Answer:
H = 31.7 cm
Explanation:
Given:
k = 52 N/m - Spring rate
L = 34 cm = 0.34 m - Spring length
m = 0.12 kg
g = 9.8 m/s²
____________
H - ?
Block weight:
F = m*g = 0.12·9.8 ≈ 1.18 N
According to Hooke's law:
F = k·ΔL
The spring is compressed by:
ΔL = F / k = 1.18 / 52 ≈ 0.023 м
Height of the block above the table:
H = L -ΔL = 0.34 - 0.023 = 0.317 m or H = 31.7 cm
An x-ray with a wavelength of 3.5 × 10^-9 m travels with a speed of 3.0 × 10^8 m/s. What is the frequency of this electromagnetic wave? A.9.52 × 10^-1 Hz B.8.57 × 10^16 Hz C.1.17 × 10^-17 Hz D.1.05 Hz
Answer. B
What is the efficiency of a block and tackle if you pull 20 m of rope with a force of 600 N to raise 200kg piano 5 m?
W out = mass * distance * gravity
E in = Force * distance
Efficiency = W out / E in = (200 kg * 5m*9.8N) / (600N * 20m) = 0.81
0.81 x 100 = 81 %
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
A maximum of about 229 repetitions of something like the exercise can be performed by that of the athlete utilizing the energy provided by each of the biscuits.
The proportion of input to produced energy can be used to define energy consumption.
A cookie, therefore, therefore has 54.0 kcal of calories. The 54.0 kcal throughout this croissant is used as power input by the athlete.
Efficiency = output energy / input energy
It can be written as:
Output energy = efficiency × input energy
Puting the values of efficiency and input energy.
Output energy = 0.25 × 54 kcal = 13.5 kcal.
The weightlifting exercise can be done n times for the output energy. This outgoing energy comes from mgh in the shape of potential energy. So,
Energy per repetition = [tex]mgh[/tex]
Put the values of m, g and h in above equation.
Energy per repetition = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J
Energy per repetition = 0.059 kcal.
So,
amount of repetitions = sum of output energy / energy per repetition
amount of repetitions = 13.5 kcal / 0.059 kcal = 229 repetitions.
Therefore, amount of repetition can be be 229.
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A person is at the top of a tower. He takes a segment of a string which measures 30 cm long when at rest and hooks his 3 kg sword at the end of it. The spring extends to 35 cm long. He will use this spring to get to the ground. What is the spring constant of the spring, and how much of the spring (measured at equibilirum) does he need in order to have a net force of 0 upon himself when he touches the ground? Assume he hangs the spring from a hook located exactly 30 m above the ground. Be certain to draw a free body diagram of the forces on him the moment he hits the ground.
The given problem can be exemplified in the following diagram:
To determine the constant of the spring we can use Hook's law, which is the following:
[tex]F=k\Delta x[/tex]Where:
[tex]\begin{gathered} F=\text{ force on the string} \\ k=\text{ string constant} \\ \Delta x=\text{ difference in length} \end{gathered}[/tex]Now, we solve for "k" by dividing both sides by the difference in length:
[tex]\frac{F}{\Delta x}=k[/tex]The force on the string is equivalent to the weight attached to it. The weight is given by:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} W=\text{ weight} \\ m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Substituting in the formula for the constant of the spring we get:
[tex]\frac{mg}{\Delta x}=k[/tex]Now, we substitute the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{35\operatorname{cm}-30\operatorname{cm}}=k[/tex]Before solving we need to convert the centimeters into meters. To do that we use the following conversion factor:
[tex]100\operatorname{cm}=1m[/tex]Therefore, we get:
[tex]\begin{gathered} 35\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.35m \\ \\ 30\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.30m \end{gathered}[/tex]Substituting in the formula we get:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.35m-0.30m}=k[/tex]Solving the operations:
[tex]588\frac{N}{m}=k[/tex]Therefore, the constant of the spring is 588 N/m.
why free-fall acceleration can be regarded as a constant for objects falling within a few hundred miles of Earth’s surface.
