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3 years ago

1

Physics

1 answer:

s344n2d4d5 [400]3 years ago

8 0

Answer:

$\large \boxed{\mathrm{40 \ Nm}}$

Explanation:

$\sf Moment \ (Nm) = Force \ (N) \times Distance \ from \ pivot \ (m)$

$M=20 \times 2$

$M=40$

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pashok25 [27]

Answer:

C: Light travels from the Sun to the grass and is then reflected to your eyes.

Explanation:

Our eyes don't produce light but detect light (so answer A is not correct)

The grass doesn't produce light unless it is burning (discard answer B)

Answer C is the correct one.

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3 years ago

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4. The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C

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Answer:

a) ΔL/L = F / (E A), b) $L_{f}$ = L (1 + L F /(EA) )

Explanation:

Let's write the formula for Young's module

E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

P = E ΔL / L

The pressure is defined as

P = F / A

Let's replace

F / A = E ΔL / L

F = E A ΔL / L

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b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

ΔL = L [F / EA]

$L_{f}$ -L = L (F / EA)

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3 years ago

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2 years ago

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Answer:

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3 years ago

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

$\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta$

∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

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3 years ago