The work done by the force on the particle is 39J and the angle between F and s is 19.7 degree.
What is Force ?An item with mass is pulled or pushed which alters its velocity. A material that has the ability to change a body's rest or motion state is referred to as an external force. It possesses a magnitude and a direction.
Briefing:Force F = (Fx, Fy)
Displacement S = (Sx,Sy)
Fx=10N
Fy=-1N
Sx=4m
Sy=1m
F=(10,-1)N and S=(4,1)m
Work done is calculated by taking dot product of force and displacement vector:
W=F·S
W=10×4 + (-1)×1
W=40+(-1)=39J
W = 39 J
To find angle between F and s:
[tex]|F|=\sqrt{10^2+(-1)^2}[/tex]
|F| = √101
|S| = √(4² + 1²)
|S| = √17
W = |F| |S| cosθ
39 = √101 √17 cosθ
θ = 19.7.
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Two protons enter a region of the transverse magnetic field. What will be the ratio of the time period of revolution if the ratio of energy is 2√2 : √3 ?
Given:
ratio of energy is 2√2 : √3
Apply:
[tex]T=2\pi\sqrt[\frac{}{}]{\frac{mr}{qBv}}[/tex]Where:
q = charge of proton
v= speed of proton
r= radius of circular path
T= time period of revolution
Kinetic energy (K)
K= 1/2mv^2
From both equations:
Tα1/k
K1:K2 = 2√2 : √3
T1:T2 = √3:2√2
Answer: √3:2√2
Question 2 of 10A motor can convert:A. electrical energy to mechanical energy.B. nuclear energy to mechanical.O C. solar energy to chemical energy.D. mechanical energy to chemical energy.SUBMIT
A motor is an electromechanic device, which is able to both receive as input electrical energy, and output mechanical energy, but also receive mechanical energy, and output electrical energy, as it happens on hydroelectrical usines. This leaves us with answer A) Electrical energy to mechanical energy.
Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a woman accidentally steps barefoot on a thumbtack. About how much time does it take the nerve impulse to travel from the foot to the brain (in s)? Assume the woman is 1.60 m tall and the nerve impulse travels at uniform speed._______ s
Given:
The speed of the nerve impulse in a human body is: v = 100 m/s
The height of the woman is: d = 1.60 m
To find:
Time taken by nerve impulses to travel from the foot to the brain.
Explanation:
We assume that the nerve impulse travels at a uniform speed.
The speed "v" is given as:
[tex]v=\frac{d}{t}[/tex]Rearranging the above equation, we get:
[tex]t=\frac{d}{v}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} t=\frac{1.60\text{ m}}{100\text{ m/s}} \\ \\ t=0.016\text{ s} \end{gathered}[/tex]Final answer:
Thus, a nerve impulse takes 0.016 seconds to travel from foot to brain.
I need help pls. Will mark brainliest.
Ek=kinetic energy (J)
m=mass of object (Kg)
v=speed of object (m/s)
We have the following formula:
Ek=(1/2)mv² ⇒ m=2(Ek)/v²
In this case:
m= mass of the child + mass of the bike
m=30 kg + m₁ (m₁=mass of the bike).
We would have to modify our formula like this:
m=2(Ek)/v²
m=30 kg + m₁
30 kg +m₁=2(Ek)/v²
m₁=2(Ek)/v²-30 kg
We have the following data:
v=0.6 m/s
Ek=12.4J
m=30 Kg + m₁
Therefore:
m₁=2(Ek)/v² - 30 kg
m₁=2(12.4 J)/(0.6 m/s)² - 30 kg
m₁=24.8 J/(0.36 m²/s²) -30 kg
m₁=68.89 Kg - 30 kg
m₁=38.89 Kg
Answer: the mass of the bike is 38.89 Kg.
19 is seven plus three times a number. find the number
Let's use the variable x to represent the missing number.
Three times a number is the same as "3x".
