The force is required to (a) just start the crate moving =340.196 N
(b) slide the crate across the dock at a constant speed =155.58 N
Given - mass m=55.9kg , μ=0.621 , μₙ=0.284
To find Horizontal pushing Force for just start the crate moving , static coefficient is applicable , while for other case kinetic coefficient is applicable.
concept used :-
F= μN ⇒μmg ⇔ For just start the crate moving
f=μₙN ⇒μₙmg ⇔For sliding the crate across the dock
Calculations :-
F= 0.621 x 55.9 x 9.8 =0.621 x 547.82
=340.196 N
f= 0.284 x 55.9 x 9.8 =0.284 x 547.82
=155.58 N
Result:-
Hence the force is required to (a) just start the crate moving =340.196 N
(b) slide the crate across the dock at a constant speed =155.58 N
To know about coefficients of static and kinetic friction :-https://brainly.com/question/17237604
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If the magnitude of the force she exerts on the suitcase is 25.0 N, and she does +1.12 X 10^3 J of work in moving the suitcase a distance of 51.0 m, at what angle above the horizontal (as shown in the figure above) is the force oriented with respect to the floor?degrees
Given data:
* The force acting on the suitcase is 25 N.
* The work done on the suitcase is,
[tex]W=1.12\times10^3\text{ J}[/tex]* The suitcase moved the distance is 51 m.
Solution:
The workdone in terms of force and displacement is,
[tex]W=Fd\cos (\theta)[/tex]where F is the force, d is the displacement,
[tex]\theta\text{ is the angle between force and displacement,}[/tex]Substituting the known values,
[tex]\begin{gathered} 1.12\times10^3=25\times51\times\cos (\theta) \\ \cos (\theta)=\frac{1.12\times10^3}{25\times51} \\ \cos (\theta)=0.000878\times10^3 \\ \cos (\theta)=0.878 \\ \theta=28.6^{\circ} \end{gathered}[/tex]Thus, the angle above the horizontal at which the force is oriented is 28.6 degree.
A lad wants to throw a bag into the open window of his friend's room 10.0 m above. Assuming itjust reaches the window, he throws the bag at 60.0° to the ground:a) At what velocity should he throw the bag? (16.2 m/s at 60.0° to the ground]b) How far from the house is he standing when he throws the bag? (11.5 m]
Given data:
* The height of the window is 10 m.
* The angle of the incident velocity with respect to the horizontal is 60 degree.
Solution:
(a). The height of the projectile in terms of the initial velocity is,
[tex]H=\frac{u^2\sin ^2(\theta)}{2g}[/tex]where u is the initial velocity, g is the acceleration due to gravity, and
[tex]\theta\text{ is the angle made by initial velocity with the horizontal}[/tex]Substituting the known values,
The initial velocity of the projectile is,
[tex]\begin{gathered} 10=\frac{u^2\sin ^2(60^{\circ})}{2\times9.8} \\ u^2=\frac{10\times2\times9.8}{\sin ^2(60^{\circ})} \\ u^2=261.33 \\ u=16.16ms^{-1} \\ u\approx16.2ms^{-1} \end{gathered}[/tex]Thus, the initial velocity of the bag is 16.2 m/s at an angle of 60 degree to the ground.
(b). The horizontal range of bag is,
[tex]H=\frac{u^2\sin (2\theta)}{g}[/tex]Substituting the known values,
[tex]\begin{gathered} H=\frac{16.16^2\times\sin(2\times60)}{9.8} \\ H=23.077\text{ m} \end{gathered}[/tex]The position of the bag from the house is,
[tex]\begin{gathered} R^{\prime}=\frac{H}{2} \\ R^{\prime}=11.53\text{ m} \\ R^{\prime}\approx11.5\text{ m} \end{gathered}[/tex]Thus, the lad is standing at the distance of 11.5 m from the house.
