What are three examples of noncontact forces

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

Answer:

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

Explanation:

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago

How dose an atom change if all of its electrons are removed

Anni [7]

Answer:

When all the electrons are removed from an atom, it becomes something as a positively charged particles also known as alpha particles.

Explanation: 

The bare nucleus which is positively charged help in scattering experiments as it has high penetrating powers. An atom is made up of electrons, protons and neutrons. We need huge energy to separate the electrons from their parent atom, still making it separated brings you a particle with a positive charge and only mass having high penetrating power.

7 0

3 years ago

PLEASE HELP I WILL GIVE MAX POINTS

Finger [1]

you can collect water and shine a light though it and record your finings

4 0

3 years ago

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A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?

Olenka [21]

To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.

7 0

3 years ago

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The instruction booklet for your pressure cooker indicates that its highest setting is 12.3 psi . you know that standard atmosph

zmey [24]

118 C The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is: Tb = 1/(1/T0 - R ln(P/P0)/Hvap) where Tb = Temperature boiling R = Ideal Gas Constant (8.3144598 J/(K*mol) ) P = Pressure of interest Hvap = Heat of vaporization of the liquid T0, P0 = Temperature and pressure at a known point. The temperatures are absolute temperatures. We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So P0 = 14.7 P = 14.7 + 12.3 = 27 T0 = 100 + 273.15 = 373.15 And for water, the heat of vaporization per mole is 40660 J/mol Let's substitute the known values and calculate. Tb = 1/(1/T0 - R ln(P/P0)/Hvap) Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol) Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660) Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660) Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660) Tb = 1/(0.002679887 1/K - 0.000124326 1/K) Tb = 1/(0.002555561 1/K) Tb = 391.3034763 K Tb = 391.3034763 K - 273.15 Tb = 118.1534763 C Rounding to 3 significant figures gives 118 C

3 0

3 years ago

What are three examples of noncontact forces​

What are three examples of noncontact forces