Increasing his acceleration will impact his velocity and rate of displacement covered in that as the speed increases due to the increased rate of acceleration, the rate of air resistance also increases.
What is air resistance?Air resistance is a force created by air. When an item moves through the air, the force operates in the opposite direction.
When a diver descends, the force of air resistance acts to counteract the force of gravity. As the skydiver falls faster and faster, the quantity of air resistance grows until it equals the magnitude of gravity's force.
A balance of forces is achieved when the force of gravity equals the force of air resistance, and the skydiver no longer accelerates. The skydiver reaches what is known as terminal velocity.
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A= 1/2 W (R+S) for W
The change of the subject in the formula is given by; W = 2A/(R + S)
How do we make a subject in formula?When we have a formula, each of the components of the formula are the parameters of the formula. In this case, we have a formula that says A= 1/2 W (R+S). The parameters that are in this formula are A, W, R and S.
According to the question, we have been asked to make W the subject of the formula.
The first step is to multiply both sides by 2
2A = W(R + S)
And secondly, we divide both sides by the sum of R and S and we have;
W = 2A/(R + S)
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Can anyone tell me the final displacement of the puck
The final displacement of the puck is 4.14i + 14.14j
What is displacement?Displacement is the change in position of an object.
How to find the final displacement of the puck?Since for the first displacement of the puck from the goal is 20 m at 45°, we have that its displacement vector is d = (20cos 45°)i + (20sin45°)j.
= (20 × 0.7071)i + (20 × 0.7071)j.
= 14.14i + 14.14j
Also, for the second displacement is 10 m to the left. Its displacement vector is d' = -10i
So, we see that the total displacement is D = d + d'
So, adding the displacements, we have D = d + d'
= 14.14i + 14.14j + (-10i)
= 14.14i - 10i + 14.14j
= 4.14i + 14.14j
So, we see that the final displacement of the puck is 4.14i + 14.14j
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I don't know what I'm doing wrong I just can't seem to get the answers right. Help please.
The current and potential difference in the lumped element model of electrical circuits are dealt with by Kirchhoff's circuit laws, which are two equalities. German scientist Gustav Kirchhoff published the first account of them in 1845. This was done before James Clerk Maxwell and generalized Georg Ohm's work.
Because the sum of the currents entering and leaving the junction is equal, Kirchhoff's first law is based on the conservation of charge. The algebraic sum of potential drops in a closed circuit must equal zero, according to Kirchhoff's second law. Therefore, it is based on energy conservation.
A stream of charged particles—such as electrons or ions—moving through a conductor for electricity or into empty space is known as an electric current. It is determined by measuring the net rate of electric charge flow through a surface or into a control volume.
The amount of resistance in an electrical circuit represents the resistance to current flow. The Greek letter omega (Ω), which stands for resistance, represents ohms.
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During which season does the north pole experience constant sunlight?
A. spring
B. winter
C. summer
D. autumn
During the summer season, the north pole gets constant sunlight.
How does the north pole experience constant sunlight?In the summer season, the north pole experience constant sunlight. Because in the summer the shadows are short and the sun remains higher in the sky. In the summertime, the sun remains above the horizon at the north pole which circles the pole every day.
In the summer the sun's rays reach the north pole while in winter season the north pole is tilted away from the incoming sunshine. Mostly the regions of the north pole remain cold because the sunlight does not directly fall on the north pole. At the north pole, there are only two types of seasons, one is summer and the other is winter season. The annual average temperature of the winter season at the north pole is -40 degrees Celsius while in summer season the average temperature is 32 F.
So we can conclude that during the months of the summer season the north pole gets maximum sunlight.
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Which measurements or observations are needed to calculate density
Answer:
mass and volume measurements are required to calculate density.
3. A parallel-plate capacitor has a capacitance of 1.35 pF. If a 12.0 V battery is connected to this capacitor, how much electrical potential energy would it store?