A car starts from rest and travels for 9.0 s with a uniform acceleration of +2.4 m/s²?. The driver then applies the brakes, causing a uniform acceleration of -2.5 m/s². If the brakes areapplied for 2.0 s, determine each of the following(a) How fast is the car going at the end of the braking period?m/s(b) How far has the car gone?m
a) At the end of the braking period, the car is moving at a speed of 16.6m/s
b) The car has traveled a total distance of 135.4m
Explanations:The car starts from rest
The initial velocity, u = 0 m/s
Uniform acceleration, a = 2.4 m/s²
time, t = 9.0 seconds
Find the final velocity when the car accelerates at 2.4m/s² using the equation
v = u + at
v = 0 + 2.4(9)
v = 21.6 m/s
The distance covered when the car accelerates at 2.4m/s²
s = ut + 0.5at²
s = 0(9) + 0.5(2.4)(9²)
s = 97.2 m
The distance covered when the car accelerates at 2.4m/s² is 97.2 m
The driver then applied a brake for 2.0 s and accelerates at -2.5m/s²
The initial velocity now becomes 21.6 m/s
That is, u = 21.6 m/s
t = 2 seconds
a = -2.5m/s²
The final speed v is calculated as:
v = u + at
v = 21.6 + (-2.5)(2)
v = 21.6 - 5
v = 16.6m/s
At the end of the braking period, the car is moving at a speed of 16.6m/s
The distance covered during the braking period is calculated as:
[tex]\begin{gathered} s\text{ = (}\frac{u+v}{2})t \\ s\text{ = }\frac{21.6+16.6}{2}\times2 \\ s\text{ = }\frac{38.2}{2}\times2 \\ s\text{ = 38.2 m} \end{gathered}[/tex]The car traveled a distance of 38.2 m during the braking period
Total distance covered = 97.2m + 38.2m
Total distance covered = 135.4m
The car has gone a distance of 135.4m
Given v = 520 sin (30t - 5π/4), what is the phase angle?Question 4 options:-225 degrees90 degrees-135 degrees-90 degrees
Given data:
The voltage can be expressed as,
v = 520 sin (30t - 5π/4) ...... (1)
Now, the general equation of the sine wave can be given as,
[tex]v=V_m\text{ sin}(\omega t+\phi)\ldots\ldots\text{ }(2)[/tex]Here,
[tex]\phi[/tex]is the phase angle,
[tex]V_m[/tex]is the maximum voltage, and
[tex]\omega[/tex]is the angular frequency.
Compare equations (1) and (2), we get:
[tex]\begin{gathered} \phi=-\frac{5\pi\text{ rad}}{4}(\frac{180\degree}{\pi}) \\ =-225\degree \end{gathered}[/tex]Thus, the phase angle is
[tex]-225\degree[/tex]and the first option (-225 degrees) is correct.
A) the frictional force F newtonsB)The resultant normal reaction of the surface on the metal block
Given:
The mass of the block is.
[tex]m=10\text{ kg}[/tex]The tension on the rope is,
[tex]T=100\text{ N}[/tex]The angle with the horizontal is,
[tex]\theta=60^{\circ}[/tex]The block is moving with constant speed.
as the block is moving with constant speed, the net force on the block will be zero.
Part (A)
we can write in the horizontal direction the component of the tension will be equal to the frictional force and we write,
[tex]\begin{gathered} T\cos 60^{\circ}=F \\ F=100\cos 60^{\circ} \\ F=50\text{ N} \end{gathered}[/tex]Hence the frictional force is 50 N.
\\
Part(B)
The resultant normal reaction will be,
[tex]\begin{gathered} N=T\sin 60^{\circ}-mg \\ =100sin60^{\circ}-10\times9.8 \\ =-11.4\text{ N} \end{gathered}[/tex]hence the resultant normal reaction is -11.4 N.
An object is dropped from a height of 65 m above ground level. A) determine the final speed in m/s, at which the object hits the ground c) determine the distance in meters, traveled during the last second of motion before hitting the ground.
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
Megan kicks a soccer ball with a mass of 2 kg. The ball leaves the ground moving 50 meters per second. What is the kinetic energy of the ball?
