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3 years ago

8

Running on a treadmill is slightly easier than running outside because there is no drag force to work against. suppose a 60 kg r

unner completes a 5.0 km race in 16 minutes.

1 answer:

professor190 [17]3 years ago

7 0

So it will take 5 minutes if a person running on a treadmill so it will be easy.

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A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago

Atoms in a molecule held together through sharing ______. (points 3)

Wittaler [7]

Covalent and ionic bonds

8 0

3 years ago

Read 2 more answers

Which word describes the maximum a point moves from its rest position when a wave passes?

Tom [10]

Answer:

Amplitude.

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest.

Hence, an amplitude is a word that describes the maximum displacement a point moves from its rest position when a wave passes.

On a graph, the vertical axis (y-axis) is the amplitude of a waveform and this simply means that, it's measured vertically.

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

6 0

2 years ago

A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning

zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

$L_{i}=L_{f}$

Here, = the system's angular momentum before the collision

$L_{i}$ = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here, = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

4 0

1 year ago

Read 2 more answers

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago