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3 years ago

What are non-contact forces please include example in your answer

Physics

2 answers:

Alina [70]3 years ago

4 0

Answer:

Gravitational force. Magnetic force. Electrostatics. Nuclear force.

Explanation:

Apple falling from a tree

raindrops falling from the sky

Katen [24]3 years ago

4 0

The most familiar non-contact force is gravity, which confers weight. In contrast a contact force is a force applied to a body by another body that is in contact with it. All four known fundamental interactions are non-contact forces: Gravity, the force of attraction that exists among all bodies that have mass.

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A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

$v = \sqrt{\frac{2GM}{R}}$

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

$v = 14.9km/h = 14900m/s$

Through this equation we can find the mass of the Planet in function of the distance, therefore

$M = \frac{v^2R}{2G}$

$M = \frac{14900^2R}{2(6.67*10^{-11})}$

$M = 16.64*10^{17}R$

The orbital velocity is

$v_o = \sqrt{\frac{GM}{R+h}}$

$9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}$

$11.1*10^7R = (R+15000*10^3)(9200)^2$

$2.64*10^7R = 12.69*10^{13}$

$R = 4.81*10^6m$

The time period of revolution is,

$T = \frac{2\pi(R+h)}{v_o}$

$T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}$

$T = 4307s$

$T = 72min = 1hour12min$

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0

3 years ago

An inductor has an inductance of 0.12 mH. Calculate the number of turns per unit length of this inductor if it has a total of 50

SSSSS [86.1K]

Answer:

Explanation:

5 0

3 years ago

What total mass must be converted into energy

Eduardwww [97]

This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
E = m c²

   1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

   Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

   = (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

   = 1.144 x 10⁻³⁰ kg . (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:

The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

   = (2,242 x 10⁶ x 86,400 x 365) joules

   = 7.0704 x 10¹⁶ joules .

How much converted mass is that ?

   E = m c²

Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s

    Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

      = 0.786 kilogram ! ! !

THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

</span>

7 0

3 years ago

Read 2 more answers

Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t

adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x (1)

Also, the T₂ = t/n₂ = t/(n₁ + x) (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0

3 years ago

Which statement best describes the importance of joints in the roadways of Bridges ?

Sergeeva-Olga [200]

Answer:

c. Joints allow the roadway to expand and contract as cars put force on the bridge

Explanation:

The reasons why joints are allowed on roadway is to accommodate the contraction and expansion of the road as cars put force on them.

Most materials used in making roadways are susceptible to expansion and contraction.
When a measure of force is applied their length either increases or decreases depending on the type of force.
To accommodate these changes, joints are placed in roadways

3 0

3 years ago