If you throw a 0.3 kg ball straight up with an initial speed of 21 m/s, how high will it climb? Round your answer to the nearest tenth and include the appropriate unit.
m = mass = 0.3 kg
v = speed = 21 m/s
g= gravity =9.8 m/s^2
h= height
Conservation of mass:
Potential energy = Kinetic energy
1/2 m v^2 = m g h
Replacing:
1/2 (0.3 ) (21)^2 = (0.3) (9.8) h
Solve for h
66.15 = 2.94h
66.15/2.94 = h
h = 22.5 m
You drive 5 miles at 30 mi/hr and then another 3 miles at 50 mi/hr. What is your average speed for the whole8-mile trip?
Given data
The initial distance covered is d1 = 5 miles
The initial speed is s1 = 30 mi/hr
The final distance is d2 = 3 miles
The final speed is s2 = 50 mi/hr
The expression for the initial time taken to travel is given as:
[tex]t_1=\frac{d_1}{s_1}[/tex]The expression for the final time taken is given as:
[tex]t_2=\frac{d_2}{s_2}[/tex]The expression for the average speed for the whole trip is given as:
[tex]\begin{gathered} \text{Average sp}eed\text{ =}\frac{total\text{ distance}}{total\text{ time}} \\ S_{Avg}=\frac{d_1+d_2}{t_1+t_2} \\ S_{Avg}=\frac{d_1+d_2}{\frac{d_1}{s_1}+\frac{d_2}{s_2}_{}} \end{gathered}[/tex]Substitute the value in the above equation.
[tex]\begin{gathered} S_{Avg}=\frac{5\text{ mi+3 mi}}{\frac{5\text{ mi}}{30\text{ mi/hr}}+\frac{3\text{ mi}}{50\text{ mi/hr}}} \\ S_{Avg}=35.3\text{ mi/hr} \end{gathered}[/tex]Thus, the average speed for the whole trip is 35.3 mi/hr.
A 5.0 m portion of wire carries a current of 8.0 A from east to west. It experiences a magnetic field of 6.0×10^−4 T running from north to south. What is the magnitude and direction of the magnetic force on the wire?1.2×10^−2N downward2.4×10^−2N upward1.2×10^−2N upward2.4×10^−2N downward
Given data
The length of the wire is L = 5 m
The magnitude of the current is I = 8 A
The magnitude of the magnetic field is B = 6 x 10-4 T
The expression for the magnitude of the magnetic force is given as:
[tex]F=(I\times L)\times B[/tex]Substitute the value in the above equation.
[tex]\begin{gathered} F=8\text{ A}\times5\text{ m}\times6\times10^{-4}\text{ T} \\ F=2.4\times10^{-2}\text{ N} \end{gathered}[/tex]Thus, the magnitude of the magnetic force is 2.4 x 10-2 N.
The direction of the magnetic field is given as:
[tex]\vec{B}=B(-\text{ }\hat{\text{j }}\text{)}[/tex]The direction of the current is given as:
[tex]\vec{I}=I(-\hat{i})[/tex]The direction of the magnetic force is given as:
[tex]\begin{gathered} \vec{F}=L(\vec{I}\times\vec{B}) \\ \vec{F}=L(I(-\hat{i})\times B(-\hat{j}) \\ \vec{F}=LIB(\hat{k}) \end{gathered}[/tex]The positive direction means the direction of the force is outside the page.
Thus, the direction of the magnetic force is upward.
Calculate the speed of an 7.7 x 10^4 kg airliner with a kinetic energy of 1.3 x 10^9 J Answer in units of m
Given that the mass of airliner is
[tex]m=\text{ 7.7}\times10^4\text{ kg}[/tex]Also, the kinetic energy is given by
[tex]KE=1.3\times10^9\text{ J}[/tex]The Kinetic energy is given by the formula
[tex]KE=\frac{1}{2}mv^2[/tex]Here, v is the speed of an airliner. The speed can be calculated as
[tex]v=\sqrt[]{\frac{2KE}{m}}[/tex]Substituting the values, speed will be
[tex]\begin{gathered} v=\sqrt[]{\frac{2\times1.3\times10^9}{7.7\times10^4}} \\ =\sqrt[]{33766.233} \\ =183.75\text{ m/s} \end{gathered}[/tex]In 20 seconds of freefall, an object will travel approximately 1.96 kilometers. What is this value expressed in centimeters, and written in scientific notation?
