Radiations associated with electric and magnetic field is called Electromagnetic radiations.
electric and magnetic field radiations = electromagnetic radiations.
Given that the charge on an electron is -1.6 x 10^-19 Coulombs, what is the magnitude of the electric force on an electron in Newtons when it is placed in an electric field of strength 12,000 Volts/m? A)1.6 x 10^-19 B)3.2 x 10^-19 C)1.92 x 10^-15 D)6.4 x 10^-19 E)4.5 x 10^22
Given,
The charge on an electron is
[tex]q=-1.6\times10^{-19}\text{ C}[/tex]The electric field strength is
[tex]E\text{ = 12,000 Volts/m}[/tex]Solution:
The electric force is calculated by the formula,
[tex]F=qE[/tex]Put the given values in the formula.
[tex]\begin{gathered} F=(1.6\times10^{-19}\text{ C\rparen}\times(12,000\text{ Volts/m\rparen} \\ F=19.2\times10^{-16}\text{ N} \\ F=1.92\times10^{-15}\text{ N} \end{gathered}[/tex]Thus, the given options c is correct.
The speed of light in an unknown medium is measured to be 2.76 x10% m/s. What is the index of refraction of the medium?
Answer:
Explanation:
The formula for calculating the index of refraction is expressed as
n = c/v
where
n is the index of refraction
v is the speed of light in the medium
c is the speed of light in vacuum and its value is 3 x 10^8 m/s
From the information given,
v = 2.76 x 10^8 m/s
Thus,
n = 3 x 10^8 m/s/2.76 x 10^8 m/s
n = 1.09
the index of refraction of the medium is 1.09
State briefly rules of drawing vector on coordinate systems
In order to draw a vector in the coordinate system, we can follow these steps:
1. Draw the initial point of the vector.
2. Draw the end point of the vector.
3. Create an arrow starting at the initial point and ending at the end point.
For example, let's use a vector that starts at (0, 0) and ends at (3, 4):
A block having a mass of m = 19.5 kg is suspended via two cables as shown in the figure. The angles shown in the figure are as follows: a = 15° and B = 35°. We will label the tension in Cable 1 as T1, and the tension in Cable 2 as T2. Solve for T1 and T2
We will have the next diagram
Then we can sum the forces in x and sum the forces in y
Forces in x
[tex]\sum ^{}_{}F_x=-T_1\sin (\alpha)+T_2\cos (\beta)=0[/tex][tex]\sum ^{}_{}F_x=-T_1\sin (15)+T_2\cos (35)=0[/tex]Forces in y
[tex]\sum ^{}_{}F_y=T_1\cos (\alpha)+T_2\sin (\beta)=mg[/tex][tex]\sum ^{}_{}F_y=T_1\cos (15)+T_2\sin (35)=19.5(9.8)[/tex]We simplify the equations found and we found the next system of equation
[tex]\begin{gathered} -T_1\sin (15)+T_2\cos (35)=0 \\ T_1\cos (15)+T_2\sin (35)=191.1 \end{gathered}[/tex]then we isolate the T2 of the first equation
[tex]T_2=\frac{T_1\sin(15)}{\cos(35)}[/tex]We substitute the equation above in the second equation
[tex]T_1\cos (15)+(\frac{T_1\sin(15)}{\cos(35)})\sin (35)=191.1[/tex]we simplify
[tex]T_1(\cos (15)+\frac{\sin (15)\sin (35)}{\cos (35)})=191.1[/tex][tex]T_1(1.147)=191.1[/tex]We isolate the T1
[tex]T_1=\frac{191.1}{1.147^{}}=166.6N[/tex]then we can substitute the value we found in T1 in the euation with T2 isolate
[tex]T_2=\frac{(166.6)_{}\sin (15)}{\cos (35)}=52.54N[/tex]
How long will it take a sample of Polonium- 194 to decay to 1/16 of its original amount the half -life of 0.7 seconds
The time it takes the Polonium-194 to decay to 1/16 of its original amount is 2.8 seconds.