ANSWER
[tex]2500J[/tex]EXPLANATION
The kinetic energy of a body is the energy it possesses due to its motion and it can be found by applying the formula:
[tex]E=\frac{1}{2}mv^2[/tex]where m = mass, v = velocity
From the question:
[tex]\begin{gathered} m=2\text{ kg} \\ v=50\text{ m/s} \end{gathered}[/tex]Therefore, the kinetic energy of the ball is:
[tex]\begin{gathered} E=\frac{1}{2}\cdot2\cdot50^2 \\ E=2500J \end{gathered}[/tex]A box is standing on a conveyor belt that is not in motion. At one point the belt starts moving with some acceleration. At that point the box starts moving too (without slipping). Which force is responsible for the acceleration of the box. a. The air resistance force. b. The force of the pull. c. The force of friction. d. The normal force.
Given that a box is standing on a conveyor belt that is not in motion.
When the belt starts moving with some acceleration, the box starts moving too without slipping.
Let's determine the force that is responsible for the acceleration of the box.
Here, since the box starts moving without slipping when the belt starts moving, there will be static friction between the box and the belt since the belt was fixed.
Now, the force which is responsible for the acceleration of the box will be the force of gravity and the normal force.
Applying the Newton's second law, if the there is only force of gravity and the normal force acting on the box, there will be zero horizontal acceleration.
In order for the box to accelerate without slipping, the force responsible will be the static frictional force.
ANSWER:
c. The force of friction.
8.1 kg of copper sits at a temperature of 64 oF. How much heat is required to raise its temperature to 743 oF? The specific heat of copper is 385 J/kg-oC. Submit your answer in exponential form.
ANSWER:
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 8.1 kg
T1 = 64 °F
T2 = 743 °F
Specific heat (C) = 385 J/kg*°C
[tex]undefined[/tex]According to the law of conservation of charge which statement can be true?A. A silk cloth gained charge. B. A metal rod lost charge.C. A peice of glass transferred electrons to felt. D. A balloon remains neutrally charged when rubbed.
Option D
Explanation:The law of conservation of charge states that the amount of charge in a system is constant
This means that as time changes, the amount of charge in a system does not change
By careful consideration of the options stated:
Each of options A to C either shows that charge is lost or gained
Only option D typifies the law of conservation of conservation of charges because charges are not lost or gained by the ballon so described.
A Tour de France cyclist at a rate of 6.5 m/s and is 333 m
ahead of a crew van with powerdrink refills. The van is
traveling at 15 m/s and is accelerating at a constant rate of
0.4 m/s2
How much time will it take for the crew van to catch up
with the cyclist?
Given data:
* The speed of the cyclist is 6.5 m/s.
* The initial distance of the cyclist from the van is 333 m.
* The initial velocity of the van is 15 m/s.
* The acceleration of the van is,
[tex]a=0.4ms^{-2}[/tex]Solution:
Let x be the distance from the van initial position at which the cyclist and van meet.
Let the cyclist meet the van at time t.
By the kinematics equation, the position of the cyclist at time t is,
[tex]x-333=u_ct+\frac{1}{2}a_ct^2[/tex]where u_c is the speed of the cyclist, a_c is the acceleration of the cyclist, t is the time taken and 333-x is the distance traveled by the cyclist at time t,
The acceleration of the cyclist is zero.
Substituting the known values,
[tex]\begin{gathered} x-333=6.5t+0 \\ x-333=6.5t \end{gathered}[/tex]By the kinematics equation, the position of the van after time t is,
[tex]x=u_vt+\frac{1}{2}a_vt^2[/tex]where u_v is the velocity of the van, a_v is the acceleration of the van, and t is the time taken,
Substituting the known values,
[tex]\begin{gathered} x=15t+\frac{1}{2}\times0.4\times t^2 \\ x=15t+0.2t^2 \end{gathered}[/tex]Substituting this value of x in the kinematics equation of the cyclist,
[tex]\begin{gathered} (15t+0.2t^2)-333=6.5t \\ 15t+0.2t^2-6.5t-333=0 \\ 0.2t^2+8.5t-333=0 \end{gathered}[/tex]By solving the quadratic equation,
[tex]\begin{gathered} t=\frac{-8.5\pm\sqrt[]{8.5^2-(4\times0.2\times(-333))}}{2\times0.2} \\ t=\frac{-8.5\pm\sqrt[]{^{}8.5^2+(4\times0.2\times333)}}{2\times0.2} \\ t=\frac{-8.5\pm18.4}{0.4} \\ t=24.8\text{ s or-67.25 s} \end{gathered}[/tex]As the value of time cannot be negative.