We want to convert 1.96 km to centimeter. Recall,
1 km = 100000 cm
Thus,
1.96 km = 1.96 x 100000
To write 1.96 x 100000 km inscientific notation, we would count the number of zeros. In this case, it is 5. We would raise 10 to the power of 5. Thus, the value in scientific notation is
1.96 x 10^5
Two hockey pucks, each 0.1 kg, slide across the ice and collide. Before the collision, 1 puck is moving 13m/s to the east and puck 2 is moving 18m/s to the west. After the collision, puck 1 is moving at 18m/s to the west. What is the velocity of puck 2.
Using conservation of momentum:
[tex]\begin{gathered} p1=p2 \\ p1=m1u1+m2v1 \\ p2=m1u2+m2v2 \end{gathered}[/tex]Where:
m1 = mass of the hockey puck 1 = m2 = mass of the hockey puck 2 = 0.1kg
u1 = Initial speed of the hockey puck 1 = 13m/s
u2 = Final speed of the hockey puck 1 = 18m/s
v1 = Initial speed of the hockey puck 2 = 18m/s
v2 = Final speed of the hockey puck 2
Therefore:
[tex]\begin{gathered} 0.1(13)+0.1(18)=0.1(18)+0.1v2 \\ 13+18=18+v2 \\ solve_{\text{ }}for_{\text{ }}v2: \\ v2=13+18-18 \\ v2=13m/s \end{gathered}[/tex]Answer:
13 m/s
Answer:
13 m/s east
Explanation:
got it right.
What will be the magnitude of the electrical field created by 4.95 * 10 ^ - 4 charge particle 7.01 * 10 ^ - 2 away?
The magnitude E of the electrical field created by a charge particle with charge q at a distance r away, is:
[tex]E=\frac{kq}{r^2}[/tex]Where k is the Coulomb constant which has a value:
[tex]k=8.99\times10^9N\cdot\frac{m^2}{C^2}[/tex]Substitute q=4.95*10^-4 C and r=7.01*10^-2 m to find the magnitude of the electrical field:
[tex]\begin{gathered} E=\frac{kq}{r^2} \\ =\frac{(8.99\times10^9N\cdot\frac{m^2}{C^2})(4.95\times10^{-4}C)^{}}{(7.01\times10^{-2}m)^2} \\ =\frac{4.45\times10^6\frac{N}{C}m^2}{4.91\times10^{-3}m^2} \\ =9.06\times10^8\cdot\frac{N}{C} \end{gathered}[/tex]Therefore, the answer is option b:
[tex]9.08\times10^6\frac{N}{C}[/tex]A transformer has 840 primary and 56 secondary windings. The primary coil is connected to a 110 V AC power supply which delivers 0.30 A of current to the transformer.a. Find the secondary voltageb. Find the secondary current
Given that the number of turns in the primary coil is
[tex]N_p=\text{ 840 }[/tex]The number of turns in the secondary coil is
[tex]N_s=\text{ 56}[/tex]The voltage provided to the primary coil is
[tex]V_p=110\text{ V}[/tex]The current delivered to the transformer is
[tex]I_p=\text{ 0.30 A}[/tex]We have to find the secondary voltage and secondary current.