What is half-life?half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay
To calculate the time it takes the sample of Polonium-196 to decay to 1/16 of its original amount, we use the formula below
Formula:
[tex]2^{n/t}[/tex] = R/R'......... Equation 1Where:
n = Total number of time it takes Polonium-194 to decay to 1/16 of its original amountt = Half-life of Polonium-194R = Original amount of Polonium-194R' = Amount of Polonium-194 after decayFrom the question,
Given:
R = 1R = 1/16t = 0.7Substitute these values into equation 1 and solve for n
[tex]2^{n/0.7}[/tex]= 1/(1/16)[tex]2^{n/0.7}[/tex] = 16[tex]2^{n/0.7}[/tex] = 2⁴Equating the base,
n/0.7 = 4n = 0.7×4n = 2.8 seconds.Hence, the time it takes the Polonium-194 to decay is 2.8 seconds.
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A rope is vibrating so as to form the standing wave pattern shown at the right. 1.How many nodes are present in the rope?2.What harmonic is shown in the pattern?3.A guitar string has a fundamental frequency of 120 Hz. What is the frequency of the third harmonic of the string?
We will have the following:
1. There are 5 nodes.
2. The harmonic shown is the 4th harmonic.
3. We will detemrine the 3rd harmonic as follows:
[tex]\lambda_3=\frac{3}{2}(120Hz)\Rightarrow\lambda_3=180Hz[/tex]Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?
Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:
f = n·fo
where n is an integer and fo is the first harmonic or fundamental.
fo is given by the length L of a string, in the following way:
fo = v/λ = v/(L/2) = 2v/L
becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.
Jason is pulling a box across the room. He is pulling with a force of 16 newtons and his arm is making a 71 angle with the horizontal, if the box weighs 15 newtons what is the netforce on the box in the vertical direction? Treat up as the positive direction, and down as the negative direction
We know that
• The force is 16 Newtons.
,• The angle of application is 71 degrees.
,• The box weighs 15 Newtons.
First, let's make a free-body diagram.
As you can observe, the vertical vectors are the weight of the box and the vertical component of the applied force.
[tex]\Sigma F_y=F_y-W[/tex]Where each force is defined as follows
[tex]\begin{gathered} F_y=F\cdot\sin \theta \\ W=mg \end{gathered}[/tex]Let's use the expressions above to find the net vertical force.
[tex]\begin{gathered} \Sigma F_y=F\cdot\sin \theta-mg=16N\cdot\sin 71-15 \\ \Sigma F_y\approx0.13N \end{gathered}[/tex]Therefore, the net vertical force is 0.13 Newtons.Derive the equation S=ut+½at²
Average velocity = (initial velocity + final velocity )/ 2
AVG v = (u+v)/2
Also,
Distance = avg v x time
s= ([u+v]/2) x t
s= distance travelled by a body in time t
u= initial velocity
a= acceleration
From the equation of motion:
v= u +at
Replacing
s= ( [u+u+at]/2)xt
s= [( 2u+ 2at ( x t ]/2
s= (2ut+at^2)/2
s= ut+1/2at^2
this is a 3 part questionplease see picture for part A73) The human brain consumes about 22 W of power under normal conditions, though more power may be required during exams.(a) For what amount of time can one Snickers bar (see the note following Problem 48) power the normally functioning brain(one bar provides 280 calories)? (b) At what rate must you lift a 3.6-kg container of milk (one gallon) if the power output of your arm is to be 22 W? (C) How much time does it take to lift the milk container through a distance of 1.0 m at this rate?
Given,
The power consumed by a brain, P=22 W
The energy per bar, E=280 calories=280×4184=1171.52 kJ
The mass of a milk container, m=3.6 kg
The power output of the arm, P₀=22 W
The distance through which the container needs to be lifted, d=1.0 m
a)
The power is given by,
[tex]P=\frac{E}{t}[/tex]Where t is the time.
