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2 years ago

>En cual de las siguientes situaciones la fuerza neta sobre el cuerpo es cero?

1 answer:

3 0

La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".

Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.

Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.

En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.

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An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr

AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

$a = \dfrac{400}{2\times 50}$

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{4}{9.81}$

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0

2 years ago

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.

inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$

= $\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

= $1.18 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

ma = qE

a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

= $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

or, d = $\frac{v^{2}}{2d}$

= $\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}$

= 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0

1 year ago

When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What

ser-zykov [4K]

Answer:

Electric field, $E=4\times 10^5\ N/C$

Explanation:

It is given that,

Magnitude of charge, $q_o=2\ nC=2\times 10^{-9}\ C$

Force experienced, $F=8\times 10^{-4}\ N$

We need to find the electric field at the origin. It is given by :

$F=q_o\times E$

$E=\dfrac{F}{q_o}$

$E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}$

$E=4\times 10^5\ N/C$

So, the electric field at the origin is $4\times 10^5\ N/C$ . Hence, this is the required solution.

3 0

2 years ago

An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in

SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

$h=-16t^2+v_ot$

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

$-16t^2+250t=500$

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0

2 years ago

Read 2 more answers

A 60 kg student in a rowboat on a still lake decides to dive off the back of the boat. The studen'ts horizontal aceleration is 2

TiliK225 [7]

As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.

So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.

Force on the student = 60 kg * 2.0 m/s^2 = 120 N.

=> horizontal force exerted by the student on the boat = 120 N

Answer: option d. 120 N. toward the back of the boat.

Of course it is toward the back because that is where the student jumped from..

4 0

2 years ago