Answer:
9.72 x 10^(-11)
Explanation:
The energy stored in a capacitor can be calculated as:
[tex]E=\frac{CV^2}{2}[/tex]Where C is the capacitance and V is the Voltage. So, replacing C by 1.35pF or 1.35 x 10^(-12) F, and V by 12.0 V, we get:
[tex]E=\frac{1.35\times10^{-12}F(12.0V)^2}{2}=9.72\times10^{-11}J[/tex]Therefore, the capacitot would store 9.72 x 10^(-11) J of energy.
The electrical potential energy that would be stored in the capacitor is 9.72 x 10^(-11) J. It is the energy which is present in the object when it is at rest.
What is Potential energy?Potential energy is the energy which is stored inside an object which is at rest.
For the parallel plate capacitors, capacitance is dependent upon its geometry, which is given by the formula:
C=ϵ⋅Ad or C = ϵ ⋅ A d ,
where, C is the value of the capacitance,
A is the area of each plate,
d is the distance between the plates, and
ϵ is the permittivity of the material between the plates of the parallel capacitor.
The energy stored in a capacitor can be calculated as:
E = CV²/2
where, C is the capacitance and V is the Voltage.
So, by replacing C with 1.35pF or 1.35 × 10⁻¹²F, and V by 12.0V, we get:
E = 1.35 × 10⁻¹² × (12)²/2
E = 9.72 × 10⁻¹¹ J
Therefore, the capacitor would store 9.72 × 10⁻¹¹ J of energy.
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A cell of a constant e. m.f is connected in series with a parallel arrangement of 8ohms coil and a uniform wire of length (Lcm). The total external resistance in the circuit is found to be 6ohms. calculate the resistance of the wire and its length. Resistivity of the wire material = 6.0×10^-5 ohm/meter, cross-sectional area of the wire = 6.0×10^-3cm².
The resistance of the wire will be 24ohm and the length of the wire will be 24×10⁴cm.
As given in the question,
Rₜₒₜₐₗ = 6ohm
Resistivity = 6×10⁻⁵ohm/m = 6×10⁻⁵/100ohm/cm
Area(A)= 6×10⁻³cm²
Let R = resistance of the wire, therefore;
Rₜₒₜₐₗ = R×8 / R+8
6 = R×8 / R+8
6R + 48 = 8R
48 = 2R
R = 24ohm
R = 24ohm
we know that resistance (R)=rho×l/A
Therefore,
rho×l/A = 24
6×10⁻⁵/100 × l/6×10⁻³ = 24
10⁻⁴ × l = 24
l = 24×10⁴cm
So, the resistance of the wire will be R=24ohm and the length of the wire will be 24×10⁴cm.
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7 The distance-time graph for a motorcyclist riding off from rest is shown in Figure 1.2.18.
a Describe the motion.
b Calculate how far the motorbike moves in 30 seconds.
c Calculate the speed.
The distance travelled by the motorbike is 600 m and the speed of the motorbike is 20 m/s.
What is the motion of the motorbike?The motion of the motorbike can be described as linear type of motion since the distance travelled by the motorbike increases with increase in the time of motion.
There is directly linear relationship between the distance travelled by the motorbike and the time of motion of the motorbike.
The distance travelled by the motorbike is calculated as follows;
from the given graph, at time 30 seconds, the corresponding the distance of the motorbike on the vertical axis is 600 m.
distance = 600 m
The speed of the motorbike is calculated as follows;
v = d/t
v = 600 m / 30 s
v = 20 m/s
Thus, the motion of the motorbike is a linear type of motion, the distance travelled by the motorbike increases with increase in the time of motion and the speed of the motorbike depends on the distance and time of motion.
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The rocket’s velocity just before it hits the ground is the same magnitude as the initial velocity. Use the appropriate kinematics equation to show that this is true.
WILL MARK BRAINLIEST LOTS OF POINTS!
