If a photon has a frequency of 5.20 x 10^14 hertz, what is the energy of the photon ? Given : Planck's constant is 6.63 x 10^-34 joule-seconds.

Answers

Answer 1

Answer: For this you would use planck's equation.

E = hv, where v = the frequency and h = planck's constant.

So E = 5.20 x10^14 x 6.63 x 10^-34
= 3.45 x 10^-19 Joules

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Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the direction of travel. Explain what your answer means in terms of the object’s energy.
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A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s 2 . What is the moment of inertia of the door about the hinges?

Answers

Answer:

Moment of inertia will be $I=2kgm^2$

Explanation:

We have given mass of the person m = 72 kg

Radius r = 0.8 m

Force is given F = 5 N

Angular acceleration $\alpha =2rad/sec^2$

Torque is given by $\tau =F* r=5* 0.8=4N-m$

We know that torque is also given by

$\tau =I\alpha$ , here I is moment of inertia and $\alpha$ is angular acceleration

So $4=I* 2$

$I=2kgm^2$

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

Answers

Answer:

$F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004$

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

$Vector (OA) = 6i + 0j + 0k \nVector (OB) = 0i + 3j + 2k \nVector (OC) = 0i - 2j + 3k$

Next step is to find unit vectors $u_(AB) ,u_(AC), u_(AD), u_(AE)$ as follows:

$u_(AB) = (vector(AB))/(magnitude(AB)) \n= \frac{OB - OA}{magnitude({vector(OB - OA))} }\n=(-6i +3j+2k)/(√(6^2 + 3^2+2^2) ) \n\n=-0.857 i +0.429j+0.286k\n\nu_(AC) = (vector(AC))/(magnitude(AC)) \n= \frac{OC - OA}{magnitude({vector(OC - OA))} }\n=(-6i -2j-3k)/(√(6^2 + 2^2+3^2) ) \n\n=-0.857 i -0.286j+0.429k\n\nu_(AD) = +1i\nu_(AC) = -1k$

Using the diagram we find the corresponding vectors Forces:

$F_(AB) = F_(AB) i + F_(AB)j +F_(AB)k\nF_(AC) = F_(AC) i + F_(AC)j +F_(AC)k\nF_(AD) = F_(AD) i + F_(AD)j +F_(AD)k\nW = -160 k$

Equation of Equilibrium:

$Sum of forces = 0\nF_(AB). u_(AB) + F_(AC).u_(AC) + F_(AD).u_(AD) + W = 0\n(-0.857F_(AB)i + 0.429F_(AB)j +0.286F_(AB)k) + (-0.857F_(AC)i - 0.286F_(AC)j +0.429F_(AC)k) + (+1F_(AD) i) + (-160k) = 0$

Comparing i , j and k components as follows:

$-0.857F_(AB) -0.857F_(AC) +1F_(AD) = 0\n+ 0.429F_(AB) - 0.286F_(AC) = 0\n+0.286F_(AB) +0.429F_(AC) = 160$

Solving Above equation simultaneously we get:

$F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004$

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B

Answers

Final answer:

The net magnetic force exerted by the external magnetic field on a current-carrying wire formed into a loop in a uniform magnetic field is absolutely zero since the individual forces on each section of the loop cancel each other out.

Explanation:

The force exerted by a magnetic field on a current carrying wire is given by Lorentz force law, which says that the force is equal to the cross product of the current and the magnetic field. However, in this case, where the wire is formed into a loop with current flowing in a counter-clockwise direction in presence of an external magnetic field, the individual forces on each infinitesimal section of the loop cancel each other out. Therefore, the net magnetic force exerted by the external field on the entire loop is zero.

Learn more about Magnetic Force here:

brainly.com/question/10353944

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Final answer:

The magnetic force exerted on a current-carrying wire loop by an external magnetic field can be calculated using the equation F = I * R * B.

Explanation:

The magnetic force exerted by the external field on the current-carrying wire loop can be determined using the equation F = I * R * B. The magnetic force is equal to the product of the current, radius, and magnetic field strength. The direction of the magnetic force can be determined using the right-hand rule, where the thumb represents the direction of the current, the fingers represent the magnetic field, and the palm represents the direction of the force.

Learn more about Magnetic force on wire loop here:

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A pair of thin spherical shells with radius r and R, r < R are arranged to share a center. What is the capacitance of the system. If a potential difference V is created between the shells, how much energy is stored between them?

Answers

Answer:

Capacitance = ( 4π×∈×r×R ) / (R-r)

energy store = ( 4π×∈×r×R )×V² / (R-r)

Explanation:

given data

radius = r

radius = R

r < R

to find out

capacitance and how much energy store

solution

we consider here r is inner radius and R is outer radius

so now apply capacitance C formula that is

C = Q/V .................1

here Q is charge and V is voltage

we know capacitance have equal and opposite charge so

V = $\int\limits^R_r {E} \, dx$

here E = Q / 4π∈k²

V = Q / 4π∈ $\int\limits^R_r {1/k^2} \, dx$

V = Q / 4π∈ × ( 1/r - 1/R )

V = Q(R-r) / ( 4π×∈×r×R )

so from equation 1

C = Q/V

Capacitance = ( 4π×∈×r×R ) / (R-r)

and

energy store is 1/2×C×V²

energy store = ( 4π×∈×r×R )×V² / (R-r)

Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by

Answers

Answer:

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Explanation:

Let $\vec A = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})$ and $\vec B = 4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$ , both measured in meters. The resultant vector $\vec R$ is calculated by sum of components. That is:

$\vec R = \vec A+\vec B$ (Eq. 1)

$\vec R = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})+4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$

$\vec R = (6\cdot \cos 30^(\circ)-4\cdot \sin 30^(\circ))\,\hat{i}+(6\cdot \sin 30^(\circ)-4\cdot \cos 30^(\circ))\,\hat{j}$

$\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

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