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2 years ago

Select the statement that correctly lists the forces from strongest to weakest. (2 points)

Physics

1 answer:

Vilka [71]2 years ago

5 0

Strong Nuclear force: it is the short range force and strongest fundamental force in all type of forces.

Electromagnetism: this is the force due to magnetic and electric behavior of the particles. It is moderate type of force and its range is more than Nuclear force.

Weak Nuclear Force: This force is also short range force which act between the nucleoside. But this force is also moderate type of force

Gravitational force: this force is between two point masses and least order of force. also the range of this force is upto infinite.

so the correct order of this fundamental force is

<em>strong nuclear, electromagnetism, weak nuclear, gravitational</em>

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3 years ago

If ?h°rxn and ?s°rxn are both negative values, what drives the spontaneous reaction and in what direction at standard conditions

irakobra [83]

The release of free energy drives the spontaneous reaction.

Spontaneity can be <span>determined using the change in </span>Gibbs free energy (the thermodynamic potencial):

delta G=delta H – T*delta S

where delta H is the enthalpy and delta S is the entropy.

The direction (the sign) of delta G depends of the changes of enthalpy and entropy. If delta G is negative then the process is spontaneous.

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3 years ago

Astronauts are trained for take-off in a high-speed centrifuge of 4.7 m radius that spins in the horizontal plane.

Leokris [45]

Answer:

a) $1.94 \frac{rad}{s}$

b) $9.12\frac{m}{s}$

c) Towards the center of the centrifuge

Explanation:

Becuse the centrifuge rotates in circular motion, there's an angular acceleration tha simulates high gravity accelerations

$a_{rad}=\omega r^{2}$

with r the radius and ω the amgular velocity, in or case $a_rad=3.5g$ so:

$3.5g=\omega r^{2}$ and g=9.8 $\frac{m}{s^{2}}$

solving for ω:

$\omega=\frac{3.5g}{r^2}=\frac{3.5*9.8}{4.2^2}$

$\omega = 1.94 \frac{rad}{s}$

b) Linear speed (v) and angular speed are related by:

$v=\omega r =(1.94)(4.7)$

$v= 9.12\frac{m}{s}$

c) The apparent weigth is pointing towards the center of the circle, becuse angular acceleration is pointing in that direction.

8 0

2 years ago

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

$A = W + \dfrac{W}{2}$

$A =\dfrac{3W}{2}$

$A =\dfrac{3mg}{2}$

When the car is at the bottom, the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

$N = m g + \dfrac{mv^2}{r}$

$\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}$

$\dfrac{mg}{2} = \dfrac{mv^2}{r}$

$v = \sqrt{\dfrac{rg}{2}}$

$v = \sqrt{\dfrac{40\times 9.8}{2}}$

v = 14 m/s

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2 years ago