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3 years ago

The two properties of an electron that cannot be known exactly at the same time are the ____________________.

1 answer:

6 0

Position and momentum.

This is Heisenberg's Uncertainty Principle:
Δx Δp ≥ h ÷ 4π, where Δx is the change in position, Δp is the change in momentum, and h is Planck's Constant.

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An object initially at rest experiences an acceleration of 1.90 m/s² for 6.60 s then travels at that constant velocity for anot

borishaifa [10]

Answer:

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

Explanation:

Hi there!

The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.

The displacement is calculated as the distance between the final position of the object and the initial position. In this problem, the displacement is equal to the traveled distance because the object travels only in one direction:

a.v = Δx/t

Where:

a.v = average velocity.

Δx = displacement = final position - initial position

t = time

So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:

x = 1/2 · a · t²

x = 1/2 · 1.90 m/s² · (6.60 s)²

x = 41.4 m

The object traveled 41.4 m during the first 6.60 s.

Now, let's find the rest of the traveled distance.

When the velocity is constant, a = 0. Then, the equation of position will be:

x = x0 + v · t

Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:

x = v · t

The velocity reached by the object during the acceleration phase is calculated as follows:

v = v0 + a · t (v0 = 0 because the object started from rest)

v = 1.90 m/s² · 6.60 s

v = 12.5 m/s

Then, the distance traveled by the object at a constant velocity will be:

x = 12.5 m/s · 8.50 s

x = 106 m

The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.

Then the displacement will be:

Δx = final position - initial position

Δx = 147 m - 0 = 147 m

and the average velocity will be:

a.v = Δx/t

a.v = 147 m / 15.1 s

a.v = 9.74 m/s

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

8 0

3 years ago

When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the fi

vova2212 [387]

Answer:

125.6 rad/s

25.12 rad/s²

Explanation:

t = time required by the fan to get up to final operating speed = 5 sec

w = final operating rotational speed = 1200 rpm

we know that :

1 revolution = 2π rad

1 min = 60 sec

w = $1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}$

w = $\frac{1200\times 2\pi }{60}\frac{rad}{s}$

w = 125.6 rad/s

w₀ = initial angular speed = 0 rad/s

α = angular acceleration

using the equation

w = w₀ + α t

125.6 = 0 + α (5)

α = 25.12 rad/s²

5 0

3 years ago

Over a period of operation, the useful work output of the fluorescent bulb was

Nadya [2.5K]

Answer:

199.0521 Will be the answer

5 0

3 years ago

A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhauste

marusya05 [52]

516.154 megawatts of heat are exhausted to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river ( $Q_{out}$ ), in megawatts:

$Q_{out} = Q_{in} - W$

$Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W$ (1)

Where:

$\eta$ - Efficiency.
$W$ - Electric power, in megawatts.

If we know that $\eta = 0.39$ and $W = 330\,MW$ , then the energy rate exhausted to the river is:

$Q_{out} = \left(\frac{1}{0.39}-1 \right)\cdot (330\,MW)$

$Q_{out} = 516.154\,MW$

516.154 megawatts of heat are exhausted to the river that cools the plant.

We kindly to check this question on first law of thermodynamics: brainly.com/question/3808473

7 0

3 years ago

Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If

trasher [3.6K]

Answer:

$d_{2}$ = 33.33 cm

Explanation:

Given:

When mass, $m_{1}$ =21 kg

distance travelled is $d_{1}$ = 140 cm

When mass, $m_{2}$ =5 kg

distance travelled is $d_{2}$ = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

$\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}$

$\frac{140}{21}=\frac{d_{2}}{5}$

$d_{2}$ = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

8 0

3 years ago