On substituting the known values in the above equation,
[tex]\begin{gathered} 22=\frac{1171.52\times10^3}{t} \\ \Rightarrow t=\frac{1171.52\times10^3}{22} \\ =53250\text{ s} \end{gathered}[/tex]That is,
[tex]\frac{53250}{3600}=14.79\text{ hr}[/tex]Therefore one snicker bar can power the brain for 14.79 hr
b)
The power output can also be calculated using the formula,
[tex]\begin{gathered} P=F\times v \\ =mg\times v \end{gathered}[/tex]Where F is the force applied by the container on the arm, v is the rate at which the container must be lifted, and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 22=3.6\times9.8\times v \\ v=\frac{22}{3.6\times9.8} \\ =0.62\text{ m/s} \end{gathered}[/tex]Thus the rate at which the milk container must be lifted is 0.62 m/s
c)
The rate at which the container must be lifted is given by,
[tex]v=\frac{d}{t}[/tex]Where t is the time it takes to lift the container at the calculated rate.
On substituting the known values,
[tex]\begin{gathered} 0.62=\frac{1}{t} \\ \Rightarrow t=\frac{1}{0.62} \\ =1.61\text{ s} \end{gathered}[/tex]Thus it takes 1.61 s to lift the container through 1 m at the given rate.
The standard exam page is 8.50 inches by 11.0 inches. Its area in cm 2 is?
We know that one inch is equal to 2.54 cm, this means that:
[tex]\begin{gathered} 8.5\text{ in=}21.59\text{ cm} \\ \text{and} \\ 11\text{ in=27.94 cm} \end{gathered}[/tex]To obtain the area of the page we multiply the width by the length then:
[tex](21.59)(27.94)=603.22[/tex]Therefore the area of the page is 603.22 squared cm
A helicopter flies at a constant altitude towing an airborne 65 kg crate as shown in the diagram. The helicopter and the crate only move in the horizontal direction and have an acceleration of 3.0 m/s21) find the vertical component of the tension in the cable (in Newtons). Ignore the effects of air resistance.2) find the magnitude of the tension in the cable (in Newtons). Ignore the effects of air resistance.3) find the angle(with respect to the horizontal) of the tension in the cable (in degrees). Ignore the effects of air resistance.
Given data:
* The weight of the crate is 65 kg.
* The acceleration of crate and helicopter is,
[tex]a=3ms^{-2}[/tex]Solution:
(1). The forces acting on the crate is represented as,
The y-component of tension is balancing the weight of the crate.
Thus, the y-component (vertical) of tension is,
[tex]\begin{gathered} T_y=W \\ T_y=mg \end{gathered}[/tex]Where m is the mass of crate and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} T_y=65\times9.8 \\ T_y=637\text{ N} \end{gathered}[/tex]Thus, the vertical component of the tesnion is 637 N.
(2). The x-component of tension in the cable is,
[tex]\begin{gathered} T_x=ma \\ T_x=65\times3 \\ T_x=195\text{ N} \end{gathered}[/tex]Thus, the tension in the cable is,
[tex]\begin{gathered} T=\sqrt[]{T^2_x+T^2_y} \\ T=\sqrt[]{195^2^{}+637^2} \\ T=666.2\text{ N} \end{gathered}[/tex]Hence, the tesnion in the cable is 666.2 N.
(3). The angle of tension with the horizontal is,
[tex]\begin{gathered} \tan (\theta)=\frac{T_y}{T_x} \\ \tan (\theta)=\frac{637}{195} \\ \tan (\theta)=3.3 \\ \theta=73.14^{\circ} \end{gathered}[/tex]Thus, the angle made by the tension with the horizontal is 73.14 degree.
If a sloth is traveling at 0.067 mps or 0.15 mph how long does it take the sloth to travel an 11.5 meter tree
Answer:
Explanation:
Given:
V = 0.067 m/s
D = 11.5 m
___________
t - ?
t = D / V
t = 11.5 / 0.067 ≈ 172 s or t ≈ 2.87 min or t ≈ 2 min 52 s
A 25.0 kg bag of dirt falls 15.0 m at a construction site. Assuming that all of the heat produced is retained by the dirt, how much will its temperature increase? (cdirt = 0.20 cal/g•°C)
Given:
• Mass of bag, m =25.0 kg
,• Height, h = 15.0 m
,• c = 0.20 cal/g•°C
Let's find by how much the temperature will increase.