Thus, the time at which the cyclist and van meet is 24.8 seconds.
10. ABC.Per=1200 NNet Force:Pit=600 NEngen-SONFrid=20 NPry=800 NPrax=800 NF50NWhich situation above would best describe free fall velocity?Which situation above would best describe a crane lifting an object?If situation C had a Fapp of 40N to the right, the net force on the object would beIf situation Chad Fapp of 20N to the right, the forces would be (balanced, unbalanced) and thehorizontal velocity would be (constant, + accelerating, - accelerating). Circle the correct terms
When the body is under free fall its apperaent weight will be zero.
Therefore
if the dolphin is moving horizontally when it goes through the hoop how high above the water is the center of the hoop
We are given that a dolphin moves describing a projectile motion. This can be represented in the following graph of position vs time:
Since the dolphin moves horizontally as he goes through the hoop this means that the hoop is at the maximum height of the motion. The maximum height of a projectile motion is given by:
[tex]h_{\max }=\frac{v^2\sin ^2\theta}{2g}[/tex]Where:
[tex]\begin{gathered} h_{\max }=\text{ max}imum\text{ height} \\ v=velocity_{} \\ \theta=\text{ initial angle} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we plug in the values:
[tex]h_{\text{max}}=\frac{(10\frac{m}{s})^2(\sin (41))^2}{2(9.8\frac{m}{s^2})}[/tex]Solving the operations:
[tex]h_{\max }=2.2m[/tex]Therefore, the hoop is at 2.2 meters above the water.
In terms of area, about how much more pizza is given if the diameter is 12 inches compared to one with a diameter of 8 inches?
B. 2.3 times more
Explanation:The pizza is circular in shape
The diameter of the large-sized pizza, d₁ = 12 inches
Tha area of the large sized pizza is calculated as:
[tex]\begin{gathered} A_1=\frac{\pi{d^2_1}}{4} \\ A_1=\frac{\pi{12^2}}{4} \\ A_1=\frac{\pi{144^{}}}{4} \\ A_1=36\pi\text{ in}^{2} \end{gathered}[/tex]The diameter of the small-sized pizza, d₂ = 8 inches
The area of the small-sized pizza is calculated as:
[tex]\begin{gathered} A_2=\frac{\pi{d^2_2}}{4} \\ A_2=\frac{\pi{8^2}}{4} \\ A_2=\frac{64\pi{}}{4} \\ A_2=16\pi\text{ in}^{2} \end{gathered}[/tex]Ratio of A₁ to A₂
[tex]\begin{gathered} \frac{A_1}{A_2}=\frac{36\pi{}}{16\pi} \\ \frac{A_1}{A_2}=2.25 \\ \frac{A_1}{A_2}=2.3(to\text{ the nearest 1 dp)} \end{gathered}[/tex]The 12 inches pizza is 2.3 times more than the 8 inches pizza
A roller coaster has a vertical loop with radius 22.8 m. With what minimum speed should the roller-coaster car be moving at the top of the loop so that the passengers do not lose contact with the seats?
Given,
The radius of the loop of the roller coaster, r=22.8 m
The forces that are acting on the roller coaster when it is at the top of the loop are the centripetal force directed upwards and the weight of the roller coaster including the passengers directed downwards.
For the passengers to stay in the seat, the centripetal force must be, at the least, equal to the weight of the passengers and the rollercoaster.
That is,
[tex]\frac{Mv^2}{r}=Mg[/tex]Where M is the combined mass of the rollercoaster and the passengers, v is the minimum speed of the roller coaster when it is at the top of the loop, and g is the acceleration due to gravity.