Let the secondary current be denoted by
[tex]I_s[/tex]Let the secondary voltage be denoted by
[tex]V_s[/tex]The secondary current can be calculated by the formula
[tex]I_s=\frac{N_pI_p_{}}{N_s}[/tex]Substituting the values, the secondary current will be
[tex]\begin{gathered} I_s=\frac{840\times0.3}{56} \\ =4.5\text{ A} \end{gathered}[/tex]The secondary voltage can be calculated by the formula
[tex]\begin{gathered} V_s=\frac{V_pI_p}{I_s} \\ =\frac{110\times0.3}{4.5} \\ =7.33\text{ V} \end{gathered}[/tex]Thus, the secondary voltage is 7.33 V and the secondary current is 4.5 A
A 30KG block is on the floor and is pushed to the right with a 60N force while the friction is 15N A) how do you know that the friction is pointed to the left? B) draw a force diagram showing all the forces present on the blockPlease help
Explanation
Step 1
Diagram
A)Friction is force that resists the sliding or rolling of one solid object over another, so the direction of the friction force must be the opposite of the applied force
Frictional force always acts in a direction opposite to the direction of applied force.
so,if the direction of the force is to the right, the direction of the friction force must be to the left
Step 2
Free body diagram:
A free body diagram is a simplified representation in a problem of an object , and the force vectors acting on it,is a graphical illustration used to visualize the applied forces , so
I hope this helps you
Number 12 multiple choice use the graph for the answer above
The acceleration of an object is the time rate of change of velocity.
That is, the acceleration is given by,
[tex]a=\frac{\Delta v}{\Delta t}[/tex]Where Δv is the change in the velocity and Δt is the time period.
Thus the acceleration is change in the velocity divided by time
Thus the correct answer is option c.
A lab cart with a mass of 521 grams has an initial speed of 1.48 meters per second. A constant force of 4.91 newtons is applied to the cart in the direction it's moving for 0.274 seconds. Friction gives the cart 0.392 newton·seconds of impulse in the opposite direction. What is the cart's final speed? Include units in your answer. Answer must be in 3 significant digits.
The net impulse acting on the cart can be given as,
[tex]J_n=Ft-j[/tex]Also, the net impulse can be expressed as,
[tex]J_n=m(v-u)[/tex]Equate both the values,
[tex]\begin{gathered} Ft-j=m(v-u) \\ v-u=\frac{Ft-j}{m} \\ v=\frac{Ft-j}{m}+u \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} v=\frac{(4.91\text{ N)(0.274 s)-0.392 Ns}}{(521\text{ g)}}+1.48\text{ m/s} \\ =\frac{0.953\text{ Ns}}{(521\text{ g)(}\frac{1\text{ kg}}{1000\text{ g}})}(\frac{1kgm/s^2}{1\text{ N}})+1.48\text{ m/s} \\ =1.83\text{ m/s+1.48 m/s} \\ =3.31\text{ m/s} \end{gathered}[/tex]Thus, the final speed of cart is 3.31 m/s
Please tell how can I get the answer for this question as 0.125 Newtons? I asked some physics teachers but they all are saying it's answer is 125.Please say how can I get the answer as 0.125 N?
The average acceleration of an object is given by:
[tex]a=\frac{v_f-v_0}{t}[/tex]In this case we know that the final velocity is zero (since it comes to rest), the initial velocity is 25 m/s and the dime is 0.03 seconds. Then the acceleration is:
[tex]\begin{gathered} a=\frac{0-25}{0.03} \\ a=-833.33 \end{gathered}[/tex]In this case the minus sign means that the ball is stopping.
Now that we have the acceleration we can apply Newton's second law:
[tex]F=ma[/tex]to determine the force; in this case the mass is 0.150 kg (we need to use kilograms) and the acceleration is 833.33 m/s^2 (since we are calculating the magnitude we drop the minus sign); then the force is:
[tex]\begin{gathered} F=(0.150)(833.33) \\ F=125 \end{gathered}[/tex]Therefore the force is 125 N.
2. Free-body diagrams for four situations are shown below. The net force is known for each situation.However, the magnitudes of a few of the individual forces are not known. Analyze cach situationindividually and determine the magnitude of the unknown forces.TB30050 l80 Nr20 NG200200 kFnet = 0 NFnet- 900 N, upet = 60 N, leftFnet = 30 N, rightClick to add speaker notes
In order to calculate the missing forces, we need to know that the net force is the sum of all forces in a direction.
For the first situation, the net force is zero, so the sum of forces in each direction is zero.
