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tino4ka555 [31]
3 years ago
9

X = v xo t. x = (10.0 m/s)(3.53 s) x = ????

Physics
2 answers:
LenaWriter [7]3 years ago
7 0

x = V<em>x</em> * t

given V<em>x</em> = 10m/s n t = 3.53s

x = 10 * 3.53

= 35.3m


labwork [276]3 years ago
7 0

The equation given is confusing: x = v xo t. But given x = (10.0 m/s)(3.53 s), I think x = (10.0 m/s)(3.53 s) = 35.3m


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In at least 150 words, discuss how Chang's use of personification, metaphor, or connotation express one his themes in "Garden of
Kay [80]
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3 years ago
I). Mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work.
Dmitry_Shevchenko [17]

Answer:

false statement : b )  For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

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differentiating both side

Δ potential energy + Δ kinetic energy = 0

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first statement is true.

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change in gravitational potential energy = change in kinetic energy + work done against friction .

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change in kinetic energy.

4 0
3 years ago
Who is the leader of the party's national committee
cestrela7 [59]
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3 0
3 years ago
Read 2 more answers
What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass
Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

8 0
2 years ago
4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2
bazaltina [42]

Answer:

1030.83\ \text{N}

Explanation:

\Delta L = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = \dfrac{\pi}{4}d^2

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}

La tensión en la cuerda es 1030.84\ \text{N}.

8 0
3 years ago
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