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tatiyna
2 years ago
8

How does an inclined plane affect the effort needed to move a load vertically?

Physics
1 answer:
VashaNatasha [74]2 years ago
5 0

If we pull an object vertically upwards then we need to apply a force which is equal in the magnitude of the weight of the object

F = mg

now when we pull the same object upwards along an inclined plane with angle then we require a force which will balance the component of weight along the inclined

so it is given as

F' = mgsin\theta

so as if we compare the two forces we can say that since the value of sine is always less than 1 for an angle less than 90 degree

so in the 2nd case when we pull the object along the inclined plane it will require less effort

so correct answer is

<em>A. reduce effort</em>

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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on
just olya [345]

Answer:

a)   v_{f} = 0.25 m / s  b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

     m₁ = 0.40 kg

     v₁₀ = 9.0 m / s

     m₂ = 14 kg

     v₂₀ = 0

Initial

     po = m₁ v₁₀

Final

     p_{f} = (m₁ + m₂) vf

     po = pf

     m₁ v₁₀ = (m₁ + m₂) v_{f}

      v_{f} = v₁₀ m₁ / (m₁ + m₂)

      v_{f} = 9.0 (0.40 / (0.40 +14)

      v_{f} = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

      u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

     K₀ = ½ m₁ v₁₀² +0

     K₀ = ½ 0.40 9 2

     K₀ = 16.2 J

Final

     K_{f}= ½ (m₁ + m₂) v_{f}2

      K_{f} = ½ (0.4 +14) 0.25 2

    K_{f} = 0.45 J

   

    ΔK = K₀ -  K_{f}

    ΔK = 16.2-0.445

    ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

      v₁₀’= v₁₀ -u

      v₁₀’= 9 -.025

      v₁₀‘= 8.75 m / s

      v₂₀ ‘= v₂₀ -u

      v₂₀‘= - 0.25 m / s

      v_{f} ‘=   v_{f} - u

      v_{f} = 0

Initial

    K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

    Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

    Ko = 15.31 + 0.4375

    K o = 15.75 J

Final

   k_{f} = ½ (m₁ + m₂) vf’²

  k_{f} = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0
2 years ago
Using an inclined plane, what happens to the amount of work that has to be done?
weeeeeb [17]
A. The amount of work increases.
Imagine you are pushing a box on a flat surface. Now imagine pushing it up a steep hill. It gets harder and that's how I remember this.
Hope this helps. 
8 0
3 years ago
Read 2 more answers
The magnitude of displacements a and b are 3m and 4m, respectively, c=a+b. What is the magnitude of c if the angel between a and
AysviL [449]

Answer:

(a) 7 m

(b) 1  m

Explanation:

Given:

The magnitude of displacement  vector 'a' is 3 m

The magnitude of displacement vector 'b' is 4 m.

The vector 'c' is the vector sum of vectors 'a' and 'b'.

(a)

Now, when the angle between the vectors is 0°, it means that the vectors are in the same direction. When vectors are in the same direction, then their resultant magnitude is simply the sum of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in same direction is given as:

|\overrightarrow c|=|\overrightarrow a|+|\overrightarrow b|\\\\|\overrightarrow c|=3 + 4 = 7\ m

Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.

(b)

When the angle between the vectors is 180°, it means that the vectors are exactly in the opposite direction. When the vectors are in opposite direction, then their resultant magnitude is the subtraction of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:

|\overrightarrow c|=||\overrightarrow a|-|\overrightarrow b||\\\\|\overrightarrow c|=|3 - 4| = 1\ m

Therefore, the magnitude of vector 'c' is 1 m when angle between 'a' and 'b' is 180°.

4 0
2 years ago
Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .
jek_recluse [69]

Explanation:

Given that,

Force, F=((-8i)+6j)\ N

Position of the particle, r=(3i+4j)\ m

(a) The toque on a particle about the origin is given by :

\tau=F\times r

\tau=((-8i)+6j) \times (3i+4j)

Taking the cross product of above two vectors, we get the value of torque as :

\tau=(0+0-50k)\ N-m

(b) Let \theta is the angle between r and F. The angle between two vectors is given by :

cos\theta=\dfrac{r.F}{|r|.|F|}

cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }

cos\theta=\dfrac{0}{50}

\theta=90^{\circ}

6 0
3 years ago
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