Seven plus three times a number is equal to "7 + 3x"
Then, this expression is equal to 19, so let's write the corresponding equation:
[tex]19=7+3x[/tex]Now, solving the equation for x, we have:
[tex]\begin{gathered} 7+3x=19 \\ 3x=19-7 \\ 3x=12 \\ x=\frac{12}{3} \\ x=4 \end{gathered}[/tex]Therefore the number is 4.
A sled is dragged along a rough horizontal path at a constant speed of 1.5 m/s by a rope that is inclined at an angle at 30° with respect to the horizontal, as shown right. The total weight of the sled is 470N. The tension force in the rope is 260 N. (a) Calculate the net force acting on the sled.
The net force acting on the sled if the tension force in the rope is 260 N and the total weight of the sled is 470N is
θ = 30°
Tension, T = 260 N
Weight, W = 470 N
Weight, W = Normal force, N = 470 N
Resolving T into its horizontal and vertical components,
[tex]T_{x}[/tex] = T cos θ
[tex]T_{y}[/tex] = T sin θ
Since velocity is constant, acceleration will be zero.
∑ [tex]F_{x}[/tex] = 0
[tex]T_{x}[/tex] - [tex]F_{k}[/tex] = 0
T cos θ = [tex]F_{k}[/tex]
[tex]F_{k}[/tex] = 260 * cos 30°
Frictional force, [tex]F_{k}[/tex] = 226.2 N
Net force, F = T - [tex]F_{k}[/tex]
F = 260 - 226.2
F = 33.8 N
Therefore, the net force acting on the sled is 33.8 N
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A trebuchet fired a tennis ball with an initial velocity. Determine the following
ANSWER
[tex]\begin{gathered} (a)\text{ }12.61\text{ }s \\ (b)\text{ }194.73m \\ (c)\text{ }608.57\text{ }m \\ (d)\text{ }78.4\text{ }m\/s \end{gathered}[/tex]EXPLANATION
Parameters given:
Initial velocity, v0 = 78.4
Angle of projectile, θ = 52 degrees
(a) To find the flight time of the tennis ball, apply the formula:
[tex]t=\frac{2v_0\sin\theta}{g}[/tex]where g = acceleration due to gravity
Hence, the total flight time of the tennis ball is:
[tex]\begin{gathered} t=\frac{2*78.4*\sin52}{9.8} \\ t=12.61\text{ }s \end{gathered}[/tex](b) To find the maximum altitude of the ball during its flight, apply the formula:
[tex]H=\frac{v_0^2\sin^2\theta}{2g}[/tex]Therefore, the maximum height attained by the tennis ball is:
[tex]\begin{gathered} H=\frac{78.4^2*\sin^2(52)}{2*9.8} \\ H=194.73\text{ }m \end{gathered}[/tex](c) To find the horizontal distance the tennis ball travels, apply the formula:
[tex]R=\frac{v_0^2\sin2\theta}{g}[/tex]Hence, the horizontal distance traveled by the tennis ball is:
[tex]\begin{gathered} R=\frac{78.4^2*\sin(2*52)}{9.8} \\ R=608.57\text{ }m \end{gathered}[/tex](d) To find the final velocity of the tennis ball, apply the formula:
[tex]v=\sqrt{v_0^2+2h(-g)}[/tex]where h = initial height = 0 m
Hence, the final velocity of the tennis ball just before impact is:
[tex]\begin{gathered} v=\sqrt{78.4^2+2(0)(-9.8)} \\ v=\sqrt{78.4^2+0}=\sqrt{78.4^2} \\ v=78.4\text{ }m\/s \end{gathered}[/tex]A skydiver accelarates at 9.81 m/s² for 1.8 s. What is her speed?
