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2 years ago

7

Một vật dao động điều hòa với phương trình x= 4cos(6πt) cm. Kể từ khi vật bắt đầu dao động, thời điểm vật qua a) vị trí cân bằng

lần thứ 1 là? b) vị trí biên x=4 lần thứ 2 là? c) vị trí x= -2 lần thứ 9 là ? d) vị trí x=2 lần thứ 2020 là?

1 answer:

dezoksy [38]2 years ago

7 0

Answer:

what are you saying.........

......

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How many mm are in a micro-m?

Dahasolnce [82]

For this case we have that by definition, a micrometer is equivalent to a thousandth of a millimeter, that is, 0.001 millimeters.

Then, in other words, we have 0.001 millimeters in a micrometer.

Answer:

In a micrometer there are 0.001 millimeters.

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3 years ago

What force is needed to accelerate an object 5 m/s2 if the object has a mass of 10 kg?

Helga [31]

We know, F = m * a
F = 10 * 5
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Hope this helps!

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3 years ago

A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

jeka57 [31]

Answer:

a. $3.1 m/s^2$

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

$mg sin \theta - \mu_k R = ma$

where :

m = 6.0 kg is the mass of the box

g = 9.8 m/s^2 the acceleration of gravity

$\theta=39^{\circ}$ is the angle of the slope

$\mu_k = 0.40$ is the coefficient of friction

R is the normal reaction

a is the acceleration

- Along the perpendicular direction:

$R-mg cos \theta =0$

From the 2nd equation, we get an expression for the reaction force:

$R=mg cos \theta$

And substituting into the 1st equation, we can find the acceleration:

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2$

6 0

3 years ago

A neutron star has more mass then what and is about the same size as what?

Alex_Xolod [135]

A neutron star has more mass than a bowling ball,
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2 years ago

Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV

forsale [732]

Answer:

(a) $\lambda=1.227\ A$

(b) $\lambda=0.388\ A$

(c) $\lambda=0.038\ A$

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$

Where

m and e are the mass of and charge on an electron

On solving,

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

V = 100 V

$\lambda=\dfrac{12.27}{\sqrt{100} }\ A$

$\lambda=1.227\ A$

(b) V = 1 kV = 1000 V

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

$\lambda=\dfrac{12.27}{\sqrt{1000} }\ A$

$\lambda=0.388\ A$

(c) If $V=100\ kV=10^5\ V$

$\lambda=\dfrac{12.27}{\sqrt{10^5} }\ A$

$\lambda=0.038\ A$

Hence, this is the required solution.

7 0

2 years ago