• ### (6 magnetism 1)help me please in physics i cant find the answer ?

The lower end of the thin uniform rod in (Figure 1) is attached to the floor by a frictionless hinge at point P. The rod has mass 0.0840 kg and length 20.0 cm and is in a uniform magnetic field 0.190 T that is directed into the page. The rod is held at an angle 59.0 ∘ above the horizontal by a horizontal string that connects the top of the rod to... show more
The lower end of the thin uniform rod in (Figure 1) is attached to the floor by a frictionless hinge at point P. The rod has mass 0.0840 kg and length 20.0 cm and is in a uniform magnetic field 0.190 T that is directed into the page. The rod is held at an angle 59.0 ∘ above the horizontal by a horizontal string that connects the top of the rod to the wall. The rod carries a current 12.0 A in the direction toward P. Part A Calculate the tension in the string. Use the fact that τ=12IBL2 for a uniform bar of length L carring a current I in a magnetic field B. Express your answer with the appropriate units. T =
1 answer · Physics · 9 months ago
• ### MIT COURSE ineed help in this question please any one can help me ?

Four point-like objects of charge −2Q , −Q , +2Q and +Q respectively are placed at the corners of a square of side sqrt(2) a as shown in the figure below. (a) Determine a vector expression for the electric field at the point P located a distance a/2 from the center of the square on the line connecting the objects with charge −2Q and +2Q as shown in... show more
Four point-like objects of charge −2Q , −Q , +2Q and +Q respectively are placed at the corners of a square of side sqrt(2) a as shown in the figure below. (a) Determine a vector expression for the electric field at the point P located a distance a/2 from the center of the square on the line connecting the objects with charge −2Q and +2Q as shown in the figure above. Clearly indicate your choice of unit vectors on the figure above and express your answer in terms of Q.a.k hati .or hatj if needed. E(p)= ???
4 answers · Physics · 9 months ago
• ### Vectors 12 what is the angle ?

If the magnitude of the cross product of two vectors is one-half the dot product of the same vectors, what is the angle between the two vectors? ∘
If the magnitude of the cross product of two vectors is one-half the dot product of the same vectors, what is the angle between the two vectors? ∘
1 answer · Physics · 10 months ago
• ### Mastring physics icant find it :5?

A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.500 miles west, and finally dS = 0.200 miles south . Part A If a bird were to start out from the origin (where the student starts) and fly directly (in a straight line) to the school, what distance db would the bird cover? Express your answer in miles. db =
A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.500 miles west, and finally dS = 0.200 miles south . Part A If a bird were to start out from the origin (where the student starts) and fly directly (in a straight line) to the school, what distance db would the bird cover? Express your answer in miles. db =
1 answer · Physics · 10 months ago
• ### Can i get help with my physics :(?

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is |F|=K|QQ′|d2, where K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -16.5 nC , is located at x1 = -1.650 m ; the second charge, q2 =... show more
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is |F|=K|QQ′|d2, where K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -16.5 nC , is located at x1 = -1.650 m ; the second charge, q2 = 39.5 nC , is at the origin (x=0.0000). Part A What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.245 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures. Force on q3 =
1 answer · Physics · 10 months ago

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.8 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.7 m from the other end. A monkey of mass 1.4 kg walks from one end of the bar to the other.... show more
To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.8 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.7 m from the other end. A monkey of mass 1.4 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar. Find T1, the magnitude of the force of tension in string 1, at the moment that the monkey is halfway between the ends of the bar. Express your answer in newtons using three significant figures. T1 = ?
1 answer · Physics · 1 year ago
• ### Can u help me please physics :(?

A nonuniform beam 4.65 m long and weighing 1.47 kN makes an angle of 25.0∘ below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 2.90 m farther down the beam and perpendicular to it in (Figure 1). The center of gravity of the beam is a distance of 1.90 m down the beam from the pivot. Lighting... show more
A nonuniform beam 4.65 m long and weighing 1.47 kN makes an angle of 25.0∘ below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 2.90 m farther down the beam and perpendicular to it in (Figure 1). The center of gravity of the beam is a distance of 1.90 m down the beam from the pivot. Lighting equipment exerts a downward force of 4.90 kN on the end of the beam. Part D Find the y component of the force exerted on the beam by the pivot. Express your answer with the appropriate units. Enter positive value if the direction of the y component of the force is upward and negative value if the direction of the y component of the force is downward. https://session.masteringphysics.com/pro... Py =
1 answer · Physics · 1 year ago
• ### Help me :)?

An object is undergoing simple harmonic motion with frequency f = 8.8 Hz and an amplitude of 0.13 m. At t = 0.00 s the object is at x = 0.00 m. How long does it take the object to go from x = 0.00 m to x = 6.00×10−2 m. An object is undergoing simple harmonic motion with frequency f = 8.8 Hz and an amplitude of 0.13 m. At t = 0.00 s the object is at... show more
An object is undergoing simple harmonic motion with frequency f = 8.8 Hz and an amplitude of 0.13 m. At t = 0.00 s the object is at x = 0.00 m. How long does it take the object to go from x = 0.00 m to x = 6.00×10−2 m. An object is undergoing simple harmonic motion with frequency f = 8.8 Hz and an amplitude of 0.13 m. At t = 0.00 s the object is at x = 0.00 m. How long does it take the object to go from x = 0.00 m to x = 6.00×10−2 m. 5.45×10−2 seconds 8.68×10−3 seconds 5.81×10−3 seconds 1.97×10−2 seconds
2 answers · Physics · 1 year ago

