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worty [1.4K]
3 years ago
7

A weightlifter lifts a 100-kg weight 1.5 meters off the floor at a constant velocity. Which statement about this situation is tr

ue?
The mechanical energy of the system increased because of friction.

The mechanical energy of the system decreased because of friction.

The kinetic energy of the weight decreased as it was lifted up.

The potential energy of the weight decreased as it was lifted up.
Physics
2 answers:
Effectus [21]3 years ago
8 0

Answer:

Give the person above me brainliest please!

Explanation:

Because they are a legend!

kirill [66]3 years ago
6 0

Answer: 1) c. The energy on the macroscopic scale decreased, and the energy on the molecular scale increased.

2) d.  Friction converts some energy on the macroscopic scale to energy on the molecular scale.

3) d. The mechanical energy of the system decreased because of friction.

4) d. The mechanical energy of the system at the beginning of the experiment is equal to the mechanical energy of the system at the end of the experiment.

5) d. The mechanical energy of the system decreased due to the transfer of energy on a macroscopic scale to energy on a molecular scale.

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a
Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

  • The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given  to the block by the spring = \mathbf{0.5  \cdot k  \cdot x^2}

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = 0.5  \cdot k  \cdot x^2 = kinetic energy of

block = 0.5  \cdot m  \cdot v^2

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = \mathbf{0.5  \cdot k  \cdot x^2}

  • The force of the weight of the block on the string, F = m \cdot g  \cdot sin(\theta)

The energy given to the block = 0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = The kinetic energy of block as it leaves the spring = \mathbf{0.5  \cdot m  \cdot v^2}

Which gives;

0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = 0.5  \cdot m  \cdot v^2

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

The graph of the equation a·x² + c·v² = b  is an ellipse

Therefore;

  • As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

  • <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
  • <em>The relationship is no longer linear and v will be more for the same value of x</em>
  • <em>The relationship is still linear, with lesser value of v</em>
  • <em>The relationship is still linear, with higher value of v</em>
  • <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

6 0
2 years ago
An increase in the magnitude of velocity is??
bazaltina [42]
It means the speed of the object is increasing
and
there is a positive acceleration in the direction of the velocity
hence
there is a force acting on the object, in the direction of the velocity
8 0
3 years ago
When 1,250^3/4 is written in simplest radical form, which value remains under the radical?
GaryK [48]

Answer:

125\sqrt[4]{8}

Explanation:

A number of the form

a^{\frac{m}{n}}

can be re-written in the radical form as follows:

\sqrt[n]{a^m}

In this problem, we have:

a = 1,250

m = 3

n = 4

So, if we apply the formula, we get

1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}

Then, we can rewrite 1250 as

1250 = 2\cdot 5^4

So we can rewrite the expression as

=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}

7 0
3 years ago
Read 2 more answers
In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of
Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, a_{Net} = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, v_y = v × sin(θ)

∴ v_y = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - a_{Net} × t

∴ t = (v₁ - v₂)/a_{Net}  = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2·a_{Net}·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0
2 years ago
What holds the positively charged metal ions together within a metallic bond?
enot [183]

Answer:

The electron cloud

Explanation:

Metallic bonds result from interaction of positively charged metal ions with free valence electrons which now forms an electron cloud around the metal ions. Electrostatic interaction between the metal ions and the electron cloud holds the metal ions together in the metallic bond.

4 0
3 years ago
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