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1 year ago

Come si compongono due forze che agiscono in diversi punti di un corpo rigido? Oof

Physics

1 answer:

bagirrra123 [75]1 year ago

7 0

Answer:

Explanation:

I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.

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Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h

ruslelena [56]

Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

$Y+F=600$ ....(I)

$X=T$ .....(II)

On taking moment about base

$X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0$

Put the value into the formula

$X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0$

$3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30$

$1.598T=450-1.5(600-F)$ ....(III)

We need to calculate the force for ladder

$2F=600\trimes 2$

$F=600\ N$

We need to calculate the tension in the rope

From equation (3)

$1.598T=450-1.5(600-600)$

$1.598T=450$

$T=\dfrac{450}{1.598}$

$T=281.60\ N$

Hence, The tension in the rope is 281.60 N.

7 0

2 years ago

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The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1

Andrei [34K]

Answer:

The expression of gravitational field due to mass $m_!$ at a distance $l_a$

Explanation:

We have given mass is $m_1$

Distance of the point where we have to find the gravitational field is $l_a$

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by $g=\frac{Gm_1}{l_a^2}$

This will be the expression of gravitational field due to mass $m_!$ at a distance $l_a$

4 0

1 year ago

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

$R=\dfrac{u^2\ \sin2\theta}{g}$

Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

$\alpha =30^{\circ}$

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0

2 years ago

An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo

lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

$\frac{d'_1}{d'_2}=\frac{f_1}{f_2}$

Magnification of first lens

$M_2=-\frac{d'_1}{d_1}$

and

$M_2=\frac{h'_1}{h_1}$

Similarly, magnification of second lens

$M_2=-\frac{d'_2}{d_1}$

and

$M_2=\frac{h'_2}{h_1}$

From the above equations we get

$\frac{M_1}{M_2}=\frac{d'_1}{d_2'}$

and

$\frac{M_1}{M_2}=\frac{h'_1}{h_2'}$

which means,

$\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}$

and

$\frac{d'_1}{d_2'}=\frac{f_1}{f_2}$

So, we get

$\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm$

∴ Focal length should this camera's lens is 24.71 mm

6 0

1 year ago

A student uses an electronic force sensor to study how much force the student’s finger can apply to a specific location. The stu

melisa1 [442]

Answer:

B. Trial 2

Explanation:

Trial 2, because the student’s finger applied the largest force to the sensor.

Because the trial 2 student finger applied to largest force.

7 0

2 years ago

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