Apply the Law of Conservation of Energy:
[tex]mc\Delta T=mgh[/tex]Where:
• m is the mass
,• c is the specific heat capacity
,• g is acceleration due to gravity
,• h is the height.
,• ΔT is the temperature change.
Thus, we have:
[tex]\begin{gathered} mc\Delta T=mgh \\ \\ Eliminate\text{ m on both sides:} \\ c\Delta T=gh \end{gathered}[/tex]Now, plug in the values and solve for ΔT:
[tex]\begin{gathered} \Delta T=\frac{gh}{c} \\ \\ \Delta T=\frac{9.8*15.0}{0.20\times10^3\times4.2} \\ \\ \Delta T=\frac{147}{840} \\ \\ \Delta T=0.175^o\text{ C} \end{gathered}[/tex]Therefore, the temperature change is 0.175° C.
• ANSWER:
0.175° C
Two skydivers jump from an airplane at an altitude of 5000m. Suppose one is 57 kg and the other is 68 kg. Using the data given in the example in class to find the amount of time each takes to get to the ground.Area = 0.18 m² air density = 1.21 kg/m³ drag coefficient C = .070
Given:
Two skydivers jump from an airplane at an altitude of: d = 5000 m
The mass of the first skydiver is: m1 = 57 kg
The mass of the second skydiver is: m2 = 68 kg
Area = 0.18 m²
Air density = 1.21 kg/m³
Drag Coefficient: C = 0.070
To find:
The amount of time each skydiver takes to get to the ground.
Explanation:
The magnitude of drag force which acts opposite in the opposite direction is equal to the weight of the skydiver. Thus, the magnitude of the drag force can be calculated as:
[tex]\begin{gathered} F_1=m_1g=57\text{ kg}\times9.8\text{ m/s}^2=558.6\text{ kg.m/s}^2 \\ \\ F_2=m_2g=68\text{ kg}\times9.8\text{ m/s}^2=666.4\text{ kg.m/s}^2 \end{gathered}[/tex]Here, F1 is the magnitude of the drag force on the first skydiver and F2 is the drag force on the second skydiver.
The expression for drag force relating to the velocity is given as:
[tex]F=\frac{1}{2}C\rho Av^2[/tex]For the first skydiver the drag force is given as:
[tex]\begin{gathered} F_1=\frac{1}{2}C\rho Av_1^2 \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ 558.6\text{ kg.m/s}^2=\frac{1}{2}\times0.070\times1.21\text{ kg/m}^3\times0.18\text{ m}^2\times v_1^2 \\ \\ 558.6\text{ kg.m/s}^2=7.623\times10^{-3}\text{ kg/m}\times v_1^2 \\ \\ v_1^2=\frac{558.6\text{ kg.m/s}^2}{7.623\times10^{-3}\text{ kg/m}} \\ \\ v_1^2=73278.24\text{ m}^2\text{/s}^2 \\ \\ v_1=\sqrt{73278.24\text{ m}^2\text{/s}^2} \\ \\ v_1=270.70\text{ m/s} \end{gathered}[/tex]The velocity v1 of the first skydiver is 270.70 m/s.
The time t1 taken by the first skydiver to get to the ground can be calculated as:
[tex]\begin{gathered} v_1=\frac{d}{t_1} \\ \\ t_1=\frac{d}{v_1} \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ t_1=\frac{5000\text{ m}}{270.70\text{ m/s}} \\ \\ t_1=18.47\text{ s} \end{gathered}[/tex]The first skydiver takes 18.47 seconds to get to the ground.
For the second skydiver, the drag force is given as:
[tex]\begin{gathered} F_2=\frac{1}{2}C\rho Av_2^2 \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ 666.4\text{ kg.m/s}^2=\frac{1}{2}\times0.070\times1.21\text{ kg/m}^3\times0.18\text{ m}^2\times v_2^2 \\ \\ 666.4\text{ kg.m/s}^2=7.623\times10^{-3}\text{ kg/m}\times v_2^2 \\ \\ v_2^2_{{}}=\frac{666.4\text{ kg.m/s}^2}{7.623\times10^{-3}\text{ kg/m}} \\ \\ v_2^2=87419.65\text{ m}^2\text{/s}^2 \\ \\ v_2=\sqrt{87419.65\text{ m}^2\text{/s}^2} \\ \\ v_2=295.67\text{ m/s} \end{gathered}[/tex]The velocity v2 of the second skydiver is 259.67 m/s.