In the vertical direction, we have:
[tex]\begin{gathered} F_{net}=B-200\\ \\ 0=B-200\\ \\ B=200\text{ N} \end{gathered}[/tex]In the horizontal direction:
[tex]\begin{gathered} F_{net}=50-A\\ \\ 0=50-A\\ \\ A=50\text{ N} \end{gathered}[/tex]For the second situation, the net force is 900 N up.
In the vertical direction, we have:
[tex]\begin{gathered} F_{net}=C-200\\ \\ 900=C-200\\ \\ C=1100\text{ N} \end{gathered}[/tex]In the horizontal direction, we have no forces.
For the third situation, the net force is 60 N to the left.
In the vertical direction, we have:
[tex]\begin{gathered} F_{net}=300-E\\ \\ 0=300-E\\ \\ E=300\text{ N} \end{gathered}[/tex]In the horizontal direction, we have:
[tex]\begin{gathered} F_{net}=D-80\\ \\ -60=D-80\\ \\ D=20\text{ N} \end{gathered}[/tex]For the fourth situation, the net force is 30 N to the right.
In the vertical direction, we have:
[tex]\begin{gathered} F_{net}=F-H\\ \\ 0=F-H\\ \\ F=H \end{gathered}[/tex]In the horizontal direction, we have:
[tex]\begin{gathered} F_{net}=G-20\\ \\ 30=G-20\\ \\ G=50\text{ N} \end{gathered}[/tex]The quark spins in the proton are shown as red arrows...notice that in each example arrangement they add to +1/2. The ∆+ has the same quark content and the spins are arranged differently...which distinguishes it from the proton.Figure 2: The dashed lines are sitting there awaiting to be turned into spin arrows to represent the ∆+ spin arrangement. Fill them in.
The spin of ∆+ is 3/2
The filling arrangement is done as
Which of the following is an example of a constant speed but a changingvelocity?A car going around a round-aboutA car merging into another lane on cruise controlA satellite orbiting EarthAll the above
To answer this question we need to remember that the speed is the distance over time, while the velocity is the change of position of the time,
Now, they are gonna be the same if the distance and change of position is the same.
We notice that in the case of a satellite oribitng earth the distance and change of position won't be the same, therefore this is an example of a constant speed but changing velocity
The Earth exerts a force on you to pull you towards the center with an acceleration of 10 m/s^ 2 Newton's 3rd law says the Force the Earth feels towards you should be: A) more than you towards the earth B) the same as you towards the earthC) less thank you towards the earthD) you exert no force on the earth
According to Newton's Third Law, every force has an equal and opposite reaction force.
Therefore, the answer is B) the same as you towards the earth.If the object is on a surface, the opposite force would be the normal force, which is opposite to the weight.
On a brisk walk, a person burns about 340 Cal/h. At this rate, how many hours of brisk walking would it take to lose 1 lb of body fat? (A pound of body fat stores an amount of chemical energy equivalent to 3,500 Cal.) answer:____h
Given:
Amount of fat stored in Cal = 3500 Cal
Rate at which the person burns = 340 Cal/h
Let's find the number of hours it would take the person to lose 1 lb of fat.
Given that the person burns at a rate of 340 Cal/h, the time it will take the person to lose 1 lb will be:
[tex]\begin{gathered} t=\frac{3500\text{ Cal}}{340\text{ Cal/h}} \\ \\ t=10.29\text{ hours} \end{gathered}[/tex]Therefore, the time it will take the person to burn 1 lb of fat is 10.29 hours.
ANSWER:
10.29 h
An engine rated at 5.0 3 104 watts exerts a constant force of 2.5 3 103 newtons on a vehicle. Determine the average speed of the vehicle.
