Which unit abbreviation is a measurement for force

I am desperate to get to Denver, Colorado to catch the Denver Broncos

Pavlova-9 [17]

Answer:

The average speed needed is 68.4 mi/h

Explanation:

Average Speed

If an object travels a distance d in a time t regardless of the direction, the average speed is the quotient of the distance over the time:

$\displaystyle V=\frac{d}{t}$

The distance to get to Denver from the Hamilton Street Campus is d=1,778 miles. An estimated time of t=26 hours should be needed to cover that distance, thus you will need to travel at an average speed of:

$\displaystyle V=\frac{1,778\ mi}{26\ h}$

V = 68.4 mi/h

The average speed needed is 68.4 mi/h

6 0

3 years ago

An electron having 500ev energy enters at right angle to a uniform magnetic field of 10^-4 Tesla. If its specific charge is 1.75

gtnhenbr [62]

Answer:

The correct answer might be $r = 2.8^{27}$ meters.

Explanation:

The Answer Given might not be correct, I just did what my brain said.

As the angle is perpendicular so Θ=90.

Putting this in the equation to calculate the magnetic force as:

F = evBsinΘ

F= evBsin90 *sin90 = 1 so,

F= evB.

Now when the electron will start to move in a circle, The necessary force that makes the electron rotate in a circle is given by Centripetal force.

So,

Magneteic Force = Centripetal Force

evB = $\frac{mv^{2} }{r}$

r = $\frac{mv}{Be}$ ......(1)

Now the problem is, We don't know " v " so we need to calculate velocity first,

Calculation of Velocity:

In order to calculate the velocity of electron, We should know the potiential difference with which the electrons are accelerated which in our case is 500ev. If "V" is the potiential difference, the energy gained by electrons during accelreation will be Ve. This appear as kinectic enrgy of electrons as,

K.E = Ve

$\frac{1}{2}$ m $v^{2}$ = Ve

v = $\sqrt{\frac{2ve}{m} }$ ................(2)

Putting value of velocity in equation 2 from 1:

r = $\frac{mv}{Be}$ $\sqrt{\frac{2ve}{m} }$

$r = \sqrt{\frac{2mev}{Be}}$

$r = \sqrt{\frac{(2)(9.1^{-31})(500) (1.6^{-19} ) }{ (1.75^{11} ) (10^{-4} ) } }$

$r = 2.8^{27}$ meters.

8 0

3 years ago

A 4.4 kg object is being pushed along a surface, causing it to accelerate at a rate of 1.5 m/s2 . Th e coeffi cient of kinetic f

emmasim [6.3K]

Friction = 0.25 * m * g = 10.8
Push = m * a + Friction = 4.4 * 1.5 + 10.8

8 0

3 years ago

An object with a mass of 3 kg has a momentum of 75 kg.m/s. What is its velocity?

tiny-mole [99]

Answer:

25m/s

Explanation:

p=mv

75 = 3v

v = 75/3

v = 25 m/s

If you want to verify your answer, just insert the value of v to the equation.

8 0

3 years ago

A guitar string is supposed to have a fundamental frequency 256 Hz. It currently has a fundamental frequency 248 Hz when the str

umka2103 [35]

Answer:

The tension to bring the guitar string into tune is 372.95 Hz.

Explanation:

Given;

current frequency, f₁ = 248 Hz

current tension, T₁ = 350 N

fundamental frequency, f₂ = 256

The tension on the string to bring the guitar string into tune is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\f\lambda = \sqrt{\frac{T}{\mu} } \\\\f^2\lambda^2 = \frac{T}{\mu} \\\\f^2 = \frac{T}{\mu \lambda^2}\\\\let \ {\mu \lambda^2} = k\\\\f^2 =\frac{T}{k} \\\\k = \frac{T}{f^2} \\\\\frac{T_1}{f_1^2} = \frac{T_2}{f_2^2}\\\\T_2 = \frac{T_1 f_2^2}{f_1^2} \\\\T_2 = \frac{350 \times 256^2}{248^2} \\\\T_2 = 372.95 \ Hz$

Therefore, the tension to bring the guitar string into tune is 372.95 Hz.

3 0

3 years ago