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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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An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet
zimovet [89]

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v²

          Em₀ = Emf

          e ΔV = ½ m v²

          v =√ 2 e ΔV / m

          v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

           v = √(1.8075 10¹⁶)

           v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

        F = 1.6 10⁻¹⁹  1.344 10⁸ 0.4

       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Emo = U = q DV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v2

          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

           .v = 1,344 108 m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

       F = 1.6 10-19 1,344 108 0.4

       F = 8.6 10-12 N

5 0
3 years ago
Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?
KATRIN_1 [288]

Answer:

  0.83 g/cm³

Explanation:

The volume of the alcohol is ...

  (33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

  (9 g)/(1 g/cm³) = 9 cm³

The total volume is ...

  42.0253 cm³ +9 cm³ = 51.0253 cm³

The total mass is ...

  33.2 g + 9 g = 42.4 g

So, the resulting density is ...

  (42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³

The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

  (33.2/46)/(33.2/46 +9/18.02) ≈ 0.59

The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

3 0
3 years ago
A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1
Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

4 0
3 years ago
Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb
Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0
3 years ago
The brakes on a 15,680 N car exert a stopping force of 640 N.
Murrr4er [49]

Answer:

1568 Kg is the answer

Explanation:

6 0
3 years ago
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