Given data:
* The power of the engine is,
[tex]P=5\times10^4\text{ watts}[/tex]* The force exerted by the engine is,
[tex]F=2.5\times10^3\text{ N}[/tex]Solution:
The power of the engine in terms of force and average speed is,
[tex]P=Fv[/tex]where v is the average speed,
Substituting the known values,
[tex]\begin{gathered} 5\times10^4=2.5\times10^3\times v \\ v=\frac{5\times10^4}{2.5\times10^3} \\ v=2\times10^{4-3} \\ v=20\text{ m/s} \end{gathered}[/tex]Thus, the average speed of the vehicle is 20 meters per second.
Particle A has 300 J of kinetic energy. Particle B has 2 times the mass and 1/2 times the speed of particle A. What is the kinetic energy of particle B?
Given data
*The particle A has kinetic energy is U_a = 300 J
*Particle B has a mass is m_b = 2 × m_a
*Particle B has a speed is v_b = 1/2 × v_a
The formula for the kinetic energy of the particle B is given as
[tex]U_b=\frac{1}{2}m_bv^2_b[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} U_b=\frac{1}{2}(2\times m_a)(\frac{1}{2}\times v_a)^2 \\ =\frac{1}{4}m_av^2_a \\ =\frac{1}{2}\times\frac{1}{2}m_av^2_a \\ =\frac{1}{2}\times300 \\ =150\text{ J} \end{gathered}[/tex]Hence, the kinetic energy of the particle B is U_b = 150 J
Can you please help me with my homework #8 please
Given that the mass of the jet is m = 20000 kg.
The tangential velocity of the jet is
[tex]\begin{gathered} v=\frac{1500\operatorname{km}}{hr} \\ =\frac{1500\times1000}{60\times60} \\ =416.67\text{ m/s} \end{gathered}[/tex]The centripetal acceleration is
[tex]a=\text{ 5 g}[/tex]This centripetal acceleration can be converted into m/s by using acceleration due to gravity, g = 9.8 m/s which is as follows
[tex]\begin{gathered} a=\frac{5}{9.8} \\ =0.51\text{ m/s} \end{gathered}[/tex]We have to find the radius, r.
It can be calculated as
[tex]\begin{gathered} a=\frac{mv^2}{r} \\ r=\frac{mv^2}{a} \end{gathered}[/tex]Substituting the values, the radius will be
[tex]\begin{gathered} r=\frac{20000\times(416.67)^2}{0.51} \\ =6.8083\times10^9\text{ m} \\ =6.8083\times10^6\text{ km} \end{gathered}[/tex]ED. 05.39 m/sE. O 0.388 m/s10. An object of mass 16 kg moving with a speedof 26 m/s to the right collides with an objectof mass 9 kg at rest. If the collision is completely inelastic,calculate the kinetic energy lost during the collision. (1 point)A. O-3302.995 JB. O-1703.969 JC. O-1205.975 JD. O-3546.668 JE. O-1946.88 JSubmit Query
Given:
The mass of one object is m1 = 16 kg
The initial speed of the object is
[tex]v_i=\text{ 26 m/s}[/tex]The mass of another object is m2 = 9 kg.
The speed of the object will be zero as it is at rest.
To calculate the kinetic energy lost.
Explanation:
The collision is inelastic.
