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2 years ago

10

What do you need to know to analyze the forces in a situation?

1 answer:

alukav5142 [94]2 years ago

6 0

Answer:

An analysis yielding the respective forces acting at any point of any member, or part of a member, of a mechanism, obtained by using relationships for dynamic equilibrium in a plane rigid body subject to external forces within this plane and to internal forces due to its motion in this plane.

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A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this

Hunter-Best [27]

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

$m = 3500lb = 1587.57kg$

The values of the velocities are,

$\text{Initial Velocity} = V_i = 0$

$\text{Final Velocity} = V_f = 60mph = 26.822m/s$

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

$a = \frac{v_f-v_i}{t}$

$a = \frac{26.822-0}{5}$

$a = 5.36m/s^2$

Now applying the Newton's second law we have,

$F= ma$

$F = (1587.57)(5.36)$

$F = 8516.36N$

Therefore the approximate magnitude is 8516.36N

5 0

3 years ago

Most of the energy released by nuclear fission is in the form of: alpha rays, beta rays, or gamma rays?

Sedbober [7]

I think it is <span>Alpha rays.</span>

3 0

3 years ago

A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizont

aev [14]

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

$F_y = F sin \theta$

$F_y = 245\times sin 55^0$

$F_y =200.69$

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

6 0

3 years ago

Read 2 more answers

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

So

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

7 0

3 years ago

A solid box made from a material whose density is rhoS floats with two thirds of its volume submerged in a liquid whose density

jok3333 [9.3K]

Answer:

$\frac{\rho_S}{\rho_L} = \frac{2}{3}$

Explanation:

density of the solid box material = $\rho_s$

density of the liquid material = $\rho_L$

Given that

solid box floats with two thirds of its volume submerged in a liquid

let V be the volume of the box

then,

$V\rho_sg= \frac{2V}{3}\rho_L g$

⇒ $\frac{\rho_S}{\rho_L} = \frac{2}{3}$

so, the ratio of densities of solid and and the liquid is 2/3

7 0

3 years ago