Answer:
17.658 m/s
Explanation:
[tex]v=v_{0} +at[/tex]
Assuming her initial speed, [tex]v_0[/tex], is zero, this give us:
[tex]v=9.81(1.8)[/tex]
[tex]v=17.658[/tex] m/s
If you drop a 2.6 kg ball from the top of a 33 m high building, how fast will it be going just before it hits the ground? Round your answer to the nearest tenth and include an appropriate unit for credit.
Given
m = 2.6 kg
h = 33m
vo = 0 m/s
g = 9.8 m/s2
Explanation
Let's solve this question using the free fall equations.
[tex]\begin{gathered} v_f^2=v_o^2+2gh \\ v_f=\sqrt{2gh} \\ v_f=\sqrt{2*9.8m/s^2*33m} \\ v_f=25.43\text{ m/s} \end{gathered}[/tex]The answer would be 25.4 m/s
Simplify the indical equation 5m^2×6ms^-2/10s
We are given the following expression:
[tex]\frac{5m^2\times6ms^{-2}}{10s}[/tex]First, we will solve the operations associated to the coeffitients:
[tex]\frac{30m^2ms^{-2}}{10s}[/tex]Solving the fraction:
[tex]3\frac{m^2ms^{-2}}{s}[/tex]Now we solve the units of measurement:
[tex]\frac{3m^3}{s^3}[/tex]A ball is thrown horizontally at a height of 1.6 metres above the ground, with initialspeed 14 m s^-1a)Find the time of flight of the ball, giving the answer as a fraction.b)Find the range of the ball.
Given:
The initial height of the ball, h=1.6 m
The initial speed of the ball, u=14 m/s
To find:
a) The time of flight of the ball.
b) The range of the ball.
Explanation:
As the ball is thrown horizontally, the ball will have no vertical component of the initial velocity. The velocity of the ball is completely horizontal.
Thus the vertical component of the initial velocity of the ball is u_y=0 m/s.
The horizontal component of the initial velocity of the ball is u_x=u=14 m/s.
a)
From the equation of motion,
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]Where g is the acceleration due to gravity and t is the time of flight of the ball.
On substituting the known values,
[tex]\begin{gathered} 1.6=0+\frac{1}{2}\times9.8\times t^2 \\ \implies t=\sqrt{\frac{2\times1.6}{9.8}} \\ =0.57\text{ s} \end{gathered}[/tex]b)
The range of the ball is given by,
[tex]R=u_xt[/tex]On substituting the known values,
[tex]\begin{gathered} R=14\times0.57 \\ =7.98\text{ m} \end{gathered}[/tex]Final answer:
a) The time of flight of the ball is 0.57 s
b) The range of the ball is 7.98 m
Two vehicles are traveling when they enter an intersection and crash and stick together. Both have a mass of 1,650 kg and both are traveling at 15 m/s. If one is headed North and the other is headed East, after the collision they end up traveling NE together at what speed (in m/s)? Please input your answer as a positive number with two decimal places.
Answer:
10.61 m/s
Explanation:
To find the final speed, we will use the conservation of momentum in each direction, so we can write the following equations:
[tex]\begin{gathered} p_{ix}=p_{fx} \\ m_1v_{1x}+m_2v_{2x}=(m_1+m_2)v_{fx} \\ \text{and} \\ p_{iy}=p_{fy} \\ m_1v_{1y}+m_2v_{2y}=(m_1+m_2)v_{fy} \end{gathered}[/tex]Where m1 and m2 are the mass of the vehicles, v1 and v2 are their respective velocities in each direction, and Vfx and Vfy are the final velocities in each direction.
