If I know it is table on ball and earth on ball what do I need to find for this question?
Answers
To solve this question, we have to know which force is applied to which object.
In this case, the force is frictional, which means it's a force from the surface of the table. The property of the surface is acting upon the book, preventing to slide.
Hence, it's table-on-book.Related Questions
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?
Answers
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]What is the net force of 10n up and 20n to the right? Draw an arrow.
Answers
Yesenia performed an experiment on the motion of a pendulum. Her data from one trial is shown on the graph below.
Answers
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.
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.
Answers
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
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
Answers
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.
An iron rod with an initial length of 12.64 m has its temperature raised from 9o C to 38.10o C. If iron has a coefficient of thermal expansion of 12x10-6 1/oC, what is the change in length of the rod in mm?
Answers
We will have the following:
First, we remember that:
[tex]\Delta L=\alpha\ast L_0\ast\Delta T[/tex]So:
[tex]\begin{gathered} \Delta L=(12\ast10^{-6}1/C)(12.64m)(38.1C-9C)\Rightarrow\Delta L=4.413888\ast10^{-3}m \\ \\ \Rightarrow\Delta L\approx4.41mm \end{gathered}[/tex]So, the change in length of the rod is approximately 4.41 mm.
What is the principle of vernier callipers?
Answers
[tex]{ \blue{ \bold{Here \: is \: your \: answer...}}}[/tex]
The vernier callipers is mainly used to measure the diameter of spherical ball and the principle states that "The magnitude of 'n' vernier scale divisions is equal to the magnitude of (n-1) number of main scale division".
A thin flashlight beam traveling in air strikes a glass plate at an angle of 38 degrees with the closet on the plane of the surface of the plate. If the index of refraction of the glass is 1.4, what angle will the refracted beam make with the normal in the glass?
Answers
The given problem can be exemplified in the following diagram:
We can use Snell's law which says the following:
[tex]n_{\text{air}}\sin i=n_{}\sin r[/tex]Where:
[tex]\begin{gathered} i=\text{ angle of incidence} \\ r=\text{ angle of refraction} \\ n=\text{ index of refraction of the material} \\ n_{\text{air}}=\text{ index of refraction of air} \end{gathered}[/tex]We will take the index of refraction of air to be 1. Now we solve for the angle of refraction:
[tex]\sin i=n\sin r[/tex]Now we divide by "n"
[tex]\frac{\sin i}{n}=\sin r[/tex]Taking the inverse sine function:
[tex]\arcsin (\frac{\sin i}{n})=r[/tex]The angle of incidence can be determined having into account that the sum of the given angle and the angle of incidence must be equal to 90, therefore:
[tex]\begin{gathered} 38+i=90 \\ i=90-38 \\ i=52 \end{gathered}[/tex]Now we substitute the values:
[tex]\arcsin (\frac{\sin 52}{1.4})=r[/tex]Solving we get:
[tex]r=34.25[/tex]Therefore, the angle of refraction is 34.25 degrees.
a force of 300n produces pressure of 20 N/m2. On what are is it acting? show your working
Answers
Answer:
Explanation:
Given:
F = 300 N
p = 20 N/m²
__________
A - ?
Valid for the area A:
A = F / p
A = 300 / 20 = 15 m²
A cannon is fired horizontally
from a 5.0 m tall tower. How much time will it take until the projectile hits the ground?
Answers
Answer:
1.0 s
Explanation:
How fast the projectile is traveling HORIZONTALLY has no effect on how long it takes to hit the ground......only the initial height (5 m) and gravity (-9.81 m/s^2) will affect how long it takes to do so...
df = final position = 0 when it hits the ground
df = do + vot + 1/2 a t^2
0 = 5 + 0 t - 1/2 (9.81) t^2
5 = 1/2 (9.81) t^2
t = 1.0 s
In Problem 6, what is the acceleration of the electron? (Recall, F=ma. The mass of an electron is 9.11.103 kg. The answer will be a large number.)