The rocket’s velocity just before it hits the ground is the same magnitude as the initial velocity because change in length in y direction will zero.
if u and v are the initial and final velocities of the rocket then from kinematic equations it may be written as
[tex]s=ut+\frac{1}{2} at^{2}\\0=ut+\frac{1}{2} at^{2}\\\frac{1}{2} at^{2} =-u t\\at=-2u[/tex]
since
v=u + at
v=u - 2u
v = - u
Hence the rocket hit the ground with same velocity as it projected.
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A golf ball is hit horizontally at 40 m/s from the top of a hill that is 2.5 m high. If the terrain around the hill is nearly flat,approximately how far will the golf ball fly? Use - 9.81 m/s2 for the acceleration caused by gravity. Ignore air resistance.Round any intermediate calculations to no less than six decimal places, and round your final answer to two decimal places.
ANSWER
28.56 m
EXPLANATION
Let's make a diagram of this situation to understand it better,
The ball has an initial velocity of 40m/s, which is horizontal. This means that there is no vertical initial velocity.
We have to find the horizontal distance the golf ball flies. Since the horizontal velocity is constant - this is because there is no acceleration in that direction, the distance the ball travels is,
[tex]\Delta x=v_{ox}\cdot t[/tex]The horizontal initial velocity is given, but we have to find the time the ball was in the air. To find it, we use the vertical distance the ball travels - which we know is the height of the hill. In this case, we do have vertical acceleration - the acceleration of gravity, so the vertical distance the ball travels, as shown in the diagram, has a parabolic form and it is given by the equation,
[tex]\Delta y=v_{oy}\cdot t+\frac{1}{2}gt^2[/tex]The initial vertical velocity is zero because the ball is hit horizontally,
[tex]\Delta y=\frac{1}{2}gt^2[/tex]Solve for t. Multiply both sides by 2/g,
[tex]\begin{gathered} \Delta y\frac{2}{g}=\frac{1}{2}g\cdot\frac{2}{g}\cdot t^2 \\ \frac{2\Delta y}{g}=t^2 \end{gathered}[/tex]And take the square root to both sides of the equation,
[tex]t=\sqrt[]{\frac{2\Delta y}{g}}[/tex]Δy is the height of the hill, 2.5m, and g = 9.81m/s²,
[tex]t=\sqrt[]{\frac{2\cdot2.5m}{9.81m/s^2}}=\sqrt[]{\frac{5m}{9.81m/s^2}}=\sqrt[]{0.509684s^2}=0.713922s[/tex]This is the time the ball was in the air for. Now we can find the distance it traveled,
[tex]\Delta x=40m/s\cdot0.713922s=28.56m[/tex]The golf ball flew 28.56 m horizontally.
A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. Howfar will the spring stretch if the mass is motionless?
A force of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.
A) What is the potential energy of the spring when it is stretched 0.300 m?
B) What is its potential energy when it is compressed 6.00 cm?
A force of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.
A) The potential energy of the spring when it is stretched 0.300 m will be 81 J
B) The potential energy when it is compressed 6.00 cm will be 3.24 J
Hooke's law, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load.
To find spring constant , using formula
k= F/x
where
F = Force
k = spring constant
x = displacement
k = 540 / .300 = 1800 N/m
A) Potential energy of the spring when it is stretched 0.300 m
U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.300^{2}[/tex] = 81 J
B ) Potential energy of the spring when it is stretched 0.06 m
U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.06^{2}[/tex] = 3.24 J
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A ball is thrown downwards from a height of 25m with an initial velocity of 10 m/s.
Find the velocity of the ball just before it hits the ground.
•V₁=10mls
h=25m
The velocity of the ball just before it hits the ground is 24.29m/s.
What is acceleration due to gravity?
The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s2 is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on earth's surface is 9.8 m/s2.