On simplifying the above equation,
[tex]v=\sqrt[]{gr}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{9.8\times22.8} \\ =14.95\text{ m/s} \end{gathered}[/tex]Thus the minimum speed that the roller coaster must have when it is at the top of the loop so that the passengers stay in contact with the seats is 14.95 m/s.
A stone is thrown vertically upwards from a height of 1.5m and lands on the ground 6s later. What was the magnitude of the initial velocity?
Answer:
2.5m/s is the correct answer
A 6 kg object is being pulled by a horizontal force F=120 N on a friction-less horizontal surface. It moved a distance of 18 m. If its initial kinetic energy was 100 Joules, what is the final kinetic energy in Joules?
The final kinetic energy is 120m.
What is Work-Energy Theorem?
The Work-Energy Theorem states that the work done is equal to the change in the K.E. i.e Kinetic Energy of the object.
W = Δ(K.E.)
In the given question we had,
Mass = 6 kg,
Force = 120 N,
Distance = 18 m,
Initial Kinetic Energy ( KE1 ) = 100 Joules
According to Work-Energy Theorem,
W = Δ(K.E.)
W = KE2-KE1
F x S = KE2 - 100
120 x 18 = KE2 - 100
2160 + 100 = KE2
2260J = KE2
So, the final kinetic energy is 2260J.
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You observe waves on the beach and measure that a wave hits the beach every 5 seconds. What is the period and the frequency of the waves ?
Answer:
Period = 5 seconds
Frequency = 0.2 Hz
Explanation:
The period is the time per cycle. So, if a wave hits the beach every 5 seconds, the period will be 5 seconds.
Additionally, the frequency is the inverse of the period, so the frequency of the waves can be calculated as:
[tex]\text{frequency = }\frac{1}{Period}=\frac{1}{5}=0.2\text{ Hz}[/tex]So, the answers are:
Period = 5 seconds
Frequency = 0.2 Hz
There is no _________ movement in a longitudinal wave.A. HorizontalB. Back and forthC. VerticalD. Parallel
Explanation
A longitudinalwave is in which the particles of the medium vibrate in the direction of the line of advance of the wave.Longitudinal waves cause the medium to move parallel to the direction of the wave.
A longitudinal wave can be set up for example in a streched spring by compressing the coils in a small region, and releasing the compressed region,
the back and forth motions of the coils of the spring is in the same direction that the wave travels
so, in a longitudinal wave there is not Vertical movement, so the answer is
C. Vertical
Which shape fits a position vs. time graph of an object that is slowing down? Which shape fits a position vs. time graph of an object that is speeding up?
1.
In a position x time graph, if the velocity is constant, so the position increases at the same rate over the time, so we have a linear relation between the position and the time.
Therefore the shape that represents this relation is C.
2.
In a velocity x time graph, if the velocity is constant, its value is always the same over the time, it doesn't change. That is represented graphically by a horizontal line, therefore the shape that represents this relation is B.
.
The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 66 kHz
Given:
The inductance is,
[tex]\begin{gathered} L=9\text{ mH} \\ =9\times10^{-3}\text{ H} \end{gathered}[/tex]The radio frequency is,
[tex]\begin{gathered} f=66\text{ kHz} \\ =66\times10^3\text{ Hz} \end{gathered}[/tex]To find:
value of the variable capacitor, in picofarads
Explanation:
The frequency of the AM is,
[tex]\begin{gathered} f=\frac{1}{2\pi\sqrt{LC}} \\ \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} 66\times10^3=\frac{1}{2\pi\sqrt{9\times10^{-3}\times C}} \\ \sqrt{9\times10^{-3}\times C}=\frac{1}{2\pi\times66\times10^3} \\ \sqrt{9\times10^{-3}\times C}=2.41\times10^{-6} \\ 9\times10^{-3}\times C=5.81\times10^{-12} \\ C=6.45\times10^{-10} \\ C=645\times10^{-12}\text{ F} \\ C=645\text{ pF} \end{gathered}[/tex]Hence, the capacitance is 645 pF.
The radioactive isotope 14C has a half-life of approximately 5715 years. Now there are 50g of 14C.(1) How much of it remains after 1600 years? (Round your answer to three decimal places.)
