positive charge of .50 are separated by .40What is the force between the charges?
Answers
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.
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.)
Answers
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 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?
Answers
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:
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
Answers
Answer: The answer is potential energy!
Explanation:
since kelly is using her own energy it would be potential.
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
Answers
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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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.
Answers
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?
Calculate the kinetic energy of a 3.15 kg bowling ball rolling down the lane at 2.5 m/s. Include the units.
Answers
Given:
The mass of the ball is: m = 3.15 kg
The velocity of the ball is: v = 2.5 m/s
To find:
The kinetic energy of the ball.
Explanation:
The kinetic energy is the energy that a particle has when is is in motion. For the particle of mass m moving with velocity v, the kinetic energy KE is given as:
[tex]KE=\frac{1}{2}mv^2[/tex]Substitute the values in the above expression, we get:
[tex]\begin{gathered} KE=\frac{1}{2}\times3.15\text{ kg}\times(2.5\text{ m/s\rparen}^2 \\ \\ KE=\frac{1}{2}\times3.15\text{ kg}\times6.25\text{ m}^2\text{/s}^2 \\ \\ KE=9.84\text{ kg.m}^2\text{/s}^2 \\ \\ KE=9.84\text{ J} \end{gathered}[/tex]Final answer:
The kinetic energy of the ball is 9.84 J.
How much work is done on a medicine ball with a force of 29 newtons when you lift it 5 meters?
Answers
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
Which of the following is needed for an electric circuit to work?A. Fusion sourceB. Nuclear sourceC. Chemical sourceD. Energy sourceVoltage is a measure of the ________ it gives to the current.A. AccelerationB. SpeedC. PullD. Push
Answers
We will have the following:
The "part" needed for an electric circuit is an energy source.
Voltage is the measure of the acceleration it gives to the current.
Harry Hand can run 100m in 20s. His daughter, Linda Hand, canrun 50m in 8.5s. Who was faster
Answers
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.
Describe
protons.
Location:
Charge:
Mass:
Answers
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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