The time t2 taken by the second skydiver to get to the ground can be calculated as:
[tex]\begin{gathered} v_2=\frac{d}{t_2} \\ \\ t_2=\frac{d}{v_2} \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ t_2=\frac{5000\text{ m}}{295.67\text{ m/s}} \\ \\ t_2=16.91\text{ s} \end{gathered}[/tex]The second skydiver takes 16.91 seconds to get to the ground.
Final answer:
The first skydiver takes 18.47 seconds to get to the ground.
The second skydiver takes 16.91 seconds to get to the ground.
An il film (n = 1.42) floating on water is84.8 nm thick. What is the wavelengthIN NANOMETERS of the brightest(m = 1) reflected color?(Hint: If you leave the wavelength in nm,the answer will be in nm. No conversionnecessary.)(Unit = nm)
Given:
n = 1.42
Thickness, t = 84.8 nm
m = 1
Let's find the wavelength in Nanometers of the brightest reflected color.
Apply the formula:
[tex]t=(m-\frac{1}{2})(\frac{\lambda}{2n})[/tex]Where:
• t = 84.8 nm
,• m = 1
,• n = 1.42
Plug in the values in the formula and solve for the wavelength λ:
[tex]\begin{gathered} 84.8=(1-\frac{1}{2})(\frac{\lambda}{2*1.42}) \\ \\ 84.8=(\frac{1}{2})(\frac{\lambda}{2.84}) \\ \\ 84.8=\frac{\lambda}{2*2.84} \\ \\ 84.8=\frac{\lambda}{5.68} \\ \\ \lambda=84.8*5.68 \\ \\ \lambda=481.67\text{ nm} \end{gathered}[/tex]Therefore, the wavelength is 481.67 nm.
• ANSWER:
481.67 nm
A coyote was holding a large bird cage a few meters above the ground. He lowers it at a slow constant speed onto the X he has drawn on the ground. Which statement best describes the change in the total mechanical energy of the Earth-bird cage system?Question 5 options:The total mechanical energy decreases, because the coyote does negative work on the bird cage by exerting a force in the direction opposite to its displacement.The total mechanical energy is unchanged, because there is no change in the bird cage’s kinetic energy as it is lowered to the table.The total mechanical energy is unchanged, because no work is done on the Earth-bird cage system while the book is lowered.The total mechanical energy decreases, because the coyote does positive work on the bird cage by exerting a force that opposes the gravitational force.
Total mechanical energy is conserved in a system.
Thus, the total mechanical energy is unchanged, because no work is done on the Earth-bird cage system while the book is lowered.
In the image below, a worker is pushing a crate with a mass of 20 kg up aramp at a constant rate. Ignoring friction, how much force must the workerapply so that the crate continues to move at the same speed? (Recall that g =9.8 m/s2WeightO A. 37.3 NOB. 62.5 NO C. 50.7NO D. 48.6N
Free diagram of the crate:
The free body equation for the crate is given as,
[tex]F=mg\sin \theta[/tex]Here, F is the force applied, m is the mass of the crate, g is the acceleration due to gravity and θ is the angle of inclination.
Substituting all known values,
[tex]\begin{gathered} F=20\text{ kg}\times9.8\text{ m/s}^2\times\sin (15^{\circ}) \\ =50.7\text{ N} \end{gathered}[/tex]T
A spring block system undergoes a simple harmonic oscillation with an amplitude A = This oscillations in one minute. The maximum acceleration of this oscillator is:
The maximum acceleration of the oscillator is 1.48 m/s²
Simple harmonic motion is a particular kind of periodic motion in which the restoring force on the moving item is inversely proportional to the size of the displacement and acts in the direction of the object's equilibrium position.
A periodic variable's amplitude is a gauge of its change over a single period.
