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2 years ago

6

Un pasager a intarziar la tren. Ajuns pe peron a observat trenul in miscare si a constatat ca unul dintre vagoane a trecut prin

fata sa in intervalul de timp t1=6,6s, iar vagonul al doilea in t2=6s. Considerand miscarea trenului rectilimie uniform variata, sa se determine:
a) intervalul de timp τ cu care a intarziat pasagerul
b) intervalul de timp t3 in care prin fata sa trece vagonul urmator

1 answer:

ExtremeBDS [4]2 years ago

3 0

b) intervalul de timp t3 in care prin fata sa trece vagonul urmator

Sper că am ajutat!

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Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?

vagabundo [1.1K]

<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

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That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

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2 years ago

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a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Verdich [7]

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

Learn more about electrostatic force:

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3 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

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3 years ago

Unbalanced contact forces between several object would result in ?

Svetlanka [38]

In a reaction with a directing vendor of the sum forces towards some object or nothing

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3 years ago

A guitar string with a length of 80.0 cm is plucked. The speed of a wave in the string is 400 m/sec. Calculate the frequency of

Sholpan [36]

Answer:

The first harmonic is: 250Hz, second harmonic 500Hz, third harmonic 750Hz.

Explanation:

Use the frequency f, speed v, and wavelentgh L relationship:

$v = f\cdot L\implies f = \frac{v}{L}$

We are given the speed v=400 m/s. The base wavelength on a string of length 80cm is twice the length of the string (a "half wave" along the full length of the string), so:

$f = \frac{400\frac{m}{s}}{2\cdot0.8 m}= 250\frac{1}{s}=250 Hz$

The fundamental frequency (first harmonic) is 250 Hz

The second harmonic is produced by one full wave across the string (adding one node in the middle), so L=80cm in this case, therefore the second harmonic frequency is: f2 = 2*250=500Hz

the third harmonic add another node (and a half wave) to the pattern and the wavelength will be 2/3 of 80cm, so f3=3*250Hz = 750Hz

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2 years ago