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Elina [12.6K]
3 years ago
6

A model train traveling at a constant speed around a circular track has a constant velocity

Physics
1 answer:
TiliK225 [7]3 years ago
3 0
A model train traveling at a constant speed around a circular track has a constant velocity. FALSE.
Hope this helps you!

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A 6.0 m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is 0.40
MrRissso [65]

Answer:

a) t = 0.0185 s = 18.5 ms

b) T = 874.8 N

Explanation:

a)

First we find the seed of wave:

v = fλ

where,

v = speed of wave

f = frequency = 810 Hz

λ = wavelength = 0.4 m

Therefore,

v = (810 Hz)(0.4 m)

v = 324 m/s

Now,

v = L/t

where,

L = length of wire = 6 m

t = time taken by wave to travel length of wire

Therefore,

324 m/s = 6 m/t

t = (6 m)/(324 m/s)

<u>t = 0.0185 s = 18.5 ms</u>

<u></u>

b)

From the formula of fundamental frquency, we know that:

Fundamental Frequency = v/2L = (1/2L)(√T/μ)

v = √(T/μ)

where,

T = tension in string

μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹

Therefore,

324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)

(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹

<u>T = 874.8 N</u>

8 0
3 years ago
Perfect symmetrical cone type volcanoes would have what type of drainage pattern
Charra [1.4K]

Perfect symmetrical cone type volcanoes would have radial drainage pattern.

A stratovolcano, also known as a composite volcano, is a conical volcano built up by many layers (strata) of hardened lava, tephra, pumice, and volcanic ash.

If you have any further questions, please don’t hesitate to ask again.

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3 years ago
HELP!!! HURRY!!!!! 100 POINTS!!!!!!!!!
bagirrra123 [75]

what happens at Point C is sublimation. the increase in temperature affects the Vapour pressure soon as you can see the curve is increasing with increasing pressure there is increase in temperature that is the sublimation Curve

7 0
3 years ago
Read 2 more answers
Action reaction forces are in which one of newtons laws
Klio2033 [76]

Answer:

the third law

Explanation:

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4 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
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