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4 years ago

8

The electromagnetic interaction _______.A. applies only to charges at rest B. applies only to charges in motion C. is responsibl

e for sliding friction and contact forces D. all of the above E. none of the above

1 answer:

Bond [772]4 years ago

8 0

Answer:

C. is responsible for sliding friction and contact forces

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(Really Need Help!) (55 points) (4 Choices) Which of the following is not a true statement?

Lena [83]

The last one, at the bottom, is false.

6 0

3 years ago

A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so

shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

$T_y+N=m.g$

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$

Recalling that

$F_r=\mu N$

$Tcos\theta-\mu N=m.a$

Replacing N from [1]

$Tcos\theta-\mu (m.g-Tsin\theta)=m.a$

Operating

$Tcos\theta-\mu m.g+\mu Tsin\theta=m.a$

Solving for T

$T(cos\theta+\mu sin\theta)=m.a+\mu m.g$

$\displaystyle T=\frac{m.a+\mu m.g}{cos\theta+\mu sin\theta}$

$\displaystyle T=m\frac{a+\mu g}{cos\theta+\mu sin\theta}$

We don't know the value of m, thus we'll plug in the rest of the data

$\displaystyle T=m\frac{0.763+0.10\cdot 9.8}{cos36.8^o+0.10 sin36.8^o}$

$T=2.0252m$

Dividing by the weight of the box m.g

$T/(m.g)=2.0252/9.8=0.2066$

Thus, the magnitude of the pulling force is 20.66% of the gravitational force acting on the box

7 0

3 years ago

A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft2/s and separated by a distance of 15 f

Rashid [163]

Answer:

0.28 ft

Explanation:

We are given that

Strength=m= $36ft^2/s$

Distance between source and sink=15 ft

Distance between the sink of the source and origin= $a=\frac{15}{2}$ ft

Uniform velocity, U=12 ft/s

We have to find the length of the oval.

Formula to find the half length of the body

$\frac{l}{a}=(\frac{m}{\pi Ua}+1)^{\frac{1}{2}}$

Where a=Distance between sink of source and origin

U=Uniform velocity

m=Strength

l=Half length

Using the formula

$\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}$

$l=\frac{2}{15}(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}$

$l=0.14$

Length of oval= $2l=2(0.14)=0.28 ft$

8 0

3 years ago

How much energy is transferred electrically by a 2000 W cooker in half an hour?

Kryger [21]

Responder:

E = 1440 kJ

Explicación:

Se da que,

La potencia de un horno de cocción es de 800 W

El voltaje al que se opera es de 230 V

Tiempo, t = 30 minutos = 1800 segundos

Necesitamos encontrar la energía eléctrica utilizada por el horno de cocción. El producto de la potencia y el tiempo es igual a la energía consumida. Entonces,

5 0

3 years ago

Read 2 more answers

Si tu velocidad al caminar es 0,3 m/s. Determina el tiempo que demorarias en llegar al sol en horas (REALIZAR TRANSFORMACIÓN DE

Contact [7]

Answer:

138,516,546.9 horas.

Explanation:

Tenemos que usar la ecuación:

Velocidad = distancia/tiempo

Acá tenemos:

Velocidad = 0.3m/s

distancia = 149597870700 m

y queremos resolver la ecuación para el tiempo:

0.3m/s = 149597870700m/tiempo.

tiempo = 149597870700m/(0.3m/s) = 498,659,569,000 s

y sabemos que una hora tiene 3600 segundos, entonces si queremos transformar de segundos a horas tenemos:

498,659,569,000 s = (498,659,569,000/3600) h = 138,516,546.9 horas.

6 0

3 years ago