Answers
Answer:
Explanation:
And the condition of problem N 6 ???
an electric bulb is rated 220vand 100w when it isoperated on 110v the power consumed will be?
Answers
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)
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.
Answers
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
For the following questions, please provide a complete step by step solution. You do notneed interaction, but you are required force diagrams.
Answers
Tennsion 1= horizontal = 484.94 N
Tension 2 = slanted string = 560 N
Explanation
Step 1
Free body diagram
Newton's first law says that if the net force on an object is zero, like in this case the mass is in rest,then that object will have zero acceleration
so
Step 1
set the equations:
a) for x-axis
[tex]\begin{gathered} \sum ^{}_{}F_x=0 \\ so \\ T_{2x}-T_1=0 \\ T_2\cos 30-T_1=0\rightarrow equation(1) \end{gathered}[/tex]b) for y -axis
[tex]\begin{gathered} \sum ^{}_{}F_y=0 \\ so \\ T_{2y}-w=0 \\ T_2\sin 30-280N=0\rightarrow equation(2) \end{gathered}[/tex]Step 2
Solve the equations
[tex]\begin{gathered} T_2\cos 30-T_1=0\rightarrow equation(1) \\ T_2\sin 30-280N=0\rightarrow equation(2) \end{gathered}[/tex]a) solve for T2in equation (2)
[tex]\begin{gathered} T_2\sin 30-280N=0\rightarrow equation(2 \\ \text{add 280 N in both sides} \\ T_2\sin 30-280N+280N=0+280\text{ N} \\ T_2\sin 30=280\text{ N} \\ \text{divide both sides by sin 30} \\ \frac{T_2\sin30}{\sin30}=\frac{280\text{ N}}{\sin30} \\ T_2=560\text{ N} \end{gathered}[/tex]b) replace the T2 value in equation (1) to find T1
[tex]\begin{gathered} T_2\cos 30-T_1=0\rightarrow equation(1) \\ 560\cos 30-T_1=0\rightarrow equation(1) \\ 484.97-T_1=0 \\ 484.97=T_1 \end{gathered}[/tex]therefore
Tennsion 1= horizontal = 484.94 N
Tension 2 = slanted string = 560 N
I hope this helps you
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.
Answers
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
19 is seven plus three times a number. find the number
Answers
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.
In what way does burning a campfire demonstrate this law?Explain how a battery operated flashlight illustrates the law of the Conservation of energy.(Please on the transfer of energy between the components of the system.)
Answers
Answer:
The law of conservation of energy says that the total energy of a system that is isolated is constant. So, in a battery-operated flashlight, we can see that the chemical energy of the battery is transformed into light. Therefore, the energy is not being created, it is just transformed into a new type. In the same way, the energy in the wood in a campfire is transformed into heat and light, so it is another way to demonstrated that the energy is conserved.
Would a person’s weight be the same or different on Mars as it is on Earth? What about his/her mass?
Answers
Given statement:
Would a person’s weight be the same or different on Mars as it is on Earth? What about his/her mass?
Explanation:
So an object or person on Mars would weigh 37.83% of its weight on earth Therefore, a person would be much lighter on mars.
Conversely, a person is 62.17% heavier on earth than on Mars.
Thus, the weight of the person on Mars would weigh 37.83% of its weight on Earth.
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 ?
Answers
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
Convert 812 Mg tog0.000 812 g812,000 g0.812 g812,000,000 g
Answers
We are asked to convert 812 Mg into g. To do that we will use the following conversion factor:
[tex]1Mg=1000000g[/tex]Applying the factor we get:
[tex]812Mg\times\frac{1000000g}{1Mg}=812000000g[/tex]Therefore, 1 Mg is 812 000 000g.
explain the two main mistakes individuals made that led to galileo's trial
Answers
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.
