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3 years ago

13

The transmission of heat requiring the movement of a liquid or a gas is A. conduction B. radiation. C. convection. D. transducti

on.

1 answer:

Llana [10]3 years ago

3 0

Answer:

convection C

Explanation:

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How does a rotating coil inside a magnetic field generate electricity?

Goryan [66]

Answer:

When an electrical current passes through a wire, a magnetic field is generated around it. Likewise, if the magnetic field around a wire is changed ( for example by rotating a coil inside a stationary manger), electricity will move through the wire.

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3 years ago

What is the smallest separation between two slits that will produce a second-order minimum for 762 nm red light?

ipn [44]

Answer:

1524 nm.

Explanation:

Angular separation of n th minima is given by the expression

Sin θ = n λ / a

Where λ is wavelength of light used and a is separation between two slits .

a = n λ / sin θ

For a to be minimum, sinθ has to be maximum .

maximum value of sin of any angle is 1 .

so the relation becomes as follows

a = n λ

n = 2 , λ = 762 nm.

a = 2 x 762 nm

= 1524 nm.

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4 years ago

A kayaker needs to paddle north across a

vova2212 [387]

Answer:

100

Explanation:

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2 years ago

Read 2 more answers

A 0.90- kg mass attached to a vertical spring of force constant 170 N/m oscillates with a maximum speed of 0.20 m/s . Find the f

Alex

Explanation:

Given that,

The mass of the object, m = 0.9 kg

Force constant, k = 170 N/m

Maximum speed of the object, v = 0.2 m/s

Solution,

(a) The angular frequency of the object is given by :

$\omega=\sqrt{\dfrac{k}{m}}$

$\omega=\sqrt{\dfrac{170}{0.9}}$

$\omega=13.74\ rad/s$

The time period is given by :

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{13.74}$

T = 0.45 seconds

(b) The maximum velocity of the object in shm is given by :

$v=A\omega$

Amplitude,

$A=\dfrac{v}{\omega}$

$A=\dfrac{0.2}{13.74}$

A = 0.0145 m

(c) The maximum acceleration of the object is given by :

$a=\omega^2A$

$a=(13.74)^2\times 0.0145$

$a=2.73\ m/s^2$

Therefore, this is the required solution.

3 0

4 years ago

A 25.0-kg box is released on a 27° incline and accelerates down the incline at 0.30 m/s2. Find the friction force impeding its m

mr Goodwill [35]

Answer:

(A) Frictional force will be 103.7276 N

(B) Coefficient of friction $\mu _k=0.4751$

Explanation:

We have given

mass of the box m = 25 kg

Angle $\Theta =27^{\circ}[/text]Acceleration [tex]a=0.3m/sec^2$

Apply Newton's second law of motion ,

Vertically :

$F_N$ = mg cosθ

= $25\times 9.8\times COS27^{\circ}=218.296N$

Horizontally :

mg sinθ - F = ma

A)

frictional force F= mg sinθ - ma

= $25\times 9.8sin27^{\circ}-25\times 0.3=103.7276N$

B)

Coefficient of friction $\mu _k=\frac{F}{F_N}=\frac{103.7276}{218.296}=0.4751$

8 0

3 years ago