The amplitude of a simple harmonic motion, A = 15 cm = 0.15 m
The frequency, f = 30 oscillations per minute = 30/60 = 1/2
The maximum acceleration of the oscillator is given as:
a = A × ω²
Now the formula for angular frequency is:
ω = 2πf
Therefore,
a = A × ( 2πf )²
Substituting the values in the equation,
a = 0.15 × ( 2 × π × 1/2 )²
a = 0.15 × π²
a = 1.48 m/s²
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As shown in Fig. B9, a block of mass m2=2.1 kg sits on an inclined plane with angle 35 degrees .The coefficients of static and kinetic friction between the block and the plane are unknown.The block is attached to a rope, which is hung over a frictionless and massless pulley and attached to a hanging mass m1.(a) The block my is on the verge of sliding, when the hanging mass is increased to m1=1.6kg.01. find the coefficient of static friction.As shown in Fig. Bob, an identical block (with mass m2) is placed on top of the block on the inclined plane. The hanging mass m1 remains unchanged. The two blocks on the inclined plane are sliding down the incline with the same acceleration a=0.31 m/s^2Find02.the coefficient of kinetic friction.03.what is the minimum value of the coefficient of static friction between the two identical blocks (m2) so that they move together?
Question 1.
The free body diagrams of the situation are shown below:
The forces applied on m1 are in only the y direction then we only have one equation of motion to get it we use Newton's Second law. Then we have:
[tex]T-W_1=m_1a[/tex]The forces applied on m2 are in the x and y direction, this means that we have two equations of motion. We don't expect this block to move on the y-direction (which we defined); this means that the acceleration in this direction has to be zero. Furthermore, since the blocks m1 and m2 are attach thorugh the rope their acceleration has to be the same; this means that the acceleration of block m2 in the x direction has to be a. Applying this and Newton's second law we have for block 2:
[tex]\begin{gathered} (W_2)_x+f_f-T=m_2a \\ N-(W_2)_y=0 \end{gathered}[/tex]where (W2)x and (W2)y denote the component of the weight in the x and y direction, respectively. Now, since block m2 is on the verge of sliding this means that the system is in equilibrium, that is, the acceleration of the system is equal to zero. Then we have that the equations above take the form:
[tex]\begin{gathered} T-W_1=0 \\ (W_2)_x+f_f-T=0 \\ N-(W_2)_y=0 \end{gathered}[/tex]To determine the coefficient of static friction we firs need to determine the force of fricction. To do this we first solve the first equation for T:
[tex]T=W_1[/tex]Now we plug this value in the second equation and solve for the force of friction:
[tex]\begin{gathered} (W_2)_x+f_f-W_1=0 \\ f_f=W_1-(W_2)_x \end{gathered}[/tex]Now we need to remember that the force of friction can be obtain by:
[tex]f_f=\mu_{}N[/tex]where mu is the coefficient of friction (this rule applies for static and kinetic friction). Then we have:
[tex]\begin{gathered} \mu N=W_1-(W_2)_x_{} \\ \mu=\frac{W_1-(W_2)_x}{N} \end{gathered}[/tex]To find the normal force we use the third equation of motion:
[tex]N=(W_2)_y[/tex]Hence we have:
[tex]\mu=\frac{W_1-(W_2)_x}{(W_2)_y}[/tex]Now we need to remember that the components of the weight on an inclined plane are given by:
[tex]\begin{gathered} W_x=W\sin \theta \\ W_y=W\cos \theta \end{gathered}[/tex]where theta is the angle of the plane.
Plugging this on the expression for the coefficient of friction:
[tex]\mu=\frac{W_1-W_2\sin \theta}{W_2\cos \theta}[/tex]Finally we plug the values given to find the coefficient:
[tex]\begin{gathered} \mu=\frac{(1.6)(9.8)-(2.1)(9.8)\sin 35}{(2.1)(9.8)\cos 35} \\ \mu=0.23 \end{gathered}[/tex]Therefore the coefficient of static friction is 0.23
Question 2
For this question we are going to look at the two block in the inclined plane as one block with mass of 4.2 kg and we are going to name the two blocks as m2.