a quantity of gas in a piston cylinder has a volume of 0.500 m^3 to the power and a pressure of 200 Pa. the Piston compresses the gas to 0.150 m^3 in an isothermal ( constant temperature) process what is the final pressure of the gas
Answers
According to Boyle's Law, an ideal gas at a constant temperature will satisfy the following relation between the initial and final conditions of pressure and volume:
[tex]P_1V_1=P_2V_2[/tex]Isolate P_2 from the equation:
[tex]P_2=\frac{_{}P_1V_1_{}_{}}{V_2}[/tex]Substitute the values for the initial pressure as well as for the two volumes to find the value of the final pressure of the gas:
[tex]\begin{gathered} P_2=\frac{(200Pa)(0.500m^3)}{(0.150m^3)} \\ =667Pa \end{gathered}[/tex]Therefore, the final pressure of the gas is:
[tex]667Pa[/tex]in the figure above, calculate the magnitude of the couple
Answers
Answer: 10^23
Explanation:10 times 10
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
Answers
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.
A sailboat runs before the wind with a constant speed of 3.2m/s2 in a direction 42 degree north of west. How far (a) west and (b) north has the sailboat traveled in 25min?
Answers
Given:
Speed = 3.2 m/s²
Direction, θ = 42 degrees north of west.
Let's solve for the following:
(a) How far west has the sailboat traveled in 25 minutes.
We have the free body diagram below:
To find the distance, let's first find the x-component and y-component of the velocity.
x-component:
[tex]\begin{gathered} Vx=-V\cos \theta \\ V_x=-3.2\cos 42 \\ V_x=-2.378\text{ m/s} \end{gathered}[/tex]y-component:
[tex]\begin{gathered} V_y=V\sin \theta \\ V_y=3.2\sin 42 \\ V_y=2.141\text{ m/s} \end{gathered}[/tex]To find the distance travelled west, we are to find the distance in the x direction.
Apply the formula:
[tex]d=|V_x|t[/tex]Where:
|Vx| = |-2.378 m/s| = 2.378 m/s
t is the time is seconds = 25 x 60 = 1500 seconds
Thus, we have:
[tex]\begin{gathered} d=2.378\ast1500 \\ \\ d=3567.1\text{ m }\approx\text{ 3.6 km} \end{gathered}[/tex]The distance traveled west in 25 minutes is 3.6 km.
• (b) How far north has the sailboat traveled in 25 minutes.
Here, we are to find the vertical distance using the y-component.
Apply the formula:
[tex]d=|V_y|t[/tex]Where:
Vy = 2.141 m/s
t = 1500 seconds
Thus, we have:
[tex]\begin{gathered} d=2.141\ast1500 \\ \\ d=3211.5\text{ m }\approx3.2\text{ km} \end{gathered}[/tex]The sailboat traveled 3.2 km in the north direction.
ANSWER:
(a) 3.6 km
(b) 3.2 km
A block of weight w = 25.0 N sits on a frictionless inclined plane, which makes an angle θ = 30.0 ∘ with respect to the horizontal, as shown in the figure. (Figure 1)A force of magnitude F = 12.5 N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.
A: The block moves up an incline with constant speed. What is the total work Wtotal done on the block by all forces as the block moves a distance L = 3.80 m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest.
B :What is Wg , the work done on the block by the force of gravity w⃗ as the block moves a distance L = 3.80 m up the incline?
C: What is WF , the work done on the block by the applied force F⃗ as the block moves a distance L = 3.80 m up the incline?
D: What is WN , the work done on the block by the normal force as the block moves a distance L = 3.80 m up the inclined plane?