According to the conservation of momentum, the speed after the collision can be calculated as
[tex]\begin{gathered} m1v_i+m2\times0=(m1+m2)v_f \\ v_f=\frac{m1v_i}{(m1+m2)} \\ =\frac{16\times26}{16+9} \\ =\text{ 16.64 m/s} \end{gathered}[/tex]The loss in kinetic energy during a collision can be calculated as
[tex]\begin{gathered} \Delta K.E\text{ =K.E.}_f-K.E._i \\ =\frac{1}{2}(m1+m2)v_f-\frac{1}{2}m1v_i^2 \\ =\frac{1}{2}\times25\times(16.64)^2-\frac{1}{2}\times16\times(26)^2 \\ =3461.12-5408 \\ =-1946.88\text{ J} \end{gathered}[/tex]Thus, the correct choice is E
A roller coaster features a near vertical drop of 140 meters. Assuming that friction and air resistance are negligible, and that the initial velocity was zero, what would be the speed at the bottom of the drop?48 m/s52 m/s42 m/s35 m/s
Answer:
52 m/s
Explanation:
By the conservation of energy, the potential energy at the top will be equal to the kinetic energy at the bottom, so
[tex]\begin{gathered} E_i=E_f_{}_{} \\ PE_i=KE_f_{} \\ \text{mgh}=\frac{1}{2}mv^2 \end{gathered}[/tex]Where m is the mass, g is the gravity, h is the height and v is the speed. Solving for the speed v, we get:
[tex]\begin{gathered} 2\text{mgh}=mv^2 \\ \frac{2\text{mgh}}{m}=v^2 \\ 2gh=v^2 \\ v=\sqrt[]{2gh} \end{gathered}[/tex]Then, replacing g = 9.8 m/s² and h = 140 m, we get:
[tex]\begin{gathered} v=\sqrt[]{2(9.8)(140)} \\ v=52.38\text{ m/s} \end{gathered}[/tex]Therefore, the answer is 52 m/s
Which describe the image formed by the convex mirror? Check all that apply.Realvirtualinverteduprightsmallerlarger
The rays after reflection meet virtually behind the mirror.
The image is upright.
The image is smaller than the object.
Thus, the image is virtual, upright and smaller.
A proton (charge +1.60 x 10-19 C) and an electron (charge -1.60 x 10-19 C) are both moving in the xy-plane with the
same speed, 5.20 x 105 m/s. The proton is moving in the +y-direction along the line x = 0, and the electron is moving in
the -y-direction along the line x = +4.00 mm. At the instant when the proton and electron are at their closest approach,
what is the magnitude of the magnetic force that the proton exerts on the electron?
The magnitude of the magnetic force that the proton of charge 1.6 * [tex]10^{-19}[/tex]exerts on the electron of charge - 1.6 * [tex]10^{-19}[/tex] is 4.33 * [tex]10^{-29}[/tex] N
B = μo q v sin θ / 4 π r²
B = Magnetic field
μo = Permeability of free space
q = Charge
v = Velocity
θ = Angle between velocity vector and resultant vector
r = Distance
B due to proton and electron will be same as they both have same magnitude and same velocities. But the direction of B due to proton will be directed into the plane and the direction of B due to electron will be directed out of the plane.
B = 4 * 3.14 * [tex]10^{-7}[/tex] * 1.6 * [tex]10^{-19}[/tex] * 5.2 * [tex]10^{5}[/tex] * sin 90 / 4 * 3.14 * ( 4 * [tex]10^{-3}[/tex] )²
B = 33.28 * [tex]10^{-21}[/tex] / 64 * [tex]10^{-6}[/tex]
B = 0.52 * [tex]10^{-15}[/tex] T
The magnitude of force exerted by proton on the electron, F = q v B
F = 1.6 * [tex]10^{-19}[/tex] * 5.2 * [tex]10^{5}[/tex] * 0.52 * [tex]10^{-15}[/tex]
F = 4.33 * [tex]10^{-29}[/tex] N
Therefore, the magnitude of the magnetic force that the proton exerts on the electron is 4.33 * [tex]10^{-29}[/tex] N
To know more about magnetic force
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A 200-turn solenoid is 20.0 cm long and carries a current of 3.25 A. 1. Find the force in [µN] exerted on a 15.0x10-6 C charged particle moving at 1050 m/s through the interior of the solenoid, at an angle of 11.5° relative to the solenoid’s axis. Find the magnetic field inside the solenoid in [mT]. = 4.08 mT
Given:
• Number ot turns, N = 200
,• Length = 20.0 cm ==> 0.2 m
,• Current = 3.25 A
,• Charge, q = 15.0 x 10⁻⁶ C
,• Speed = 1050 m/s
,• Angle = 11.5 degrees
Let's solve for the following:
• (a), The force exerted on the particle.
