Which of the following causes the efficiency of a machine to be less than 100%?

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

kipiarov [429]

Answer:

$2.24 \times 10^{-14}$ Newtons

Explanation:

Magnitude of charge on the electron = q = $1.6 \times 10^{-19}$ Coulombs

The negative sign in the question statement indicates that the charge is negative.

Magnitude of Electric Field experienced by the electron = E = $1.4 \times 10^{5}$ Newtons/Coulomb

Magnitude of Force on the electron = F = ?

The relation between the charge, electric field and the force on the charge because of electric field is given by:

$E=\frac{F}{q}$

From here we can write:

F = qE

Using the values, we get:

$F = 1.6 \times 10^{-19} \times 1.4 \times 10^{5}\\\\ F = 2.24 \times 10^{-14}$

Thus, the magnitude of the electric force experience by the electron would be $2.24 \times 10^{-14}$ Newtons

8 0

3 years ago

A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc

OLga [1]

Answer:

5 N

Explanation:

The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N

7 0

3 years ago

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A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

$E=U_f = mg h_f$

where $h_f$ is the maximum height reached. Solving for this quantity, we find

$h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m$

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

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5 0

3 years ago

Which statement is true?

iogann1982 [59]

B
Think of inertia of getting into a car accident without a seat belt although the car stops you will not you would likely fly out the window

7 0

3 years ago

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A monochromatic light passes through a narrow slit and forms a diffraction pattern on a screen behind the slit. As the slit widt

maw [93]

Answer:

shrinks with all the fringes getting narrower

Explanation:

As the light passes through the slit, the diffraction pattern shrinks, as the waves have more opening to penetrate, and the fringes becomes more narrow as a result of that, The opposite happens as the conditions are reversed.

8 0

3 years ago