Given data
*The given mass is m = 0.520 kg
*The spring stretches at a distance is x = 18.7 cm = 0.187 m
*The value of the acceleration due to gravity is g = 9.8 m/s^2
(a)
The formula for the spring constant of the spring is given as
[tex]\begin{gathered} F=kx \\ mg=kx \\ k=\frac{mg}{x} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} k=\frac{(0.520)(9.8)}{(0.187)} \\ =27.2\text{ N/m} \end{gathered}[/tex]Hence, the spring constant of the spring is k = 27.2 N/m
(b)
The formula for the frequency of its
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
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.
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
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.
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
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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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.
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?
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 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:
The diagram below represents a 2.0 kg toy car moving at across the speed of 3.0 meters per 2nd counter clockwise in a circular path with a radius of 2.0 meters.At the Instant shown in the diagram, the direction of the centripetal force acting on the car is_____.
Given data
The mass of the toy car is m = 2 kg
The speed of the car is v = 3 m/s
The radius of the circular track is r = 2 m
The centripetal force is always in the same direction as that of the centripetal acceleration.
The centripetal acceleration direction is towards the center of the circle, towards west.
Therefore, the direction of the centripetal force points to the west direction.
Thus, the direction of the centripetal force at this instant is towards the west.
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]
Kinetic energy differs from potential energy inA. Kinetic energy can be created or destroyed, while potential energy can not be created and destroyedB. Kinetic energy can be converted into various forms of energy, whereas potential energy can only be transformed into heat energy.C. Kinetic energy is energy of a moving object, whereas potential energy is energy possessed by matter as a result of its location or structure.D. Kinetic energy is stored energy that has the capacity to do work, and potential energy is the energy of motion.
The kinetic enerfy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]And the potential energy is given by:
[tex]U=mgh[/tex]Where:
v = Velocity
h = height
m = mass
g = gravitational accceleration of earth
As we can see kinetic energy is associated to the movement and the potential energy is associated to the location, therefore the answer is:
C. Kinetic energy is energy of a moving object, whereas potential energy is energy possessed by matter as a result of its location or structure.
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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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
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
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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B. MULTIPLE CHOICE. Choose the letter of the best answer2. the equation to calculate the momentum isa. p = mgb. p = mvc. p = mghd. p = mt
Answer:
b. p = mv
Explanation:
The momentum is the mass in motion, so it is calculated as the mass times the velocity. It means that the equation to calculate the momentum is
p = mv
Where m is the mass and v is the velocity of the object.
So, the answer is
b. p = mv
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°.
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.
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.
i am not sure the best way to solve this problem
ANSWER
14.11 s
EXPLANATION
We know that in total, the runner will run a distance of 100m. He runs at constant acceleration for a while and then his velocity gets constant until the end of the track - this means that in the last part, his acceleration is zero.
So we have two parts:
For the first part, we have the acceleration and time. If we set that the initial position is zero, as shown in the diagram above, and that the runner starts from rest - therefore, his initial velocity is zero - we can find the distance of the first part of the path, which we'll call x1:
[tex]x_1=x_0+v_0t+\frac{1}{2}at^2[/tex]Since x0 and v0 are both zero, then those terms get cancelled:
[tex]x_1=\frac{1}{2}\cdot a\cdot t^2=\frac{1}{2}\cdot1.5\cdot6^2=27m[/tex]So the first part of the track, where the runner is speeding up, has a distance of 27m. Therefore, the rest of the track where the runner runs at constant acceleration is:
[tex]100-27=73[/tex]73m.
We want to find the time it took the runner to run the whole 100m. We know that he did the first part in 6 seconds. To find the time of the second part, we can use the distance we just found. Let's call it xf:
[tex]x_f-x_1=\frac{1}{2}at^2+v_0t[/tex]We know that the acceleration in this part of the track is zero and the initial velocity for this part is the velocity the runner had when he reached 6 seconds - i.e. 27m:
[tex]73m=v_1\cdot t[/tex]We don't know the time and we don't know the velocity, but we can find the second one using the formula for velocity for the first part of the track with t = 6s:
[tex]\begin{gathered} v_1=a\cdot t+v_0 \\ v_1=1.5\cdot6 \\ v_1=9m/s \end{gathered}[/tex]Now we can find the time for the second part of the track:
[tex]\begin{gathered} 73m=9m/s\cdot t \\ t=\frac{73m}{9m/s} \\ t\approx8.11s \end{gathered}[/tex]Therefore, the total time it took the runner to run 100m was:
[tex]\begin{gathered} t=6s+8.11s \\ t=14.11s \end{gathered}[/tex]14.11 s
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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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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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
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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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 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
itial height at Two balls are thrown vertically from the same • Ball I is launched upward with an initial velocity voj = + 10m/s. Ball 2 is launched downward with an initial velocity vo2 = - 10m/s. same The distance between the two balls after I second from the beginning of motion is:
Given
vo1 = +10 m/s
vo2 = -10 m/s
Procedure
Using the free fall equations, we have:
[tex]\begin{gathered} x1=v_{o1}t-\frac{1}{2}gt^2 \\ x1=10*1-\frac{1}{2}9.8*1 \\ x1=5.1m \end{gathered}[/tex][tex]\begin{gathered} x2=v_{o2}t-\frac{1}{2}gt^2 \\ x2=-10*1-\frac{1}{2}9.8*1 \\ x2=-14.9m \end{gathered}[/tex][tex]\begin{gathered} x1-x2=5.1-\lparen-14.9) \\ x1-x2=20 \end{gathered}[/tex]The distance between the balls would be 20m