To find the force exerted on the particle, apply the formula:
[tex]\begin{gathered} F=qV\times B \\ \\ F=\text{qVBsin}\theta \end{gathered}[/tex]Where:
B is the magnetic field strength. = 4.08 mT
Thus, we have:
[tex]\begin{gathered} F=(15\times10^{-6})\times1050\times(4.08\times10^{-3})\times\sin 11.5 \\ \\ F=1.28\times10^{-5}N \end{gathered}[/tex]Therefore, the force exterted on the charged particle is 1.28 x 10⁻⁵ N.
ANSWER:
1.28 x 10⁻⁵ N
In Europe the standard voltage in homes is 240 V insteadof the 120 V used in the United States. Therefore a “200-W” Euro-pean bulb would be intended for use with a 240-V potential difference.(a) If you bring a “200-W” European bulb home to the United States,what should be its U.S. power rating? (b) How much current will the“200-W” European bulb draw in normal use in the United States?
In Europe, the standard voltage in homes is 240 V and in the United States, it is 120 V.
(a) If you bring a “200-W” European bulb home to the United States, what should be its U.S. power rating?
Recall that the power of a bulb is given by
[tex]P=VI[/tex]Where V is the voltage and I is the current.
In Europe, the current flowing through a 200 W bulb is
[tex]I=\frac{P}{V}=\frac{200}{240}=0.833\; A[/tex]So, a current of 0.833 A will be flowing through the bulb in Europe.
The equivalent power rating of the bulb in the United States is
[tex]P=VI=120\cdot0.833=100\; W[/tex]So, the equivalent power rating of the bulb should be approximately 100 W.
(b) How much current will the “200-W” European bulb draw in normal use in the United States?
The rating of the bulb is 200 W and the voltage in the United States is 120 V
So, the current drawn by the bulb is given by
[tex]I=\frac{P}{V}=\frac{200}{120}=1.667\; A[/tex]As you can see, the bulb will draw more current if the voltage is reduced to supply the same power.
Therefore, the bulb will draw a current of 1.667 A
How much energy is required to completely boil away 0.350 kg of ice at 0oC?
We are asked to determine the amount of energy required to turn ice into water and then that water to boil completely.
First, we calculate the energy required to turn ice into water at 0°C. We will use the following formula:
[tex]Q_1=mh_{fg}[/tex]Where "m" is the mass and hfg is the heat of fusion which is a constant equivalent to 334 kJ/kg. Replacing the values we get:
[tex]Q_1=(0.350\operatorname{kg})(334\frac{kJ}{\operatorname{kg}})[/tex]Solving the operation:
[tex]Q_1=116.9kJ[/tex]Now we determine the amount of heat required to convert the water at 0°C into water at 100°C which is the boiling point. We will use the following formula:
[tex]Q_2=mc_p(T_f-T_0)_{}[/tex]Where "cp" is the specific heat of water which is a constant equivalent to 4.18 kJ/kg°C. Replacing the values:
[tex]Q_2=(0.350\operatorname{kg})(4.18\frac{kJ}{kgC})(100^0C-0^0C)[/tex]Solving the operation
[tex]Q_2=146.3kJ[/tex]Now we determine the amount of heat required to turn water at 100°C into steam. To do that we will use the following formula:
[tex]Q_3=mh_v[/tex]Where "hv" is the heat of vaporization and is equivalent to 2260 kJ/kg. Replacing we get:
[tex]Q_3=(0.350\operatorname{kg})(2260\frac{kJ}{\operatorname{kg}})[/tex]Solving the operations:
[tex]Q_3=791kJ[/tex]Now we add all the energy to determine the total energy required:
[tex]Q=Q_1+Q_2+Q_3[/tex]Replacing the values:
[tex]Q=116.9kJ+146.3kJ+791kJ[/tex][tex]Q=1054.2kJ[/tex]Therefore, the total amount of heat is 1054.2 kJ.
how does specular reflection differ from diffuse reflection?
Specular reflection is the reflection of the light from a smooth surface. This occurs at a definite angle.
Diffuse reflection is produced when the light reflects from a rough surface. It does not have one definite angle. Scattering is a diffuse reflection.