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Umnica [9.8K]
2 years ago
12

What is the derivative of x^2 - x + 3 at the point x = 5?​

Physics
1 answer:
Hunter-Best [27]2 years ago
5 0

Answer:

9

Explanation:

d/dx (x² - x + 3)

= 2x - 1

when x = 5,

2x-1

= 2(5) - 1

= 10 - 1

= 9

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The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of th
Bezzdna [24]

Answer:

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r................................ Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track

7 0
3 years ago
Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen
Ghella [55]

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

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5 0
1 year ago
What is the magnitude of the magnetic dipole moment of the bar magnet
Annette [7]

The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²

<h3> Magnetic dipole moment of the bar magnet</h3>

The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;

B = \frac{2\mu_0m}{4\pi r^3} \\\\m = \frac{4\pi r^3 B}{2\mu_0}

where;

  • B is magnetic field
  • m is dipole moment
  • μ is permeability of free space

m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)

m = 1.2 Am²

The complete question is below:

What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.

Learn more about dipole moment here: brainly.com/question/27590192

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6 0
1 year ago
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take
vivado [14]

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT   ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)}  )\\T=1.38*10^{6} seconds

3 0
3 years ago
Please help fast i need urgent
Sloan [31]
Is there more information ?
5 0
3 years ago
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