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2 years ago

The following is regarding Momentum Change and Force. help is needed please!

Physics

1 answer:

mixer [17]2 years ago

4 0

The distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.

<h3>Velocity of the object at the end of the time</h3>

F = mv/t

where;

m is mass of the object
v is velocity of the object
t is time

Ft = mv

v = Ft/m

v = (50 x 10)/250

v = 2 m/s

<h3>Distance traveled by the object</h3>

v² = u² + 2as

where;

u is initial velocity = 0

a is acceleration

a = F/m

a = 50 N/ 250 kg

a = 0.2 m/s²

v² = 0 + 2as

s = v²/2a

s = (2²)/(2 x 0.2)

s = 10 m

Thus, the distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.

Learn more about distance here: brainly.com/question/2854969

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What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

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Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

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Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

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A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?

forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

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$Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}$

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Now similar way we can say

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