A car travels at a velocity of 8m/s for 16s . How far did it travel?

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

Please help This very important for me you will get 40 points if you help me and explain in deatil please.

posledela

Well, I think mostly it is oceanic crust well because <span>The scientist noticed a symmetrical pattern of positive and negative magnetic lines as they moved along the ocean floor, and the line of symmetry was at the mid-ocean ridge. that why it is seen as the youngest on our earth</span>

4 0

3 years ago

Read 2 more answers

A. Write 2 -3 sentences describing the mass number, charge, and penetrating ability of gamma rays.

Leya [2.2K]

Answer:

a. Mass number is the sum of number of protons and neutrons present in the nucleus of an atom.

A = p + n

Charge is a unit of matter that tells how more or less the number of electrons are present as compared to number of protons in an atom. A positive charge means the atom lacks electrons.

Gamma rays are highly energetic electromagnetic rays. They have high penetrating power and are very difficult to shield owing to high energy.

b. The compete nuclear equation is:

$^{210} _{89} Ac \rightarrow ^4 _2 He + ^{206} _{87}Fr$

This is a alpha decay process because alpha particle which is Helium atom is being produced. Francium is being formed after alpha particle is emitted.

c. The complete nuclear reaction is:

$^{80} _{38} Sr \rightarrow ^0 _{+1}e + ^{80} _{37}Rb$

This is a positron emission process where in Rubidium is being formed.

8 0

3 years ago

Read 2 more answers

The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the ce

NISA [10]

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

$V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}$ -------equation (1)

The electric field on the sphere is also given as;

$E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}$

$|q |= 4 \pi \epsilon _o} R^2E$

Substitute in the value of q in equation (1)

$V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V$

Therefore, the potential at the center of the sphere is -924 V

8 0

3 years ago

A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms

babymother [125]

Answer:

The intensity of the EM wave is 5.729 W/m²

The rms value of the electric field is 46.452 V/m

Explanation:

Given;

power of the electromagnetic wave, P = 1800 W

The intensity of light is given by;

I = P / A

Where;

A is the surface area of the sphere = 4πr²

A is the surface area of the sphere = 4π(5)²

A is the surface area of the sphere = 314.2 m²

The intensity is calculated as;

I = P / A

I = (1800) / (314.2)

I = 5.729 W/m²

The rms value of the electric field is given by;

$E_{rms} = \sqrt{\frac{I}{c \epsilon _o}}\\\\E_{rms} = \sqrt{\frac{5.729}{(3*10^8)( 8.85*10^{-12})}}\\\\E_{rms} = 46.452 \ V/m$

8 0

4 years ago