Given:
Initial velocity u = 10m/s
height h = 25m
we know that the acceleration due to gravity g is : 9.8m/s
Therefore using the formula,
v² - u² = 2gs where s is the height in this case and v is the final velocity.
v² - 10² = 2 x 9.8 x 25
=> v² = 490 + 100
=> v = √590
=> v = 24.29m/s
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When you drop a 0.36 kg apple, Earth ex- erts a force on it that accelerates it at 9.8 m/s² toward the earth's surface. According to New- ton's third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98 x 1024 kg, what is the magnitude of the earth's acceleration toward the apple? Answer in units of m/s². Answer in units of m/s^2
The magnitude of the earth's acceleration toward the apple is 5.7x10^-25 m/s^2
Given:
Mass if the apple (m)= 0.36
Mass of the earth (M)=5.98x10^24
Gravitational acceleration (g)= 9.8
we know that the formula is:
Ma=mg
So, according tot the formula
a=(m/M)g
=(0.36/5.98x10^24)9.8
=5.7x10^-25 m/s^2
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if the camera left his hands at 21.9 m/s and took 2.009s to hit the ground, how far down the cliff did it fall?
The camera falls down 25.835 meters under gravity.
While a frame falls, it happens because of gravity, and the force of air will increase with speed. This continues till the force of air equals the weight. Now the accelerates but falls at a constant speed known as the terminal speed.
A motion under gravity refers back to the motion of an item whose vertical movement is laid low with the presence of gravity. The pressure that attracts gadgets downwards is gravity. In fact, gravity works towards the center of the Earth.
S = ut - 1/2at²
= 21.9 × 2.009s - 1/2 × 9.8 × (2.009)²
= 43.997 - 18.162
= 25.835 meter.
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thunder can make objects inside a room vibrate explain what causes the objects to vibrate
Thunder can make objects inside a room vibrate by which our house columns in underground that causes the objects to vibrate.
What causes the sound of thunder?Thunder is caused by the rapid expansion of air around the path of lightning. It only takes a few millionths of a second for lightning to burst through the air from a cloud to a nearby tree or roof. The loud thunder that follows a lightning strike is generally said to come from the lightning itself. However, the rumbling and roaring heard during thunderstorms is actually due to the rapid expansion of air around the lightning.
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Please help me
how law of motion applied in basketball explain
Explanation:
According to the first law of motion, the basketball is always moving in one direction, unless acted on by another force.
Jason is pulling a box across the roompulling with a force of 16 newtons and his arm is making a 68 anglet with the horizontal what is the horizontal component of the he is pulling with?
The horizontal component of a force is defined as follows
[tex]F_x=F\cdot\cos \theta[/tex]Where F = 16 N, and theta = 68. Let's use these values to find the horizontal component of F.
[tex]F_x=16N\cdot\cos 68\approx6N[/tex]Therefore, the horizontal component of the pulling force is 6 Newtons.An Earth satellite moves in circular orbit 602 km above the earths surface with a period of 96.53 min. What are the speed and the magnitude of the centripetal acceleration of the satellite.
The speed and the magnitude of the centripetal acceleration of the satellite are 7.568 km/s and 8.205 x 10^-3 m / s^2
Here this problem we are dealing with the speed and centripetal acceleration where the speed of the satellite is the speed required to attain adjustment between gravity's pull on the satellite and the inertia of the satellite's motion whereas centripetal acceleration is referred to as the acceleration of a body navigating a circular way.Since we are given the orbital radius which is 602 km and the period which is 96.53 min.
The radius of the satellite's orbit, r = earth radius + orbit radius
=>r =6378 km + 602 km = 6980 km
Since the formula for the orbital circumference is :
2πR
=>2 • 3.14 •6980 = 43,834.4 km
Now divide the circumference by the time period we get the orbital speed which is,
43,834.4 km / (96.53 min • 60) = 7.568 km/s
the formula we are referring for calculating the centripetal acceleration is:
a = v²/ r ,
=> a = (7.568 ^2) / 6980 = 8.205 x 10^-3 m / s^2
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16. On the moon, Bob weighs 160 N while on earth Fred weighs 882 N Who has thegreater mass?