If one of the vehicles is headed north, its horizontal velocity Vx = 0 m/s. In the same way if the other is headed east, its vertical velocity Vy = 0m/s
Therefore, we can replace the values and solve the first equation as:
[tex]\begin{gathered} 1650(0)+1650(15)=(1650+1650)v_{fx} \\ 24750=3300v_{fx} \\ \frac{24750}{3300}=v_{fx} \\ 7.5m/s=v_{fx} \end{gathered}[/tex]In the same way, we can solve the second equation as:
[tex]\begin{gathered} 1650(15)+1650(0)=(1650+1650)v_{fy} \\ 24750=3300v_{fy} \\ \frac{24750}{3300}=v_{fy} \\ 7.5m/s=v_{fy} \end{gathered}[/tex]Now, we know that the vertical and horizontal speed was 7.5 m/s. So, we can calculate the final speed using the Pythagorean theorem as:
[tex]\begin{gathered} v=\sqrt[]{(v_{fx})^2+(v_{fy})^2} \\ v=\sqrt[]{(7.5)^2+(7.5)^2} \\ v=\sqrt[]{56.25+56.25} \\ v=\sqrt[]{112.5}=10.61\text{ m/s} \end{gathered}[/tex]So, the speed after the collision was 10.61 m/s
What is the angular speed and the number of revolutions per minute?
Given data:
* The linear speed of the cyclist is v = 19 miles per hour.
* The diameter of the circular motion is d = 30 inches.
Solution:
The radius of the circular motion is,
[tex]\begin{gathered} r=\frac{d}{2} \\ r=\frac{30}{2} \\ r=15\text{ inches} \end{gathered}[/tex]The radius of the circular path in terms of feet is,
[tex]\begin{gathered} r=\frac{15}{12}\text{ ft} \\ r=1.25\text{ ft} \end{gathered}[/tex]The linear velocity of the cyclist in terms of feet per minute is,
[tex]\begin{gathered} v=19\times\frac{5280}{60}\text{ ft/min} \\ v=19\times\frac{5280}{60}\text{ ft/min} \\ v=1672\text{ ft/min} \end{gathered}[/tex]The angular speed of the cyclist is,
[tex]\begin{gathered} v=r\omega \\ \omega=\frac{v}{r} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \omega=\frac{1672}{1.25} \\ \omega=1337.6\text{ rad/min} \\ \omega=1338\text{ rad/min} \end{gathered}[/tex]Thus, the angular speed of the cyclist is 1338 radian per minute.
(b). The number of revolutions in one minute is,
[tex]\begin{gathered} \omega=2\pi f \\ f=\frac{\omega}{2\pi} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} f=\frac{1338}{2\pi} \\ f=213\text{ rev/min} \end{gathered}[/tex]Thus, the number of revolutions per minute is 213 rev/min.
What is the net force of 10n up and 20n to the right? Draw an arrow.
how is constant speed different from no speed on a motion graph
Answer:
No speed is when there is a line on the horizontal axis where speed=0. Constant speed is when there is a horizontal line above the horizontal axis.
Explanation:
explain the two main mistakes individuals made that led to galileo's trial
The first mistake individuals made was to believe that Earth was the center of everything, as the Catholic Church believed. Galileo's idea about Earth revolving arount the Sun was heretical and enough to me condemned.
The second mistake individuals made was to believe that all the physical things was comanded by God only, denying the possibility of having formal science to study the physical world.
Two protons are 11.86 fm apart. (1 fm= 1 femtometer = 1 x 10-15 m.) What is the ratio of the electric force to the gravitational force on one proton due to the other proton?
This is the answer tab
Firstly, we need to write the formulas for both the gravitational force and electric force. Our gravitational force is:
[tex]F_g=G\frac{m_1m_2}{d^2}[/tex]And the electric force is:
[tex]F_e=k\frac{q_1q_2}{d^2}[/tex]We can see that these forces have almost equal formulas. What we want is Fe/Fg. Before this, we can simplify the forces, as both particles have the same charge and mass. We're left with the following:
[tex]F_g=G\frac{m^2}{d^2}[/tex]And
[tex]F_e=k\frac{q^2}{d^2}[/tex]By dividing both, we get
[tex]\frac{F_e}{F_g}=\frac{(\frac{kq^2}{d^2})}{(\frac{Gm^2}{d^2})}=\frac{kq^2}{d^2}*\frac{d^2}{Gm^2}[/tex]We have d^2 on the numerator and denominator. We can elimante the distance then, as it is different from zero. We have the following:
[tex]\frac{F_e}{F_g}=\frac{kq^2}{Gm^2}[/tex]We can then replace our values with the constants. k is Coulomb's constant, q is the charge of a proton, G is Newton's constant, and m is the mass of a proton. We finally get
[tex]\frac{F_e}{F_g}=\frac{(9*10^9)*(1.6*10^{-19})^2}{(6.67*10^{-11})*(1.67*10^{-27})^2}=1.2386*10^{36}[/tex]So, the electric force is 1.2386*10^36 times higher than the gravitational. The most interesting about this, is that it doesn't depend on the distance the two of them are apart.
