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kirill115 [55]
3 years ago
11

On a farm, you are pushing on a stubborn pig with a constant horizontal force with magnitude 30.0 n and direction 37.0° counterc

lockwise from the +x-axis. how much work does this force do during a displacement of the pig that is (a) s s = 15.00 m2nd; (b) s s = ?16.00 m2ne; (c) s s = ?12.00 m2nd + 14.00 m2ne?
Physics
1 answer:
sashaice [31]3 years ago
8 0

<span>Since the force is applied at an angle from the horizontal, we will use the horizontal component of this force in calculating for the displacements.
From derivation, the Fx is:</span>

Fx = F cos φ

Where:

Fx = is the horizontal component of the force

F = total force

φ = angle in radian = 37 * pi / 180 = 0.645 rad

Calculating: Fx = 30.0 N * cos(0.645)

Fx = 23.97 N = 24 N

Calculating for Work: W = Fx * d

A. W = 24 N * 15 m = 360 N

B. W = 24 N * 16 m = 384 N

C. W = 24 N * 12 m = 288 N

D. W = 24 N * 14 m = 336 N 

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The main requirement for a good conductor of electricity is to have a lot of valence electrons. Valence electrons are the electrons of the outer shells of atoms not bound with other atoms (for example through covalent bounds). These electrons are "free to escape" as soon as an electric field with enough intensity is applied to the material, and therefore these electrons will be free to move in the material producing an electric current.
3 0
4 years ago
A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that
kherson [118]

Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

Find the Speed of the Proton

The formula for Kinectic Energy =

K.E = 1/ 2 mv²

Where

m = mass of the proton

v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

v = 43772331.227m/s

v = 4.3772331227 × 10^7m/s

Approximately = 4.4 × 10^7 m/s

Step 3

Find the Magnetic Field of that region of space

The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

4 0
3 years ago
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Answer:

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4 0
3 years ago
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You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

F_f = \mu m g

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now it is given that

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g = 9.81 m/s^2

so here we have

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4 0
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Answer:Water Only

Explanation:

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Heat released by hot water is equal to heat gain by cold water .

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