Which of the following would increase the output force of a lever?

An airplane is flying overhead at a constant elevation of 3000ft. A man is viewing the plane from a position 4000ft from the bas

ziro4ka [17]

Answer: The distance between the man and the plane increasing at a rate of 400ft/s

Explanation: Please see the attachments below

4 0

3 years ago

An HR manager is asked to meet with upper management to discuss the

amm1812

Answer:

the answer to this question perhaps is service

3 0

3 years ago

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A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago

The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

$X_f = X_i +\frac{1}{2}(V_i+V_f)t$

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

$X_f = \frac{1}{2} (V_i+V_f)t$

Clearing for time,

$t = \frac{2X_f}{(V_i+V_f)}$

$t = \frac{2*1.2}{24+0}$

$t = 0.1s$

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

$F = m\frac{V_f-V_i}{t}$

$F = m\frac{-V_i}{t}$

$F = \frac{(1600kg)(-24m/s)}{(0.1s)}$

$F = -384000N$

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

$V_f^2=V_i^2-2ax$

$0 = (24m/s)^2-2*a(1.2m)$

$a = \frac{(24m/s)^2}{1.2m}$

$a=480m/s^2$

The gravity is equal to 0.8, then the acceleration is

$a = 480*\frac{g}{9.8}$

$a = 53.3g$

3 0

3 years ago

How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners

Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

$U =\frac{Kq_{1}q_{2}}{d}$

So, the total potential energy of teh system is

$U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}$

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

$U =3\times \frac{Kq^{2}}{d}$

K = 9 x 10^9 Nm^2/C^2

By substituting the values

$U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}$

U = 1.38 x 10^-18 J

6 0

3 years ago