We are given the weight of two people, one on the moon and the other on earth. To determine the mass we will use the following formula:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} W=\text{ weight,}\lbrack N\rbrack \\ m=\text{ mass,}\lbrack kg\rbrack \\ g=\text{ acceleration of gravity, }\lbrack\frac{m}{s^2}\rbrack \end{gathered}[/tex]Now, we solve for the mass "m" by dividing both sides by "g":
[tex]\frac{W}{g}=m[/tex]Now, for the case of the moon we have that the acceleration of gravity is:
[tex]g_{moon}=1.62\frac{m}{s^2}[/tex]Plugging in the values:
[tex]\frac{160N}{1.62\frac{m}{s^2}}=m[/tex]Solving the operations:
[tex]98.77kg=m[/tex]Now, for the case of the earth we have:
[tex]g_{\text{earth}}=9.8\frac{m}{s^2}[/tex]Plugging in the values:
[tex]\frac{882N}{9.8\frac{m}{s^2}}=m[/tex]Solving the operations:
[tex]90kg=m[/tex]Therefore, the greater mass is the mass of Bob.
A railroad locomotive is at rest with its whistle shrieking and then it starts moving toward you. Describe what happens to the frequency and wavelength of the sound that hits your ear when:A) It is approaching you.B) It pulls away from you.What happens to the amplitude in both of these scenarios?
a) Frequency increases , the time between waves decreases , wavelenghts decrease
b) Frequency decreases , wavelenght increases.
The doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. AS the source approaches the listener , the sound waves get closer together, increasing their frequency and decreasing their wavelenghts.
Amplitude remains the same.
7. Determine the resistance of a lamp that draws 3 amperes when connected to a120-volt supply
A lamp draws 3 amperes of current and is connected to a 120 volt supply.
We are asked to find the resistance of the lamp.
Recall the Ohm's law is given by
[tex]V=IR[/tex]Where V is the voltage, I is current, and R is the resistance.
Re-arrange the equation for resistance
[tex]R=\frac{V}{I}[/tex]Substitute the given values
[tex]\begin{gathered} R=\frac{120}{3} \\ R=40\; \Omega \end{gathered}[/tex]Therefore, the resistance of the lamp is 40 ohms.
Graphing is a way to represent data visually to represent the motion of a rolling ball students craft data on A-line graph which of the following would accurately label the X ax
Such an object lacks a velocity since it lacks a slope. Such an object must have a variable velocity. It's quickening.
Which kind of graph is utilized for motion?Starting off, let's graph a few instances of motion with constant speed. The graph on the right shows three alternative curves, each with a zero initial point. The first thing to note is how straight each graph is. (On a graph, a curve can be any type of line that is drawn.
It is not possible to determine the velocity from a position-time graph's slope when the graph is curved. Only straight lines have slope as a characteristic. Such an object lacks a velocity since it lacks a slope. Words like "the" and "a" are highlighted in this text to emphasize that there isn't a single velocity for these conditions. Such an object must have a variable velocity. It's quickening.
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a 10 kg block of ice slides a ramp 20 m long. inclined at 10 degrees to the horizontal, if the ramp frictionless what is the acceleration of the ice
The acceleration of the ice is obtained as 0.97 m/s².
What is the acceleration?We have to recall that the acceleration has to do with the rate at which the velocity is change per unit time. Now, we know that from the parameters that have been given in the question that is before us here;
μmgcos(θ) = ma
μgcos(θ) = a
Hence;
μ = coefficient of friction
g = acceleration due to gravity
θ = the angle of inclination of the plane
We can now find the acceleration by the use of the formula as shown above hence we have;
a = (0.1)(9.8) cos(10)
a = 0.97 m/s²
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1) speed of bob?2)Show calculation for the speed of the bob:(write formula first, then insert value with unit, then find the result with unit)
Speed of bob:
[tex]\begin{gathered} \omega=2\pi f \\ _{\text{ }}where\colon \\ f=\frac{1}{T} \\ \omega=\frac{2\pi}{T} \end{gathered}[/tex]so:
[tex]\begin{gathered} \omega=\frac{2\pi}{1.293} \\ \omega=4.83\frac{rad}{s} \end{gathered}[/tex][tex]\begin{gathered} v=\omega\cdot r \\ so\colon \\ v=4.83\cdot0.22 \\ v=1.06\frac{m}{s} \end{gathered}[/tex]
Provide an example of a situation that can be explained by each of Newton's Three Laws of Motion.
