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Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches

aleksandrvk [35]

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass $m=380kg$

Height of supporting Rock $X=85cm$

Length of Board $L_r=4.5m$

Mass of board $M_b=22kg$

Mass of adult $M_a=62$

Generally the moment of balance about wedge part about is mathematically given by

$N -Q + R = Mg + mg$

$0.85*N - Mg*2.25 - mg*(2.25 + x) = 0$

$0.85*N = + Mg*2.25 + mg*(2.25 + x)$

where

$N+R=4547$

therefore

$N = 570.70588 + 1608.3529 + 714.823 x$

if N=0 at fallen person

$x=3.04m$

Therefore it is save to carry a 62kg adult

8 0

3 years ago

The resistivity of a silver wire with a radius of 5.04 × 10–4 m is 1.59 × 10–8 ω · m. if the length of the wire is 3.00 m, what

Alexxx [7]

<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>

4 0

3 years ago

The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?

SIZIF [17.4K]

The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
$I(r)= \frac{1}{r^2}$
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put
$I= \frac{1}{r^2}= \frac{1}{(1)^2}=1$
The intensity at r=3 m will be
$I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}$
Therefore, the sound intensity has decreased by a factor $1/9$ .

8 0

3 years ago

Read 2 more answers

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts

Nata [24]

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

∑F is net force
m is mass of the third child
a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

the wagon
the children outside the wagon

<h3>Free body diagram</h3>

→ → Ф ←

1st child friction wagon 2nd child

<h3>Acceleration of the child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

#SPJ1

7 0

2 years ago

12 POINTS PLEASE HELP.

bazaltina [42]

<span>Are programs that basically want to publicize the lives of other people sopadamente to help them</span>

3 0

4 years ago