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3 years ago

If there no motion then there is no force acting? true or false

Physics

1 answer:

goblinko [34]3 years ago

4 0

Answer:

false.

Explanation:

If a object is at rest it does not means that no force is acting on the object.

There can be a scenario that all to forces acting on the object balance each other and the net force required for motion is zero.

So, the given statement is false.

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One of the checks that you could do for problem 1) would be to check the output resistance of the Wheatstone bridge to make sure

Ray Of Light [21]

Answer:

Explanation:

Let the four resistances of th wheat stone bridge is

P, Q, R and S and the value of each is 350 ohm.

Here, P and Q are in series.

R' = P + Q = 350 + 350 = 700 ohm

Then R and S are in series

R' = R + S = 350 + 350 = 700 ohm

Now R' and R'' are in parallel.

So, the equivalent resistance is

Req = R' x R'' / ( R' + R'')

Req = 700 / 2 = 350 ohm

Thus, the reading of ohmmeter is 350 ohm.

6 0

3 years ago

Un globo de aire caliente tiene un volumen de 560 mL a la presión atmosférica normal y una temperatura del aire de 400 ºC. Cuand

dalvyx [7]

Explanation:

charles law V1/T1 =V2/T2

560 x 673 =V2/973

376880 = V2/973

V2 = 376880 x 973 = 366704240mL

4 0

3 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

3 years ago

A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0

3 years ago

Model rocket A is launched at an angle of 70° above the x axis, with an initial velocity of 40 m/s. An identical rocket B is lau

amid [387]

<span>A. Rocket A will travel farther horizontally than rocket B.

This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.</span>

3 0

3 years ago