an electric bulb is rated 220vand 100w when it isoperated on 110v the power consumed will be?
Answer:
P = I V = I * 220 = 100
I = .45 amps bulb rating
P(110) = .45 * 110 = 49.5 W at 110 V (1/2 its value at 220V)
In the diagram, A(-7; 4), B(-6; 6), C(0; 3) and D(-1; t) are the vertices of a rectangle. Calculate the: a) length of the diagonal AC.
Explanation:
150 km North, 50 km West,
Which is true about gravitational force?
A. Gravitational force can be repulsive or attractive.
B. Gravitational force acts only on charged particles.
C. Gravitational force has infinite range.
D. Gravitational force is a contact force.
Gravitational force acts only on charged particles. Option B.
Gravity is the force between two bodies whether or not there is a body between them. All bodies in the universe attract all other bodies with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The universal attractive force acting between objects is called gravity. It is one of the fundamental forces of the universe. The magnitude of this force depends on the mass of each object and the distance between the centers of the two objects. Mathematically, we say that gravity depends directly on the mass of an object and is inversely proportional to the square of the distance between them.
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The correct answer is highlighted. I need and explanation on how to arrive at that answer please and thank you.
ANSWER
Wbottom = Fadhesion
EXPLANATION
The system is formed by the two blocks together. The tension force is holding these two blocks, so the tension force is equal to the total weight of the system - in other words, to the sum of the weights of the blocks.
The adhesion force is the force that is keeping the two blocks stuck together, so it is not an external force, but an internal force of the system. Since the blocks are hanging, the adhesion force must be such that the bottom block doesn't fall off. Hence, the adhesion force is equal to the weight of the bottom block.
Determine the applied force (in newtons, N) required to accelerate a 3.46-kg object rightward with a constant acceleration of 1.86 m/s^2 if the force of friction opposing the motion is 16.4 N.
Explanation
The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
[tex]\begin{gathered} F=ma \\ where\text{ F is the force} \\ a\text{ is the acceleration} \\ m\text{ is the mass} \end{gathered}[/tex]
Step 1
Free body diagram
so
the sum of force in x direction
[tex]\begin{gathered} sum\text{ of forces=required force-force of friction} \\ replace \\ sum=required\text{ force-16.4 N} \end{gathered}[/tex]now, replace in the formula
[tex]\begin{gathered} F=ma \\ required\text{ force-16.4 N=3.46 kg*1.86}\frac{m}{s^2} \\ required\text{ force-16.4 N=6.4354 N} \\ add\text{ 16.4 N in both sides} \\ requ\imaginaryI red\text{force-16.4N+16.4 N=6.435,4N+16.4 N} \\ requ\mathrm{i}red\text{ force=22.83 N} \end{gathered}[/tex]therefore, the required force is 22.83 N
I hope this helps you
An applied force of 75.0 N accelerates a 10.0 kg block at 3.5 m/s2 along a horizontal surface. a. How large is the frictional force? b. What is the coefficient of friction?
Answer:
a. 40.0 N
b. 0.41
Explanation:
By the second law of Newton, the net force is equal to the mass times the acceleration. In this case, the net force is the applied force less the frictional force, so
[tex]\begin{gathered} F_{\text{net}}=ma \\ F-F_f=ma \end{gathered}[/tex]Where F is the applied force, Ff is the friction force, m is the mass and a is the acceleration.