The free body diagram in this case is:
Applying Newton's second law in each mass lead to the system of equations:
[tex]\begin{gathered} T-W_1=m_1a \\ (W_2)_x-f_f-T=m_2a \\ N-(W_2)_y=0 \end{gathered}[/tex]From the first equation of motion we have:
[tex]T=W_1+m_1a[/tex]Plugging this in the second equation:
[tex](W_2)_x-f_f-W_1-m_1a=m_2a[/tex]but we know that:
[tex]f_f=\mu_{}N[/tex]and:
[tex]N=(W_2)_y[/tex]Then we have:
[tex](W_2)_x-\mu(W_2)_y-W_1-m_1a=m_2a[/tex]Solving for the coefficient and using the expression for the wight on an inclined plane we have:
[tex]\begin{gathered} (W_2)_x-\mu(W_2)_y-W_1-m_1a=m_2a \\ \mu=\frac{W_2\sin \theta-W_1-m_1a-m_2a}{W_2\cos \theta} \end{gathered}[/tex]Plugging the values given we have:
[tex]\begin{gathered} \mu=\frac{(4.2)(9.8)\sin 35-(1.6)(9.8)-(1.6)(0.31)-(4.2)(0.31)}{(4.2)(9.8)\cos 35} \\ \mu=0.18 \end{gathered}[/tex]Therefore the coefficient of kinetic friction is 0.18.
Question 3.
To find the coefficient of static friction between the two blocks on the inclined plane we need that the upper block to be stationary with respect to the lower block, that means that the friction has to be equal to the weight in the x-direction, that is:
[tex]\begin{gathered} f_f=W_2\sin 35 \\ \mu W_2\cos 35=W_2\sin 35 \\ \mu=\frac{\sin 35}{\cos 35} \\ \mu=0.7 \end{gathered}[/tex]Therefore the coefficient between the blocks has to be at least 0.7
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
A. the speed of the object
Explanation:
The color wouldn't make any difference to the potential energy and knowing the mass the way the object is shaped or the height of it would make no difference
2. A student claims that the normal force acting on the cart is equal in magnitude to the weight of the cart.Is the student correct?A. Yes, the student is correct because the normal force is directly proportional to the mass of the cart,as described in Newton's Second Law of Motion.B. No, the student is incorrect because the normal force is not the equal and opposite reaction to thecart's weight being applied on a surface, as described in Newton's Third Law of Motion.C. Yes, the student is correct because the normal force is the equal and opposite reaction to thecart's weight being applied on a surface, as described in Newton's Third Law of Motion.D. No, the student is incorrect because the normal force is inversely proportional to the mass of the cart,as described in Newton's Second Law of Motion.
Explanation:
The first law of Newton's says that the velocity of an object will remain unless there is a force acting on the object, the second law of newton says that the force is proportional to the mass and the acceleration, and the third law of newton says that there is always an equal an opposite forces acting on two objects that interact.
So, the normal force can be described as a result of the third law of Newton. It means that it is equal in magnitude to the weight of the cart.
Therefore, the answer is:
C. Yes, the student is correct because the normal force is the equal and opposite reaction to the cart's weight being applied on a surface, as described in Newton's Third Law of Motion.
A student is sitting in a chair. The student's mass is 55 kg. Find the normal force exerted on the student by thechair, in N.Round your answer to one decimal place
We have the next diagram
Where N is the normal and W is the weight
Therefore
N=W
The formula to calculate the W is
[tex]W=mg[/tex]where m is the mass and g is the gravity
In our case,
m=55
g=9.8 m/s^2
[tex]W=55(9.8)=539N[/tex]Therefore the normal force
N=539N
ANSWER
the normal force is 539N
Which of the following inventions came first?fiber opticsthe radiothe televisionthe telegraph
Telegraph is the invention that came first.
