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3 years ago

7

A normal walking speed is around 2.0 m/s . how much time t does it take the box to reach this speed if it has the acceleration 5

.54 m/s2 that you calculated above?

1 answer:

creativ13 [48]3 years ago

6 0

Given:

u(initial velocity)=0

a=5.54m/s^2

v(final velocity)=2 m/s

v=u +at

Where v is the final velocity.

u is the initial velocity

a is the acceleration.

t is the time

2=0+5.54t

t=2/5.54

t=0.36 sec

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The magnetic dipole moment of the current loop is 0.025 Am².

The magnetic torque on the loop is 2.5 x 10⁻⁴ Nm.

<h3>What is magnetic dipole moment?</h3>

The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.

Mathematically, magnetic dipole moment is given as;

μ = NIA

where;

N is number of turns of the loop
A is the area of the loop
I is the current flowing in the loop

μ = (1) x (25 A) x (0.001 m²)

μ = 0.025 Am²

The magnetic torque on the loop is calculated as follows;

τ = μB

where;

B is magnetic field strength

B = √(0.002² + 0.006² + 0.008²)

B = 0.01 T

τ = μB

τ = 0.025 Am² x 0.01 T

τ = 2.5 x 10⁻⁴ Nm

Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.

Learn more about magnetic dipole moment here: brainly.com/question/13068184

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1 year ago

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate

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Explanation:

It is given that,

Average power per unit mass generated by Lance, $\dfrac{P}{m}=6.5\ W/kg$

$P=6.5\times 75=487.5\ W$

(a) Distance to cover race, $d = 160\ km =160\times 10^3\ m$

Average speed of the person, v = 11 m/s

If t is the time taken to cover the race.

$t=\dfrac{d}{v}$

$t=\dfrac{160\times 10^3\ m}{11\ m/s}$

t = 14545.46 s

Let W is the work done. The relation between the work done and the power is given by :

$P=\dfrac{W}{t}$

$W=P\times t$

$W=487.5\times 14545.46$

W = 7090911.75 J

(b) Since, $1\ J=2.389\times 10^{-4}\ calories$

So, in 7090911.75 J, $W=7090911.75 \times 2.389\times 10^{-4}$

W = 1694.01 J

Hence, this is the required solution.

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3 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

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We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

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Hence, the temperature on a spring day 28 degree C.

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3 years ago

Calculate the kinetic energy in joules of a ball of mass 40g moving at a velocity of 4 metres per second

AfilCa [17]

Given : A ball of mass 40 g moving at a velocity of 4 m/s.
To find : Calculate the kinetic energy in joules ?
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where, v is the velocity v=4 m/s
m is the mass m=40 g
Convert g into kg,

Substitute the values,

Therefore, the kinetic energy is 0.32 Joules.

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2 years ago

A 2,200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.40 m before coming into contact

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Answer:

The magnitude of Force is 8.58×10⁵N and direction is upwards

Explanation:

The work beam does on the pile driver is given by

W=(FCos180°)Δx= -F(0.088m)

From work energy theorem

$W_{nc}=(KE_{f}-KE_{i})+(PE_{f}-PE_{i})\\W_{nc}=1/2m(vf^{2}-vi^{2})+mg(yf-yi)$

Choosing y=0 at the the level where the driver first contacts the beam and vi=0 at yi=+3.40m and comes to rest again vf=0 at yf= -0.088m

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The magnitude of Force is 8.58×10⁵N and direction is upwards

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3 years ago