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3 years ago

A 3,273N force is applied to a 223kg mass. What is the acceleration of the mass?

Physics

1 answer:

Alexxandr [17]3 years ago

8 0

Answer:

acceleration =

$14.67ms {}^{ - 2}$

Explanation:

force = mass × acceleration
acceleration = force / mass
a = f / m

we know that f = 3273 N, M = 223 kg

a = 3273 / 223
a = 14.67 m/s^2

 i hope it helped.....

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A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i

andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93

note : w = $\sqrt{k/m}$ = $\sqrt{337.93}$ = 18.38 rad/sec

Period of oscillation = $2\pi / w$

= 0.34 sec

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3 years ago

A 160.-kilogram space vehicle is traveling along a straight line at a constant speed of 800. Meters per second. The magnitude of

natulia [17]

Answer:

Zero

Explanation:

As force acting on the body is equal to the product of mass and acceleration.

Acceleration is equal to rate of change in velocity.

Here velocity is constant so acceleration is zero.

It means the net force acting on the vehicle is zero.

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3 years ago

Read 2 more answers

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

Scrat [10]

Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$ , hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$ (5)

Finally:

$\theta_{2}=1.8\°$ (6)

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3 years ago

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kenny6666 [7]

Answer:

Explanation:

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3 years ago

How can you verify the archimedes principle?

Ipatiy [6.2K]

Answer:

It is found that W1 - W2 loss in weight of solid when immersed in water is equal to the weight of the water displaced by the body. This verifies Archimedes' principle.

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2 years ago