A force F1 of magnitude 4.80 units acts on an object at the origin in a direction = 27.0° above the positive x-axis. (See the figure below.) A second force F2 of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force F1 + F2.magnitude unitsdirection ° counterclockwise from the +x-axis
We will determine the magnitude of the final component as follows:
[tex]\begin{gathered} m=\sqrt{(5cos(27))^2+(5sin(27)+4.8)^2}\Rightarrow m=8.356527029... \\ \\ \Rightarrow m\approx8.36 \end{gathered}[/tex]So, the magnitude is approximately 8.36.
And the direction will be approximately 57.8° counter-clockwise.
This can be seeing as follows:
What causes water to freeze solid when placed in a freezer? Takes heat away from the waterAdds heat to the waterPuts electricity in the waterDoes not let light shine on the water
Takes heat away from the water.
Water freezes solid when placed in a freezer.
If heat is added water melts.
Electricity in the water doesn't freeze it.
Light doesn't freeze water.
So, the correct answer is:
Takes heat away from the water
A box of mass m = 2 kg is kicked on a rough horizontal plane with an initial velocity v_0 . If the net work done on the crate during its entire motion , until it comes to rest , is -36 J , then the initial velocity v_0 of the box is equal to :
Given:
The mass of the box is m = 2 kg
The work done is W = -36 J
The final velocity of the object is
[tex]v_f=\text{ 0 m/s}[/tex]To find the initial velocity of the box.
Explanation:
The initial velocity of the box can be calculated as
[tex]\begin{gathered} Work\text{ done = change in kinetic energy} \\ W=\frac{1}{2}m(v_f^2-v_o^2) \\ -36=\frac{1}{2}\times2(0^2-v_o^2) \\ -v_{_0}^2=-36 \\ v_o=6\text{ m/s} \end{gathered}[/tex]Hence, the initial velocity of the box is 6 m/s.
What is the current produced by a potential difference of 240 Volts through a resistance of .28 ohms
Given:
The potential difference is V = 240 Volts
The resistance is R = 0.28 Ohms
To find the current produced.
Explanation:
The current produced can be calculated by the formula
[tex]I=\text{ }\frac{V}{R}[/tex]On substituting the values, the current produced will be
[tex]\begin{gathered} I=\frac{240}{0.28} \\ =\text{ 857.14 A} \end{gathered}[/tex]Thus, the current produced is 857.14 A
Select the correct answerA car moves with an average speed of 45 miles/hour. How long does the car take to travel 90 miles?O A2 minutesВ.2 hoursOC.45 minutesΟ Ο ΟD45 hoursE90 minutesResetNext
Given data:
The average speed of the car is
[tex]S=45\text{ miles/hour}[/tex]The distance traveled by the car is
[tex]D=90\text{ miles}[/tex]The average speed can be expressed as,
[tex]\begin{gathered} S=\frac{D}{T} \\ T=\frac{D}{S} \end{gathered}[/tex]Here, T is the time taken by the car.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} T=\frac{90\text{ miles}}{45\text{ miles/hour}} \\ =2\text{ hours} \end{gathered}[/tex]Thus, the time taken by the car is 2 hours, and option (B) is correct.
A 8,707 newton car is initiallyat rest. How much force (inNewton) is required to movethe car by 16.73 meters, with afinal velocity of 5.84 m/s?
Given the weight of the car, W = 8707 N, and the car moves a distance, s= 16.73 m, and final velocity, v = 5.84 m/s
Let the mass of the car be m and acceleration due to gravity g = 9.8 m/s^2
Also, weight is given by the formula,
[tex]W=mg\text{ }[/tex]Then, the mass of the car will be
[tex]\begin{gathered} m=\frac{W}{g} \\ =\frac{8707}{9.8} \\ =888.46\text{ kg} \end{gathered}[/tex]The acceleration, a can be calculated by the formula
[tex]\begin{gathered} v^2-u^2=2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]Here, u is the initial velocity, u=0.
[tex]\begin{gathered} a=\frac{(5.84)^2}{2\times16.73} \\ =\frac{34.10}{33.46} \\ =1.019m/s^2 \end{gathered}[/tex]The force will be
[tex]\begin{gathered} F=\text{ ma} \\ =888.46\times1.019 \\ =\text{ 905.34 N} \end{gathered}[/tex]Thus, the force is 905.34 N