We know that the amount of matter is given by:
[tex]N=N_0e^{-\lambda t}[/tex]where λ is the decay constant. The decay constant is related to the half-life of the element by the equation:
[tex]\lambda=\frac{\ln2}{t_{\frac{1}{2}}}[/tex]Then we can express our first equation as:
[tex]N=N_0e^{-\frac{\ln2}{t_{\frac{1}{2}}}t}[/tex]Plugging the initial amount, 50 g, the half-life of 5715 years and the time we want to know we have that:
[tex]\begin{gathered} N=50e^{-\frac{\ln2}{5715}(1600)} \\ N=41.181 \end{gathered}[/tex]Therefore, after 1600 years there are 41.181 g
Please help with Question(ii). I don't understand the shown step of calculating the momentum of ball B. Especially after the third line 12+Pb=15.
Given:
m1 = mass 1 = 1kg
v1= initial velocity 1 = 12 m/s
m2= mass 2 = 3 kg
P after = momentum after collision = 15 kgm/s
(i)
Momentum of Ball A before collision
Momentum = mass x velocity
Pa = m1 v1
Replacing with the values given:
Pa = (1 kg) (12 m/s) = 12 kg m/s
(ii)
Momentum before = momentum after
Pa + Pb = P after
12 + Pb = 15
Since The ball B is travelling North, the distances travelled form a right triangle:
Apply pythagorean theorem:
c^2 = a^2 + b^2
Where c is the hypotenuse= P after = 15
a & b are the other 2 legs of the triangle = Pa and Pb
Replacing:
15^2 = 12^2 + Pb^2
Solve for Pb
15^2 - 12^2 = pb^2
√15^2 -12^2 = Pb
pb= 9 kgms^2
8) If the volume of the liquid in graduated cylinder B is 90 mL, then whatis the volume of the rock?AYour answer8060B100180
Answer:
20 mL
Explanation:
The volume of the rock is equal to the difference of volume of A and B. So, it is equal to
90 mL - 70 mL = 20 mL
Because 90 mL is the volue in cylinder B and 70 mL is the volume in cylinder A.
Therefore, the volume of the rock is 20 mL
Energy transformations always produce a wasteful amount of energy called ?
Energy transformation always produce a wasteful amount of energy called heat energy.
Hence, the answer is heat energy.
Need help with this question Short straight forward answers please :)
We will have the following:
a. The gravitational potential energy will be:
[tex]P_C=(15kg)(9.8m/s^2)(6m)\Rightarrow P_C=882J[/tex]So, the gravitational potential energy of C is 882 J.
b. The velocity of C right before it hits the ground will be:
[tex]\begin{gathered} 882J=\frac{1}{2}(15kg)v^2\Rightarrow\frac{1764J}{15kg}=v^2 \\ \\ \Rightarrow v=\frac{14\sqrt{15}}{5}m/s\Rightarrow v\approx10.84m/s \end{gathered}[/tex]So, the velocity will be approximately 10.84 m/s.
c.
1. We will have that Eg at the initial position will be: B < C
2. Vfinal upon impact with ground: B = C
3. Ek right before hitting he ground: B < C
f.
1. Eg: A > B
2. V final: A > B
3. Ek: A > B
4. V at 2 meters above the ground: A > B
5. Total energy at 2 m above the ground: A > B.
5. Draw a transverse wave with two wavelengths and label amplitude, crest, trough, and
equilibrium position.
A wave is considered to be transverse if its oscillations run counterclockwise to the wave's direction of advance. A longitudinal wave, on the other hand, moves in the direction of its oscillations. Transverse waves include water waves.
A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm).
The largest displacement or distance made by a point on a wave or vibrating body relative to its equilibrium position is its amplitude. It is equal to one-half of the vibration path.
The crest and trough of a wave, respectively, are its highest and lowest surface portions. The wave height is the vertical distance between the peak and trough. The wavelength is the horizontal separation between two consecutive crests or troughs.
The horizontal line at the wave's centre stands in for balance. A period is the length of time it takes to complete a cycle, which includes travelling from one peak to another, from one trough to another, or from one equilibrium point to another (both equilibrium points same direction).
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