Solving for Ff, we get
[tex]Ff=F-ma[/tex]Now, we can replace the values
[tex]\begin{gathered} Ff=75.0N-(10.0\operatorname{kg})(3.5m/s^2) \\ F_f=75.0N-35.0N \\ F_f=40N \end{gathered}[/tex]So, the frictional force is 40N
On the other hand, the force of friction is equal to the normal force times the coefficient of friction. Where the normal is equal to its weight, so
[tex]\begin{gathered} F_f=\mu F_n \\ F_f=\mu mg \end{gathered}[/tex]Where μ is the coefficient of friction and g is the gravity and it is equal to 9.8 m/s². Solving for μ
[tex]\mu=\frac{F_f}{mg}[/tex]Now, we can replace the values to get
[tex]\mu=\frac{40.0N}{10.0\operatorname{kg}(9.8m/s^2)}=0.41[/tex]Therefore, the coefficient of friction is 0.41
Then, the answers are
a. 40.0 N
b. 0.41
Sally has a mass of 45.9 kilograms. Earth has a mass of 5.98 x 10^24 kilograms and an average radius of 6.38 x 10^6 meters.What is the force due to gravity between Sally and Earth? Include units in your answer. Answer must be in 3 significant digits.
Newton's law of universal Gravity:
F = G * (M1 * M2)/ r^2
Where:
G = gravitational constant = 6.674 x10^-11 Nm^2kg^2
M1 = mass 1 = 45.9 kg
M2 = mass 2 = 5.98 x 10 ^24 kg
r = Distance between the 2 objects = 6.38 x 10 ^6 m
Replacing;
[tex]F=6.674x10^{-11}Nm^2kg^2\cdot\frac{45.9\operatorname{kg}\cdot5.98x10^{24}\operatorname{kg}}{(6.38x10^6m)^2}[/tex]F = 450.048 N
4. Explain A girl is jumping on a trampoline. When she is at the top of her jump, her mechanical energy is in what form? Explain why.
Mechanical energy is of two types,
1. Kinetic energy
2. Potential energy
When the girl is at the top of her jump, the speed of the girl becomes zero.
Thus, the kinetic energy of the girl at the top of her jump is,
[tex]K=\frac{1}{2}mv^2[/tex]where m is the mass of the girl and v is her speed,
Substituting the known values,
[tex]K=0\text{ J}[/tex]Thus, the kinetic energy of the girl is zero at the top of her jump.
The potential energy of the girl at the top of her jump is,
[tex]U=\text{mgh}[/tex]where g is the acceleration due to gravity and h is the height of the girl,
Thus, the potential energy of the girl at the maximum height is maximum.
Hence, the mechanical energy of the girl at the top is in potential energy form because her kinetic energy is zero at the top of her jump.
What is the potential difference across a computer power supply with a resistance of 50 Ω, if the power supply draws a current of 2.2 A?
Given data:
* The current through the computer power supply is I = 2.2 A.
* The value of resistance is R = 50 ohm.
Solution:
According to Ohm's law, the voltage (potential difference) across the computer power supply is,
[tex]V=IR[/tex]Substituting the known values,
[tex]\begin{gathered} V=2.2\times50 \\ V=110\text{ volts} \end{gathered}[/tex]Thus, the potential difference across the computer supply is 110 volts.
Assume no loss of energy to friction. If a 270 kg roller coaster is 96.8 meters tall and has an initial velocity of zero, what will the velocity be at the bottom?
