All categories

3 years ago

What is the change in its velocity, v, during this 0.80-s interval?

Physics

1 answer:

lyudmila [28]3 years ago

7 0

This link should help you out https://quizlet.com/40330134/biomechanics-questions-flash-cards/

You might be interested in

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! Which number marks a crest?

lyudmila [28]

number 2 because the curve demstrates the crest GOOD LUCK i hope i got you the correct answer if not sorry

5 0

3 years ago

S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to

Gre4nikov [31]

By calculation, the diameter of the wire is 2.8 * 10^-3 m.

<h3>How do we obtain the length?</h3>

The following data are given in the question;

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

Area of the wire = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

6 0

1 year ago

Pls help me... taking a quiz.

saw5 [17]

Guess I recommend doing that

7 0

3 years ago

Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. h

Dvinal [7]

Answer:

734.16 kg m/ $s^{2}$

Explanation:

The problem is asking for the Force of pushing off the ground.

The formula of Force is: F = mass x acceleration

Given = Mass: 600 newtons (N)

Acceleration: 12 m/ $s^{2}$

We have to convert the mass into kg first. Remember that 1 kg is equal to 9.80665 newtons.

Let x be the mass in newtons.

Let's convert: $\frac{1 kg}{9.80665 N}$ x $\frac{x}{600 N}$ = $\frac{600}{9.80665}$ = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/ $s^{2}$

F = 734.16 kg m/ $s^{2}$

7 0

2 years ago

An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 5.5 mm. If th

stepan [7]

Answer:

Number of photons travel through pin hole= $6.4*10^{17}$

Explanation:

First we will calculate the energy of single photon using below formula:

$E=\frac{h*c}{λ}$

Where :

h is plank's constant with value $6.626*10^{-34} J.s$

c is the speed of light whch is $3*10^{8}$

λ is the wave length = 532nm

$E=\frac{6.6268*10^{-34}* 3*10^{8}}{532nm}$

E= $3.73*10^{-19}$ J

Number of photons emitted per second:

$\frac{5J/s}{1 photon/3.73*10^{-19} }$

Number of photons emitted per second= $1.34*10^{19} photons/s$

$\frac{A-hole}{A-beam}$ = $\frac{\frac{pie*d-hole^2}{4}}{\frac{pie*d-beam^2}{4} }$

Where:

A-hole is area of hole

A-beam is area of beam

d-hole is diameter of hole

d-beam is diameter if beam

$\frac{A-hole}{A-beam}$ = $\frac{d-hole^2}{d-beam^2}$

$\frac{Ahole}{A-beam}$ = $\frac{1.22^2}{5.5^2}$

$\frac{A-hole}{A-beam}$ = $\frac{144}{3025}$

Number of photons travel through pin hole= $1.34*10^{19} *\frac{144}{3025}$

Number of photons travel through pin hole= $6.4*10^{17}$

7 0

3 years ago