Answers
a ) The total work Wtotal done on the block by all forces = 0
b ) The work done on the block by the force of gravity = - 47.5 J
c ) The work done on the block by the applied force F = 47.5 N
a ) The total work Wtotal done on the block by all forces,
Since, only the work done after the block has started is considered, the total work done is the change in kinetic energy. Since, the block is moving at a constant speed there is no change in kinetic energy and so total work Wtotal done is zero.
b ) The work done on the block by the force of gravity,
Wg = Fg * h * cos θ
Fg = W = 25 N
sin 30 = h / 3.8
h = 1.9 m
Wg = 25 * 1.9 * cos 180
Wg = - 47.5 J
c ) The work done on the block by the applied force F,
WF = F * d * cos θ
WF = 12.5 * 3.8 * cos 0
WF = 47.5 N
d ) The work done on the block by the normal force,
Normal force acts perpendicular to the surface. The movement is towards the right side of the surface. So cos θ will be cos 90 which is equal to zero. So the work done will be zero.
Therefore,
a ) The total work Wtotal done on the block by all forces = 0
b ) The work done on the block by the force of gravity = - 47.5 J
c ) The work done on the block by the applied force F = 47.5 N
d ) The work done on the block by the normal force = 0
To know more about work done
https://brainly.com/question/28439201
#SPJ1
if the current density in a wire is given by j=alpha*r,where alpha is a constant and r is the distance from the center of the wire, 0
Answers
Given:
The current density is,
[tex]J=ar[/tex]a is a constant, and the radius of the wire is R.
To find:
The current in the wire
Explanation:
The current in the wire is,
[tex]\begin{gathered} I=\int JdA \\ =\int ardA \end{gathered}[/tex]We know,
[tex]\begin{gathered} A=\pi r^2 \\ dA=2\pi rdr \end{gathered}[/tex]So,
[tex]\begin{gathered} I=\int_0^Rar\times2\pi rdr \\ =2\pi a\int_0^Rr^2dr \\ =2\pi a\times\frac{R^3}{3} \end{gathered}[/tex]Hence, the required current is,
[tex]\frac{2\pi aR^3}{3}[/tex]I need help pls. Will mark brainliest.
Answers
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.
A skydiver accelarates at 9.81 m/s² for 1.8 s. What is her speed?
Answers
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
In Exercises 39 and 40, use the following information.A shot put is thrown a distance of 54.5 feet at a high school track and field meet.The shot put was released from a height of 6 feet and at an angle of 43°.39. Write a set of parametric equations for thepath of the shot put.Ans:x = (v*cos43°)t, y = -16t^2 + (v*sin43°)t + 6 -or-x = 0.731vt, y = -16t^2 + 0.682vt + 640. Use the equations to determine the speedof the shot put at the time of release.Must be solve without calc, but without t, how do you find v?I solved the quadratic and found t in terms of v, plugged that into the x-EQ, but it became a quartic, it’s an Alg 2 class so there must be a simpler solution …
Answers
We are asked to determine a set of parametric equations for the parabolic motion of an object. The parametric equations for such a motion is given by:
[tex]\begin{gathered} x=v\cdot\cos 43\degree t,(1) \\ y=-16t^2+v\cdot\sin 43\degree t+6,(2) \end{gathered}[/tex]To determine the velocity we will solve for "t" in equation (1):
[tex]t=\frac{x}{v\cos 43}[/tex]Now, we will replace this in equation (2):
[tex]y=-16(\frac{x}{v\cos43})^2+v\cdot\sin 43\degree(\frac{x}{v\cos43})+6[/tex]Simplifying:
[tex]y=-\frac{16x^2}{v^2\cos^243}+x\tan 43+6[/tex]Now we replace the values x = 54.5 and y = 0:
[tex]0=-\frac{16(54.5)^2}{v^2\cos^243}+(54.5)\tan 43+6[/tex]Simplifying:
[tex]0=-\frac{88850.1}{v^2}+56.82[/tex]Now we solve for "v":
[tex]\frac{88850.1}{v^2}=56.82[/tex][tex]88850.1=56.82v^2[/tex][tex]\frac{88850.1}{56.82}=v^2[/tex]Solving the operation:
[tex]1563.71=v^2[/tex]taking the square root:
[tex]\begin{gathered} \sqrt[]{1563.71}=v \\ 39.54=v \end{gathered}[/tex]Therefore, the velocity is 39.54 ft/s.