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3 years ago

What does the force of friction between two objects in context depend on?

1 answer:

7 0

The amount of friction depends on the force pushing the surfaces together. If this force increases, the hills and valleys of the surfaces can come into closer contact. The close contact increases the friction between the surfaces.

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Which two of the following reduce the background noise in magneto- medicine?

horrorfan [7]

Answer:

The answer is b.) and d.)

Explanation:

The options to reduce the background noise in magneto-medicine are given as follows:

a.) Orienting the heart parallel to the Earth's field

- This will have no significant effect on the measurements.

b.) Taking the difference of two nearby sensor measurements (gradiometer).

-This answer is TRUE.

c.) Placing the heart in a perpendicular fashion to the Earth's magnetic field.

_ This answer also will not have any significant effect on measurements.

d.) Using physical means to shield environmental fields.

- This answer is TRUE.

4 0

3 years ago

Why should we convert units within the metric system

irga5000 [103]

It is often necessary to convert one metric measurement to another unit—this happens frequently in the medical, scientific, and technical fields, where the metric system is commonly used.

7 0

4 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

4 years ago

Calculate the change internal energy (Î´e) for a system that is giving off 45.0 kj of heat and is performing 855 j of work on th

adelina 88 [10]

The change in internal energy of a system is given by (second law of thermodynamics)
$\Delta U = Q + W$
where Q is the heat absorbed by the system and W is the work done on the system.

In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings

In our problem, the heat released by the system is $Q=-45 kJ=-45000 J$ (with negative sign since it is released by the system), and the work done is $W=-855 J$ still with negative sign because it is performed by the system on the surrounding, so the change in internal energy is
$\Delta U = Q +W=-45000 J - 855 J=-45855 J$

3 0

3 years ago

The mechanical advantage of a simple machine is 3 and its velocity ratio is 4 find its efficiency

LiRa [457]

Answer:

75%

Explanation:

Efficiency=MA/VR*100

=3/4*100

=0.75*100

=75%

6 0

2 years ago