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3 years ago

Q1) After applying a force of 1000 N an object of mass 2000 kg will achieve whatacceleration?

Physics

2 answers:

wlad13 [49]3 years ago

7 0

Answer:

F=mass x acceleration

1000= 2000 x a

a=1000/2000

a=1/2 m/sec^2

mark it as brainliest!!!

Gwar [14]3 years ago

5 0

First comment is absolutely correct

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How do you calculate the resultant velocity of the object in this image?

alexgriva [62]

Answer:

the resultant velocity is Zero

Explanation:

by the rule of adding and subtracting factors, we know that; when the force acting on an object is from east & north we add and with forces acting from South & west we minus.

Therefore:

1) List the forces down:

the 100m/s acting west is (-) while

the other 100m/s is acting in easterly direction

so it is (+)

2) Add the forces:

-100+100=0

therefore the answer is 0m/s for the the resultant velocity

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3 years ago

What object best represents a true scale model of the shape of earth?

BabaBlast [244]

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4 years ago

A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted

Snezhnost [94]

We can solve the problem by using Snell's law, which states
$n_i \sin \theta_i = n_r \sin \theta_r$
where
$n_i$ is the refractive index of the first medium
$\theta_i$ is the angle of incidence
$n_r$ is the refractive index of the second medium
$\theta_r$ is the angle of refraction

In our problem, $n_i=1.00$ (refractive index of air), $\theta_i = 28.0^{\circ}$ and $n_r=1.63$ (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_r = \frac{1.00}{1.63} \sin 28.0^{\circ} = 0.288$
From which we find
$\theta_r = \arcsin (0.288)=16.7^{\circ}$

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3 years ago

What is the expermintal example of Zeeman effect?

Mrac [35]

Answer:

When the spectral lines are absorption lines, the effect is called inverse Zeeman effect.

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3 years ago

A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west

Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

$\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}$

(see the attached graphic)

We have

ducks (relative to wind) = 7.0 m/s in some direction θ relative to the positive horizontal direction, or

$\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

wind (relative to Earth) = 5.0 m/s due East, or

$\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)$

ducks (relative to earth) = some speed v due South, or

$\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

Then by setting components equal, we have

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0$

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

We only care about the direction for this question, which we get from the first equation:

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}$

$\cos\theta=-\dfrac57$

$\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)$

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0

3 years ago

Q1) After applying a force of 1000 N an object of mass 2000 kg will achieve whatacceleration?​

Q1) After applying a force of 1000 N an object of mass 2000 kg will achieve whatacceleration?