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3 years ago

If the net force on a block is zero

Physics

1 answer:

amm18123 years ago

8 0

If the net force on a block is zero, the block will move at constant velocity

Explanation:

We can answer this question by applying Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:

$\sum F = ma$ (1)

where

$\sum F$ is the net force on the object

m is its mass

a is its acceleration

In this problem, we have a block, and the net force on it is zero:

$\sum F = 0$

According to eq.(1), this also implies that

$a=0$

So, the acceleration of the block is zero.

However, acceleration is the rate of change of velocity of a body:

$a=\frac{\Delta v}{\Delta t}$

where $\Delta v$ is the change in velocity in a time of $\Delta t$ . Since the acceleration is zero, this means that $\Delta v=0$ , and therefore the velocity of the object is constant.

Learn more about Newton's second law:

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3 years ago

Unpolarized light with an average intensity of 845 W/m2 moves along the x-axis when it enters a Polarizer A with a vertical tran

horsena [70]

Answer:

θ = 36.2º

Explanation:

When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law

I = I₀ cos² tea

The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization

I₁ = I₀ / 2

I₁ = 845/2

I₁ = 422.5 W / m²

In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of

I = 275 W / m ²

Cos² θ = I / I₁

Cos θ = √ I / I₁

Cos θ = √ (275 / 422.5)

Cos θ = 0.80678

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3 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

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3 years ago

3 bulbs are in series and the same 3 bulbs are in parallel with the same battery. Which bulbs will be dimmer?

Marina86 [1]

3 bulbs are in series and if the same 3 bulbs are in parallel with the same battery then the bulbs that are connected in parallel will be dimmer

<h3>What is power?</h3>

The rate of doing work is known as power. The Si unit of power is the watt.

Power =work/time

The mathematical expression for the electric power is as follows

P = VI

The same current flows through both bulbs when they are connected in series. A greater voltage drop across the bulb with the higher resistance will result in higher power dissipation and brightness. In the case of the parallel combination, the bulb will be dimmer

Thus, If the same three bulbs are connected in series and parallel with the same battery, the parallelly connected bulbs will be dimmer, therefore the correct option is A

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