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Inessa05 [86]
3 years ago
15

The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C, where U will be in joules w

hen x is in meters. A, B, and C are constants. show answer No Attempt 50% Part (a) Enter an expression for the x-component of the force as function of position F(x), in terms of the constants A, B, and C.
Physics
1 answer:
satela [25.4K]3 years ago
3 0

Answer:

F = - 2 A x - B

Explanation:

The force and potential energy are related by the expression

      F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^

Where i ^, j ^, k ^ are the unit vectors on the x and z axis

The potential they give us is

     U (x) = A x² + B x + C

Let's calculate the derivatives

    dU / dx = A 2x + B + 0

The other derivatives are zero because the potential does not depend on these variables.

Let's calculate the strength

      F = - 2 A x - B

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For an object like a planet, with a typical temperature of a few hundred kelvin, what kind of blackbody radiation would it princ
navik [9.2K]

Answer:

Low-temperature blackbody

Explanation:

There are 3 types of blackbody temperatures.

Low-temperature blackbody

High temperature extended area blackbody

High-temperature cavity blackbody

A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.

6 0
3 years ago
A car is accelerating to the right. In which direction is the applied force?
Natasha_Volkova [10]

Answer:

Left

Explanation:

The force is applied opposite of the acceleration.

8 0
3 years ago
During intercourse, Lydia constantly worries whether her partner thinks she is
Misha Larkins [42]
Answer is c spectatoring
6 0
3 years ago
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
3 years ago
A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spe
saw5 [17]

Answer:

Mass of ion will be 22\times 10^{-13}kg                

Explanation:

We have given ion is triply charged that is q=3\times 1.6\times 10^{-19}=4.8\times 10^{-19}C

Radius r = 36 cm = 0.36 m

Velocity of the electron v=7\times 10^6m/sec

Magnetic field B = 0.55 T

We know that radius of the path is given by r=\frac{mv}{qB}

m=\frac{rqB}{v}=\frac{0.36\times 4.8\times 10^{-19}\times 7\times 10^6}{0.55}=22\times 10^{-13}kg

6 0
3 years ago
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