ANSWER
43.56 m/s
EXPLANATION
Given:
• The roller coaster's mass, m = 270 kg
,• The roller coaster's height, h = 96.8 m
,• The roller coaster's initial velocity, v₀ = 0 m/s
Unknown:
• The roller coaster's final velocity at the bottom of the hill, v
By the law of conservation of energy, the total mechanical energy of the roller coaster at the top of the hill must be equal to its total mechanical energy at the bottom. If no energy is lost to friction, then we only have gravitational potential energy and kinetic energy,
[tex]PE_o+KE_o=PE_f+KE_f[/tex]The height where the roller coaster starts its motion is with respect to the bottom of the hill. This means that at the bottom of the hill, we could say that its height is zero and, therefore, it has no gravitational potential energy.
Also, we know that the roller coaster starts from rest, so its initial kinetic energy is zero. The equation above reduces to,
[tex]PE_o=KE_f[/tex]Replace with the expressions for each of these kinds of energy,
[tex]m\cdot g\cdot h=\frac{1}{2}\cdot m\cdot v^2[/tex]The mass is on both sides of the equality, so it cancels out,
[tex]g\cdot h=\frac{1}{2}v^2[/tex]Solving for v,
[tex]v=\sqrt[]{2\cdot g\cdot h}[/tex]Replace with the known values. The acceleration due to gravity is 9.8 m/s,
[tex]v=\sqrt[]{2\cdot9.8m/s\cdot96.8m}\approx43.56m/s[/tex]Hence, the velocity of the roller coaster at the bottom is 43.56 m/s.
A woman bungee jumper jumps from the top of a bridge with a height of 75m. The length of the bungee elastic is 25m and the woman has a mass of 60kg.
(a) Calculate the potential energy of the woman as she stands on top of the bridge.
(b) Calculatethemaximumspeedthatthewomanreachesandstateat what height this will happen.
Answer:
Explanation:
Given:
H = 75 m
m = 60 kg
g = 9.8 m/s
__________
Wp - ?
V - ?
a) The potential energy:
Wp = m·g·H = 60·9.8·75 = 44 100 J
b) The maximum speed:
V = √ (2·g·H) = √ (2·9.8·60) = √ 1176 ≈ 34 m/s
The Sun radiates energy at the rate of 3.80 ✕ 1026 W from its 5500°C surface into dark empty space (a negligible fraction radiates onto Earth and the other planets). The effective temperature of deep space is −270°C. (Due to the sensitive nature of the calculations, use T(K) = T(°C) + 273.15.(a) What is the increase in entropy (in J/K) in one day due to this heat transfer?
Given:
The sun radiates energy at a rate of
[tex]W=3.80\times10^{26}\text{ W}[/tex]The temperature in the empty space is,
[tex]\begin{gathered} T_H=5500+273 \\ =5773\text{ K} \end{gathered}[/tex]The temperature at the deep space is,
[tex]\begin{gathered} T_c=-270+273 \\ =3\text{ K} \end{gathered}[/tex]To find:
the increase in entropy (in J/K) in one day
Explanation:
The value of heat is,
[tex]\begin{gathered} Q_H=Q_c=Q=3.80\times10^{26}\times24\times3600 \\ =3.28\times10^{31}\text{ J} \end{gathered}[/tex]The change in entropy is,
[tex]\begin{gathered} \Delta s=-\frac{Q_H}{T_H}+\frac{Q_c}{T_c} \\ =-\frac{3.28\times10^{31}}{5773}+\frac{3.28\times10^{31}}{3} \\ =1.08\times10^{31}\text{ J/K} \end{gathered}[/tex]Hence, the increase in entropy is
[tex]1.08\times10^{31}\text{ J/K}[/tex]Yesenia performed an experiment on the motion of a pendulum. Her data from one trial is shown on the graph below.
As the experiment of the motion of the pendulum is performed. The motion of the pendulum is periodic.
This periodic nature can be represented the sinusoidal functions.
In the given graph, the position of the pendulum is repeated after a regular interval of time which can be represented by the sinusoidal function whose value oscillates.
Thus, the sinusoidal function can best fit the data represented in the figure.
Hence, the second option is the correct answer.