A vertical scale on a spring balance reads from 0 to 225 N . The scale has a length of 13.0 cm from the 0 to 225 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.10 Hz . Part A Ignoring the mass of the spring, what is the mass m of the fish? Express your answer in kilograms. m = kg
A vertical scale on a spring balance reads from 0 to 225 N . The scale has a length of 13.0 cm from the 0 to 225 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.10 Hz . Part A Ignoring the mass of the spring, what is the mass m of the fish? Express your answer in kilograms. m = kg
1 answer · Physics · 1 year ago
• ### Help me in my exam please ineeed full solution :(?

Two forces, of magnitudes F1 = 90.0 N and F2 = 45.0 N , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. (Figure 1)Initially, the center of the block is at position xi = -6.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 5.00 cm . Part... show more
Two forces, of magnitudes F1 = 90.0 N and F2 = 45.0 N , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. (Figure 1)Initially, the center of the block is at position xi = -6.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 5.00 cm . Part A Find the work W1 done on the block by the force of magnitude F1 = 90.0 N as the block moves from xi = -6.00 cm to xf = 5.00 cm . Express your answer numerically, in joules. W1 = J Part B Find the work W2 done by the force of magnitude F2 = 45.0 N as the block moves from xi = -6.00 cm to xf = 5.00 cm . Express your answer numerically, in joules. W2 =J Part C What is the net work Wnet done on the block by the two forces? Express your answer numerically, in joules. Wnet = J Part D Determine the changeKf−Ki in the kinetic energy of the block as it moves from xi = -6.00 cm to xf = 5.00 cm . Express your answer numerically, in joules. Kf−Ki = J
1 answer · Physics · 1 year ago
• ### Help me with this please for my exam?

A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 56 m and acquired a velocity of 60 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls... show more
A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 56 m and acquired a velocity of 60 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance. The speed of the rocket upon impact on the ground is closest to A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 56 m and acquired a velocity of 60 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance. The speed of the rocket upon impact on the ground is closest to 84 m/s 77 m/s 63 m/s 69 m/s 56 m/s
2 answers · Physics · 1 year ago
• ### HELP Me please exammss?

A 0.24-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 310 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x... show more
A 0.24-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 310 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 2.6×10−2 m, what is the kinetic energy of the block? A 0.24-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 310 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 2.6×10−2 m, what is the kinetic energy of the block? 0.94 J 0.89 J 1.0 J 0.83 J 0.99 J
1 answer · Physics · 1 year ago
• ### Please hep me fast for my exam :)?

If the fastest you can safely drive is 65 mi/h, what is the longest time you can stop for dinner if you must travel 488 mi in 9.8 h total? If the fastest you can safely drive is 65 mi/h, what is the longest time you can stop for dinner if you must travel 488 mi in 9.8 h total? 1.8 h 2.3 h 2.8 h You can't stop at all.
If the fastest you can safely drive is 65 mi/h, what is the longest time you can stop for dinner if you must travel 488 mi in 9.8 h total? If the fastest you can safely drive is 65 mi/h, what is the longest time you can stop for dinner if you must travel 488 mi in 9.8 h total? 1.8 h 2.3 h 2.8 h You can't stop at all.
1 answer · Physics · 1 year ago
• ### Ineed fast help for my exam please?

When a flea (m = 450 μg) is jumping up, it extends its legs 0.5 mm and reaches a speed of 0.40 m/s in that time. Part A How much work did the flea do during that time? Use g = 10 m/s2 How much work did the flea do during that time? Use = 10 3.8×10−2 mJ 3.8×10−5 mJ 3.6×10−5 mJ 2.3×10−6 mJ
When a flea (m = 450 μg) is jumping up, it extends its legs 0.5 mm and reaches a speed of 0.40 m/s in that time. Part A How much work did the flea do during that time? Use g = 10 m/s2 How much work did the flea do during that time? Use = 10 3.8×10−2 mJ 3.8×10−5 mJ 3.6×10−5 mJ 2.3×10−6 mJ
1 answer · Physics · 1 year ago
• ### Help me please anyone know physics :(?

The displacement of an oscillating object of mass, 10.0 kg , and total energy, −5.40×10−4 J , as a function of time is shown in the figure(Figure 1). help me please Part A What is the frequency? f = Hz Part B What is the amplitude? A = cm Part C What is the period? T = s Part D ... show more
The displacement of an oscillating object of mass, 10.0 kg , and total energy, −5.40×10−4 J , as a function of time is shown in the figure(Figure 1). help me please Part A What is the frequency? f = Hz Part B What is the amplitude? A = cm Part C What is the period? T = s Part D What is the angular frequency of this motion? ω = rad/s Request Answer Part E What is the spring constant? k = N/m Part F What is the maximal speed? vmax = cm/s Part G What is the maximal force? Fmax = N Part H What is the minimal potential energy? Umin = J Part I What is the maximal power of the spring on the attached mass? pmax = Watt Part J What is the position of equilibrium? (¯x) = m Part K What is the speed when t=10 s ? v = cm/s
1 answer · Physics · 1 year ago