ANSWER and EXPLANATION
First, let us state Newton's three laws of motion.
The first law of motion states that:
The second law of motion states that:
The third law of motion states that:
Let us take an example of a ball that is stationary on a frictionless floor. As long as no force acts on the ball, it will continue to remain at rest but once a person kicks the ball, it begins to move and stay in motion (since the floor is frictionless, hence, no resistive force).
That is an application of the first law of motion.
The acceleration of this ball depends on the force with which it was kicked and the mass of the ball, as shown in the force equation:
[tex]F=ma[/tex]where m = mass, a = acceleration
That is an application of the second law of motion.
Assuming that this ball then comes in contact with a wall and stops moving. The force that the ball exerts on the wall is equal to the force that the wall exerts on the ball, but both forces act opposite one another.
That is an application of the third law of motion.
A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle α (Figure 1). Initially the truck is moving downhill at speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β. The truck ramp has a soft sand surface for which the coefficient of rolling friction is μr.
What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.
Express your answer in terms of m , α , v0 , L , g , β and μr .
((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ) is the distance that the truck moves up the ramp before coming to a halt.
Let the distance the truck moves up the ramp be by x.
The kinetic energy of the truck on an icy road is given by,
K1 = (1÷2)mv²
The potential energy of the truck on an icy road is given by,
U1 = mgLSinα
The kinetic energy of the truck on the tuck ramp is given by,
K2 = 0
The potential Energy of the truck-on-truck ramp is given by,
U2 = mgxSinβ
Work done is given by,
W(others) = -µ×mg×cosФ
Hence, by using the work-energy theorem,
W(others) = (K2 + U2)(K1 + U1)
Therefore, by putting the values we get,
((1÷2)mv² + mgLSinα)(0 + mgxSinβ) = -µ×mg×cosФ
x = (K1 + mgLSinα) ÷ (mg(Sinβ + µcosβ))
x = ((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ)
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What would happen if an unbalanced force was acting on an object? Would it move at a constant speed? Explain why or why not.
According to Newton's Second Law of Motion, if a net force acts on an object, the object accelerates acording to the equation:
[tex]\Sigma F=ma[/tex]Where m is the mass of the object.
Then, if an unbalanced force is acting on an object, the net force is different from 0, which means that the object will accelerate in the direction of the net force.
If an object has a uniform acceleration, it won't have a constant velocity, but it may have a constant speed anyway if the force is always perpendicular to the velocity, in that case the object would be moving in a circular trajectory.
In other words, an unbalanced force will cause an acceleration, this can either change the speed of an object or the direction of its velocity (or both).
Then, it might be possible for an object to have a constant speed under the influence of an unbalanced force if that object follows a circular trajectory.
At room temperature, most metals are
A. Solids
B. Liquids
C. Gases
D. Plasma
Answer:
A. Solids
Explanation:
only Mercury is liquid soo .... solid
Which is true of the force pair of Newton's third law?a.)The two forces are in the same direction.b.)The two forces never produce an acceleration. c.)The two forces act on different objects.
Given:
Force pair of Newton's third law.
Required:
To find the true statement for the force pair of Newton's third law.
Explanation:
According to Newton's third law
This can be understood by taking an example.
Suppose a person hits the wall with his hand, the person hitting the wall is the action.
This action results in pain in the person's hand which is the reaction.
Here, the hand exerted force on the wall and the wall exerted an equal force in opposite direction.
The conclusion can be drawn as
Action and reaction act on two different objects.
The forces act in opposite direction.
Final Answer: The two forces act on different objects.