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3 years ago

A spring with a force constant of 120 N/m is used to push a 0.27-kg block of wood against a wall

Physics

1 answer:

kvv77 [185]3 years ago

4 0

Answer

given,

Spring force constant = 120 N/m

mass of the wooden block = 0.27 Kg

coefficient of friction = 0.46

a) we know,

μ N = mg

0.46 x N = 0.27 x 9.8

N = 5.75 Newton

This should be the normal force on the block by the wall. Thus minimum spring force should be 5.75 newton.

F = k x d

k = spring constant,

d = compression in spring)

5.75 = 120 x d

d = 0.048 m

d = 4.8 cm

b) answer will change if the mass of block is doubled.

mass becomes = 0.54 kg

the normal force needed = 11.4 newton.

the compression becomes = 9.6 cm.

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A mass of M-kg rests on a frictionless ramp inclined at 30°. A string with a linear mass density of μ=0.025" kg/m" is attached t

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Answer:

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Explanation:

a) Draw a free body diagram of the mass M. There are three forces:

Weight force mg pulling down,

Normal force N pushing perpendicular to the ramp,

and tension force T pulling parallel up the ramp.

Sum of forces in the parallel direction:

∑F = ma

T − Mg sin 30° = 0

T = Mg sin 30°

T = Mg / 2

Draw a free body diagram of the hanging mass m. There are two forces:

Weight force mg pulling down,

and tension force T pulling up.

Sum of forces in the vertical direction:

∑F = ma

T − mg = 0

T = mg

Substitute:

mg = Mg / 2

m = M / 2

M = 2m

b) Velocity of a standing wave in a string is:

v = √(T / μ)

T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N. Therefore:

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v = 44.3 m/s

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3 years ago

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3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

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(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

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2 years ago

A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going

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Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
After another half-life, 500 are left.
After another half-life, 250 are left.
After another half-life, 125 are left.

That was FOUR half-lifes.

X = 4 .

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