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Vadim26 [7]
3 years ago
11

Which requires more work, lifting a 10.0kg load a vertical distance of 2m or lifting a 5.0kg load a distance of 4m?

Physics
1 answer:
Minchanka [31]3 years ago
6 0

For each load,  Work = (mass) x (gravity) x (distance .

Bigger load:      Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .

Smaller load:    Work = (5 kg)  x  (9.8 m/s²)  x  (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of  2:1  is exactly balanced by
the height ratio of  1:2 .

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Picture a long, straight corridor running east-west, with a water fountain located somewhere along it. Starting from the west en
Brut [27]

Answer:

The corridor's distance is "90 m".

Explanation:

  • She heads in the east directions but creates the first pause, meaning she crosses the distance 'x' in step 1.
  • Now, provided that perhaps the distance by her to another fountain or waterfall just after the first stop is twice as far away she traveled.
  • Because she moved the distance of 'x,' then, therefore, her distance towards the fountain of '2x.' She casually strolls and once again pauses 60 m beyond her stop.
  • The gap about her to the waterfall during that time approximately twice the distance and her to the eastern end of the hallway.
  • Assume her gap from either the east end of the platform seems to be 'y' at either the second stop, after which '2y' may become the distance between the 2nd pause and the waterfall.

Now,

⇒  2x + 2y = 60

⇒  x + y = 30

The total distance of the corridor will be:

=  x + 2x + 2y + y

=  3\times (x + y)

=  3\times 30

=  90 \ m

4 0
3 years ago
1. Plot the following graphs:
VLD [36.1K]

Answer:

(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram

(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time

Explanation:

8 0
3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
4 years ago
a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(1
tekilochka [14]

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>m</em> <em>a</em>

∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N

Let <em>f</em> be the unknown force. Then

∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>

=>   <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N

8 0
3 years ago
What is its maximum altitude above the ground? The answer is the maximum height above the ground
Kisachek [45]

Answer:

Maximum altitude above the ground = 1,540,224 m = 1540.2 km

Explanation:

Using the equations of motion

u = initial velocity of the projectile = 5.5 km/s = 5500 m/s

v = final velocity of the projectile at maximum height reached = 0 m/s

g = acceleration due to gravity = (GM/R²) (from the gravitational law)

g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)

g = -9.82 m/s² (minus because of the direction in which it is directed)

y = vertical distance covered by the projectile = ?

v² = u² + 2gy

0² = 5500² + 2(-9.82)(y)

19.64y = 5500²

y = 1,540,224 m = 1540.2 km

Hope this Helps!!!

3 0
3 years ago
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